p^{\prime}_{10}<p_{x}$

To see this, note that

$\displaystyle
p_{11}-p_{y_{0}}<\frac{p_{y_{0}}((1-p_{x})-p^{\prime}_{10})}{p_{x}-(1-p_{x})}+p_{11}$
$\displaystyle\implies-1<\frac{((1-p_{x})-p^{\prime}_{10})}{p_{x}-(1-p_{x})}$
$\displaystyle\implies p_{x}>p^{\prime}_{10}$

This concludes the proof. ∎

######  Lemma 61.

Assume $p_{x}>1-p_{x}$. If $p^{\prime}_{10}<p_{x}$, then

$\frac{p_{11}p_{x}-p^{\prime}_{10}p_{y_{0}}}{p_{x}}<\frac{p_{y_{0}}((1-p_{x})-p^{\prime}_{10})}{p_{x}-(1-p_{x})}+p_{11}$

###### Proof.

Following a proof by contra-positive, the contra-positive is given as

$\frac{p_{11}p_{x}-p^{\prime}_{10}p_{y_{0}}}{p_{x}}\geq\frac{p_{y_{0}}((1-p_{x})-p^{\prime}_{10})}{p_{x}-(1-p_{x})}+p_{11}\implies
p^{\prime}_{10}\geq p_{x}$

To see this, note that

$\displaystyle\frac{p_{11}p_{x}-p^{\prime}_{10}p_{y_{0}}}{p_{x}}\geq\frac{p_{y_{0}}((1-p_{x})-p^{\prime}_{10})}{p_{x}-(1-p_{x})}+p_{11}$
$\displaystyle\implies\frac{-p^{\prime}_{10}}{p_{x}}\geq\frac{((1-p_{x})-p^{\prime}_{10})}{p_{x}-(1-p_{x})}$
$\displaystyle\implies-p^{\prime}_{10}p_{x}+p^{\prime}_{10}(1-p_{x})\geq(1-p_{x})p_{x}-p^{\prime}_{10}p_{x}$
$\displaystyle\implies p^{\prime}_{10}\geq p_{x}$

This concludes the proof. ∎

######  Lemma 62.

Assume $p_{x}>1-p_{x}$. If $p^{\prime}_{1i}\geq p_{x}$ for
$i\in\\{1,\dots,N\\}$, then

$\displaystyle 1\leq\frac{p^{\prime}_{1i}-(1-p_{x})}{p_{x}-(1-p_{x})}$

###### Proof.

A similar proof by contra-positive can be carried out to see this. ∎

######  Lemma 63.

Assume $p_{x}>1-p_{x}$. If $p^{\prime}_{1i}<x$ for $i\in\\{1,\dots,N\\}$, then

$\displaystyle\frac{p^{\prime}_{11}}{p_{x}}>\frac{p^{\prime}_{1i}-(1-p_{x})}{p_{x}-(1-p_{x})}$

###### Proof.

Following a proof by contra-positive, the contra-positive is given as

$\frac{p^{\prime}_{11}}{p_{x}}\leq\frac{p^{\prime}_{1i}-(1-p_{x})}{p_{x}-(1-p_{x})}\implies
p^{\prime}_{1i}\geq x$

To see this, note that

$\displaystyle\frac{p^{\prime}_{11}}{p_{x}}\leq\frac{p^{\prime}_{1i}-(1-p_{x})}{p_{x}-(1-p_{x})}$
$\displaystyle\implies p^{\prime}_{11}p_{x}-p^{\prime}_{11}(1-p_{x})\leq
p^{\prime}_{1i}p_{x}-p_{x}(1-p_{x})$ $\displaystyle\implies
p^{\prime}_{11}\geq p_{x}$

This concludes the proof. ∎

######  Lemma 64.

Assume $p_{x}<1-p_{x}$. If $p^{\prime}_{10}\geq 1-p_{x}$, then

$\frac{(1-p_{x})p_{y_{0}}+p_{11}p_{x}-p^{\prime}_{10}p_{y_{0}}}{p_{x}}\leq\frac{p_{y_{0}}(p^{\prime}_{10}-(1-p_{x}))}{(1-p_{x})-p_{x}}+p_{11}$

###### Proof.

Following a proof by contra-positive, the contra-positive is given as

$\displaystyle\frac{(1-p_{x})p_{y_{0}}+p_{11}p_{x}-p^{\prime}_{10}p_{y_{0}}}{p_{x}}>\frac{p_{y_{0}}(p^{\prime}_{10}-(1-p_{x}))}{(1-p_{x})-p_{x}}+p_{11}\implies
p^{\prime}_{10}<1-p_{x}$

To see this, note that

$\displaystyle\frac{(1-p_{x})p_{y_{0}}+p_{11}p_{x}-p^{\prime}_{10}p_{y_{0}}}{p_{x}}>\frac{p_{y_{0}}(p^{\prime}_{10}-(1-p_{x}))}{(1-p_{x})-p_{x}}+p_{11}$
$\displaystyle\implies\frac{(1-p_{x})-p^{\prime}_{10}}{p_{x}}>\frac{p^{\prime}_{10}-(1-p_{x})}{(1-p_{x})-p_{x}}$
$\displaystyle\implies(1-p_{x})((1-p_{x})-p_{x})-p^{\prime}_{10}((1-p_{x})-p_{x})>p^{\prime}_{10}p_{x}-(1-p_{x})p_{x}$
$\displaystyle\implies(1-p_{x})(1-p_{x})-p^{\prime}_{10}(1-p_{x})>0$
$\displaystyle\implies(1-p_{x})>p^{\prime}_{10}$

This concludes the proof. ∎

######  Lemma 65.

Assume $p_{x}<1-p_{x}$. If $p^{\prime}_{10}<1-p_{x}$, then

$p_{11}>\frac{p_{y_{0}}(p^{\prime}_{10}-(1-p_{x}))}{(1-p_{x})-p_{x}}+p_{11}$

###### Proof.

Following a proof by contra-positive, the contra-positive is given as

$p_{11}\leq\frac{p_{y_{0}}(p^{\prime}_{10}-(1-p_{x}))}{(1-p_{x})-p_{x}}+p_{11}\implies
p^{\prime}_{10}\geq 1-p_{x}$

To see this, note that

$\displaystyle
p_{11}\leq\frac{p_{y_{0}}(p^{\prime}_{10}-(1-p_{x}))}{(1-p_{x})-p_{x}}+p_{11}$
$\displaystyle\implies
0\leq\frac{p_{y_{0}}(p^{\prime}_{10}-(1-p_{x}))}{(1-p_{x})-p_{x}}$
$\displaystyle\implies p^{\prime}_{10}\geq 1-p_{x}$

This concludes the proof. ∎

######  Lemma 66.

Assume $p_{x}<1-p_{x}$. If $p^{\prime}_{1i}\geq 1-p_{x}$ for
$i\in\\{1,\dots,N\\}$, then

$\displaystyle\frac{p^{\prime}_{1i}-(1-p_{x})}{p_{x}}\geq\frac{(1-p_{x})-p^{\prime}_{1i}}{(1-p_{x})-p_{x}}$

###### Proof.

Following a proof by contra-positive, the contra-positive can be written as

$\frac{p^{\prime}_{1i}-(1-p_{x})}{p_{x}}<\frac{(1-p_{x})-p^{\prime}_{1i}}{(1-p_{x})-p_{x}}\implies
p^{\prime}_{1i}<1-p_{x}$

To see this, note that

$\displaystyle\frac{p^{\prime}_{1i}-(1-p_{x})}{p_{x}}<\frac{(1-p_{x})-p^{\prime}_{1i}}{(1-p_{x})-p_{x}}$
$\displaystyle\implies p^{\prime}_{1i}(1-p_{x})-(1-p_{x})(1-p_{x})<0$
$\displaystyle\implies p^{\prime}_{1i}(1-p_{x})<1-p_{x}$

This concludes the proof. ∎

######  Lemma 67.

Assume $p_{x}<1-p_{x}$. If $p^{\prime}_{1i}<1-p_{x}$ for
$i\in\\{1,\dots,N\\}$, then

$\displaystyle 0<\frac{(1-p_{x})-p^{\prime}_{1i}}{(1-p_{x})-p_{x}}$

###### Proof.

A similar proof by contradiction can be followed to prove the Lemma. ∎

######  Lemma 68.

Let $X$, $Z$ and $Y$ be binary random variables obeying the causal structure
in 1. Then, if the following conditions hold,

$\displaystyle x$ $\displaystyle>1-x$ $\displaystyle p^{\prime}_{11}$
$\displaystyle\geq x$ $\displaystyle p^{\prime}_{10}$ $\displaystyle\geq x$

Then $\implies\Phi_{X}>0$.

###### Proof.

Since $P(Z,X)$ and $P(Z,Y)$ are marginals of $P(Z,X,Y)$, the following holds

$p_{11}x+p_{10}(1-x)=p^{\prime}_{11}y_{1}+p^{\prime}_{10}y_{0}$

Since $x>1-x$

$\displaystyle\implies
p_{11}x+p_{10}x>p^{\prime}_{11}y_{1}+p^{\prime}_{10}y_{0}$ Since
$p^{\prime}_{11}\geq x$ and $p^{\prime}_{11}\geq x$ $\displaystyle\implies
p_{11}x+p_{10}x>xy_{1}+xy_{0}$ $\displaystyle\implies
p_{11}x+p_{10}x>x\implies p_{11}+p_{10}>1$ $\displaystyle\implies\Phi_{X}>0$

∎

######  Lemma 69.

Let $X$, $Z$ and $Y$ be binary random variables obeying the causal structure
in 1. Then, if the following conditions hold,

$\displaystyle 1-x$ $\displaystyle>x$ $\displaystyle p^{\prime}_{11}$
$\displaystyle\geq 1-x$ $\displaystyle p^{\prime}_{10}$ $\displaystyle\geq
1-x$

Then $\implies\Phi_{X}>0$.

###### Proof.

Since $P(Z,X)$ and $P(Z,Y)$ are marginals of $P(Z,X,Y)$, the following holds

$p_{11}x+p_{10}(1-x)=p^{\prime}_{11}y_{1}+p^{\prime}_{10}y_{0}$

Since $x>1-x$

$\displaystyle\implies
p_{11}(1-x)+p_{10}(1-x)>p^{\prime}_{11}y_{1}+p^{\prime}_{10}y_{0}$ Since
$p^{\prime}_{11}\geq 1-x$ and $p^{\prime}_{11}\geq 1-x$ $\displaystyle\implies
p_{11}(1-x)+p_{10}(1-x)>(1-x)y_{1}+(1-x)y_{0}$ $\displaystyle\implies
p_{11}(1-x)+p_{10}(1-x)>1-x$ $\displaystyle\implies\Phi_{X}>0$

∎

######  Lemma 70.

Let $X$, $Z$ and $Y$ be binary random variables obeying the causal structure
in 1. Then, if the following conditions hold,

$\displaystyle x$ $\displaystyle>1-x$ $\displaystyle p^{\prime}_{11}$
$\displaystyle\leq 1-x$ $\displaystyle p^{\prime}_{10}$ $\displaystyle\leq
1-x$

Then $\implies\Phi_{X}<0$.

###### Proof.

Since $P(Z,X)$ and $P(Z,Y)$ are marginals of $P(Z,X,Y)$, the following holds

$p_{11}x+p_{10}(1-x)=p^{\prime}_{11}y_{1}+p^{\prime}_{10}y_{0}$

Since $p^{\prime}_{11}\leq 1-x$ and $p^{\prime}_{11}\leq 1-x$

$\displaystyle\implies p_{11}x+p_{10}(1-x)\leq(1-x)y_{1}+(1-x)y_{0}$ Since
$1-x<x$ $\displaystyle\implies p_{11}(1-x)+p_{10}(1-x)<(1-x)y_{1}+(1-x)y_{0}$
$\displaystyle\implies p_{11}+p_{10}<1\implies\Phi_{X}<0$

∎

######  Lemma 71.

Let $X$, $Z$ and $Y$ be binary random variables obeying the causal structure
in 1. Then, if the following conditions hold,

$\displaystyle 1-x$ $\displaystyle>x$ $\displaystyle p^{\prime}_{11}$
$\displaystyle\leq x$ $\displaystyle p^{\prime}_{10}$ $\displaystyle\leq x$
$\displaystyle\implies$ $\displaystyle\Phi_{X}<0$

###### Proof.

A similar proof as above can be followed to obtain the result of this Lemma. ∎

######  Lemma 72.

Assume $x>1-x$. Denote $\Phi_{X}=p_{11}+p_{10}-1$, and denote $a$ and $b$ as

$\displaystyle a$
$\displaystyle=\frac{y_{0}((1-x)-p^{\prime}_{10})}{y_{1}(x-(1-x))}+\frac{p_{11}}{y_{1}}$
$\displaystyle b$ $\displaystyle=\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$

Then, $\Phi_{X}\geq 0\implies a\leq b$

###### Proof.

We utilize a proof of contrapositive, where the contrapositive of the above
statement is

$a>b\implies\Phi_{X}<0$
$\displaystyle\implies\frac{y_{0}((1-x)-p^{\prime}_{10})}{y_{1}(x-(1-x))}+\frac{p_{11}}{y_{1}}$
$\displaystyle>\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$
$\displaystyle\implies\frac{y_{0}((1-x)-p^{\prime}_{10})}{y_{1}(x-(1-x))}+\frac{p_{11}(x-(1-x))}{y_{1}(x-(1-x))}$
$\displaystyle>\frac{p^{\prime}_{11}y_{1}-y_{1}(1-x)}{y_{1}(x-(1-x))}$
$\displaystyle\implies y_{0}((1-x)-p^{\prime}_{10})+p_{11}(x-(1-x))$
$\displaystyle>p^{\prime}_{11}y_{1}-y_{1}(1-x)$
$\displaystyle\implies(1-x)+p_{11}(x-(1-x))$
$\displaystyle>p^{\prime}_{11}y_{1}+p^{\prime}_{10}y_{0}$
$\displaystyle\implies 0>\Phi_{X}$

∎

######  Lemma 73.

Assume $x>1-x$. Denote $\Phi_{X}=p_{11}+p_{10}-1$, and denote $a$ and $b$ as

$\displaystyle a$
$\displaystyle=\frac{y_{0}((1-x)-p^{\prime}_{10})}{y_{1}(x-(1-x))}+\frac{p_{11}}{y_{1}}$
$\displaystyle b$ $\displaystyle=\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$

Then, $\Phi_{X}<0\implies a>b$

###### Proof.

The contrapositive has been shown to hold from the Lemma above, hence the
Lemma holds.

∎

Proof for Lemma 27: This lemma is proved in two steps, first using
mathematical induction performed on the the Reduced Row Echelon Form (RREF) of
the matrix representation of the system of equations that Equation XXXX, then,
simplifying the final RREF based on properties of probability distributions.

First, the equations for invariances provided in Eq. XXX are given as

$\displaystyle p^{T}_{10}$ $\displaystyle=\sum_{j=0}^{N}P^{T}(Z=1\mid
X=0,Y=j)p_{y_{j}}$ $\displaystyle p^{T}_{11}$
$\displaystyle=\sum_{j=0}^{N}P^{T}(Z=1\mid X=1,Y=j)p_{y_{j}}$ $\displaystyle
p^{E}_{10}$ $\displaystyle=P^{T}(Z=1\mid X=0,Y=0)(1-p_{x}-\delta^{i}_{X})$
$\displaystyle\hskip 14.45377pt+P^{T}(Z=1\mid X=1,Y=0)(p_{x}+\delta^{i}_{X})$
$\displaystyle\vdots$ $\displaystyle p^{E}_{1N}$ $\displaystyle=P^{T}(Z=1\mid
X=0,Y=N)(1-p_{x}-\delta^{i}_{X})$ $\displaystyle\hskip
14.45377pt+P^{T}(Z=1\mid X=1,Y=N)(p_{x}+\delta^{i}_{X})$

The above equations can be represented in matrix form $A\mathbf{x}=b$ as

$\displaystyle A$
$\displaystyle=\begin{bmatrix}p_{y_{0}}&0&p_{y_{1}}&0&\dots&p_{y_{N-1}}&0&p_{y_{N}}&0\\\
0&p_{y_{0}}&0&p_{y_{1}}&\dots&0&p_{y_{N-1}}&0&p_{y_{N}}\\\
1-p_{x}-\delta_{X}&p_{x}+\delta_{X}&0&0&\dots&0&0&0&0\\\
0&0&1-p_{x}-\delta_{X}&p_{x}+\delta_{X}&\dots&0&0&0&0\\\
\vdots&\ddots&&&&&&&\vdots\\\
0&0&\dots&0&0&1-p_{x}-\delta_{X}&p_{x}+\delta_{X}&0&0\\\
0&0&\dots&0&&0&0&1-p_{x}-\delta_{X}&p_{x}+\delta_{X}\\\ \end{bmatrix}$
$\displaystyle b$ $\displaystyle=\begin{bmatrix}p^{T}_{10}\\\ p^{T}_{11}\\\
p^{E}_{10}\\\ p^{E}_{11}\\\ p^{E}_{12}\vdots\\\ p^{E}_{1(N-1)}\\\
p^{E}_{1N}\\\ \end{bmatrix}\hskip
72.26999pt\mathbf{x}=\begin{bmatrix}P^{T}(Z=1\mid X=0,Y=0)\\\ P^{T}(Z=1\mid
X=1,Y=0)\\\ P^{T}(Z=1\mid X=0,Y=1)\\\ P^{T}(Z=1\mid X=1,Y=1)\\\ \vdots\\\
P^{T}(Z=1\mid X=0,Y=N)\\\ P^{T}(Z=1\mid X=1,Y=N)\\\ \end{bmatrix}$

Next, denote the augmented matrix $A|b$ as

$\displaystyle\begin{bmatrix}[ccccccccc|c]p_{y_{0}}&0&p_{y_{1}}&0&\dots&p_{y_{N-1}}&0&p_{y_{N}}&0&p_{10}\\\
0&p_{y_{0}}&0&p_{y_{1}}&\dots&0&p_{y_{N-1}}&0&p_{y_{N}}&p_{11}\\\
1-p_{x}-\delta_{X}&p_{x}+\delta_{X}&0&0&\dots&0&0&0&0&p^{E}_{10}\\\
0&0&1-p_{x}-\delta_{X}&p_{x}+\delta_{X}&\dots&0&0&0&0&p^{E}_{11}\\\
\vdots&\ddots&&&&&&&\vdots&\vdots\\\
0&0&\dots&0&0&1-p_{x}-\delta_{X}&p_{x}+\delta_{X}&0&0&p^{E}_{1(N-1)}\\\
0&0&\dots&0&&0&0&1-p_{x}-\delta_{X}&p_{x}+\delta_{X}&p^{E}_{1N}\\\
\end{bmatrix}$

First, we utilize mathematical induction to provide a general form for the
RREF of $A|b$ for any $N$. Formally, let $P(N)$ denote the following
proposition.
Proposition: Let $P(N)$ denote the proposition that the support of $Y$ has a
cardinality of $N+1$, then $A|b$ will be a matrix with $N+3$ rows and
$2(N+1)+1$ columns. The following steps when executed will put $A|b$ in RREF,
where $R_{i}$ denotes the $i$-th row of $A|b$

  1. 1.

$\displaystyle R_{1}$ $\displaystyle=R_{1}/p_{y_{0}}$ (56) $\displaystyle
R_{3}$ $\displaystyle=R_{3}-(1-p_{x}-\delta_{X})R_{1}$ (57)

  2. 2.

$\displaystyle R_{2}$ $\displaystyle=R_{2}/p_{y_{0}}$ (58) $\displaystyle
R_{3}$ $\displaystyle=R_{3}-(p_{x}+\delta_{X})R_{2}$ (59)

  3. 3.

$\displaystyle R_{3}$
$\displaystyle=R3/(-\frac{(1-p_{x}-\delta_{X})p_{y_{1}}}{p_{y_{0}}})$ (60)
$\displaystyle R_{1}$ $\displaystyle=R_{1}-\frac{p_{y_{1}}}{p_{y_{0}}}R3$ (61)
$\displaystyle R_{4}$ $\displaystyle=R4-(1-p_{x}-\delta_{X})R3$ (62)

  4. 4.

If $N\geq 2$, then loop through the following steps for $i\in\\{2,\dots,N\\}$:

$\displaystyle
R_{2+i}=R_{2+i}/(-\frac{p_{y_{i}}(1-p_{x}-\delta_{X})}{p_{y_{i-1}}})$ (63)
$\displaystyle R_{1+i}=R_{1+i}-\frac{p_{y_{i}}}{p_{y_{i-1}}}R_{2+i}$ (64)
$\displaystyle R_{3+i}=R_{3+i}-(1-p_{x}-\delta_{X})R_{2+i}$ (65)

And the corresponding matrix will evaluate to

$\displaystyle\begin{bmatrix}1&0&0&-\frac{(p_{x}+\delta_{X})p_{y_{1}}}{(1-p_{x}-\delta_{X})p_{y_{0}}}&\dots&0&-\frac{(p_{x}+\delta_{X})p_{y_{N-1}}}{(1-p_{x}-\delta_{X})p_{y_{0}}}&0&-\frac{(p_{x}+\delta_{X})p_{y_{N}}}{(1-p_{x}-\delta_{X})p_{y_{0}}}&\frac{p^{E}_{10}p_{y_{0}}-p^{T}_{11}(p_{x}+\delta_{X})}{(1-p_{x}-\delta_{X})p_{y_{0}}}\\\
0&1&0&\frac{p_{y_{1}}}{p_{y_{0}}}&\dots&0&\frac{p_{y_{N-1}}}{p_{y_{0}}}&0&\frac{p_{y_{N}}}{p_{y_{0}}}&\frac{p^{T}_{11}}{p_{y_{0}}}\\\
0&0&1&\frac{p_{x}+\delta_{X}}{1-p_{x}-\delta_{X}}&\dots&0&0&0&0&\frac{p^{E}_{11}}{1-p_{x}-\delta_{X}}\\\
\vdots&&&&&&&\ddots\\\
0&0&0&0&\dots&0&0&1&\frac{p_{x}+\delta_{X}}{1-p_{x}-\delta_{X}}&\frac{p^{T}_{10}(1-p_{x}-\delta_{X})+p^{T}_{11}(p_{x}+\delta_{X})-\sum_{i=0}^{N-1}p^{E}_{1i}p_{y_{i}}}{p_{y_{N}}(1-p_{x}-\delta_{X})}\\\
0&0&0&0&\dots&0&0&0&0&\frac{\sum_{i=0}^{N}p^{E}_{1i}p_{y_{i}}-p^{T}_{10}(1-p_{x}-\delta_{X})-p^{T}_{11}(p_{x}+\delta_{X})}{p_{y_{N}}}\end{bmatrix}$
(66)

An identical proof used in Lemma 1 can be applied to prove this induction.
Next, to complete this proof for this lemma, we prove the following
identities.

$\frac{\sum_{j=0}^{N}p^{E}_{1j}p_{y_{j}}-p^{T}_{10}(1-p_{x}-\delta_{X})-p^{T}_{11}(p_{x}+\delta_{X})}{p_{y_{N}}}=0$

To prove this, note that

$p^{T}_{10}(1-p_{x}-\delta_{X})&=\sum_{j=0}^{N}P^{T}(Z=1\mid
X=0,Y=j)p_{y_{j}}(1-p_{x}-\delta_{X})$

Along with

$p^{T}_{11}(p_{x}+\delta_{X})=\sum_{j=0}^{N}P^{T}(Z=1\mid
X=1,Y=j)p_{y_{j}}(p_{x}+\delta^{i}_{X})$

Consequently,

$\displaystyle
p^{T}_{10}(1-p_{x}-\delta^{0}_{X})+p^{T}_{11}(p_{x}+\delta^{i}_{X})=$
$\displaystyle\sum_{j=0}^{N}p_{y_{j}}\Bigg{(}P^{T}(Z=1\mid
X=0,Y=j)(1-p_{x}-\delta^{0}_{X})+P^{T}(Z=1\mid
X=1,Y=j)(p_{x}+\delta^{i}_{X})\Bigg{)}$
$\displaystyle=\sum_{j=0}^{N}p_{y_{j}}\Bigg{(}P^{E}(Z=1\mid
X=0,Y=j)P^{E}(X=0)+P^{E}(Z=1\mid X=1,Y=j)P^{E}(X=1)\Big{)}$
$\displaystyle=\sum_{j=0}^{N}p_{y_{j}}p^{E}_{1j}$

And this will be subtracted from $\sum_{j=0}^{N}p_{y_{j}}p^{E}_{1ji}$, hence
it equals $0$.

$\frac{p^{T}_{10i}(1-p_{x}-\delta^{i}_{X})+p^{T}_{11i}(p_{x}+\delta^{i}_{X})-\sum_{j=0}^{N-1}p^{E}_{1ji}p_{y_{j}}}{p_{y_{N}}(1-p_{x}-\delta^{i}_{X})}=$

Following a similar approach as before, this will simplify to

$\displaystyle\frac{\sum_{j=0}^{N}p_{y_{j}}p^{E}_{1j}-\sum_{j=0}^{N-1}p_{y_{j}}p^{E}_{1j}}{p_{y_{N}}(1-p_{x}-\delta^{i}_{X})}=\frac{p^{E}_{1N}}{1-p_{x}-\delta^{i}_{X}}$

This concludes the proof of the Lemma.

######  Lemma 74.

Let $X$ and $Z$ be binary random variables, and let $C$ and $Y$ be discrete
random variables taking values in $\\{0,\dots,M\\}$ and $\\{0,\dots,N\\}$
respectively. Let $\delta^{C}_{X}$ represent the link between the treatment
assignment mechaism between the $P^{E}$ and $P^{T}$ as follows

$P^{E}(X=1\mid C)=P^{T}(X=1)+\delta^{C}_{X}$

For notational ease, we introduce the following terms. Let $P^{T}(Z=1\mid
X=j,C=i)$ be denoted by $p^{T}{1ji}$ and let $P^{E}(Z=1\mid Y=j,C=i)$ be
denoted as $p^{E}_{1ji}$. Given distributions $P^{T}(Z,X,C)$ and
$P^{E}(Z,Y,C)$ obeying the following contraints

$\displaystyle p^{T}_{10i}$ $\displaystyle=\sum_{j=0}^{N}P(Z=1\mid
X=0,Y=j,C=i)p_{y_{j}}$ $\displaystyle p^{T}_{11i}$
$\displaystyle=\sum_{j=0}^{N}P(Z=1\mid X=1,Y=j,C=i)p_{y_{j}}$ $\displaystyle
p^{E}_{10i}$ $\displaystyle=P(Z=1\mid X=0,Y=0,C=i)(1-p_{x}-\delta^{i}_{X})$
$\displaystyle\hskip 14.45377pt+P(Z=1\mid X=1,Y=0,C=i)(p_{x}+\delta^{i}_{X})$
$\displaystyle\vdots$ $\displaystyle p^{E}_{1Ni}$ $\displaystyle=P(Z=1\mid
X=0,Y=N,C=i)(1-p_{x}-\delta^{i}_{X})$ $\displaystyle\hskip
14.45377pt+P(Z=1\mid X=1,Y=N,C=i)(p_{x}+\delta^{i}_{X})$

Then this system of equations can be expressed in terms of free parameters
$P(Z=1\mid X=1,Y=i,C)$ for $i\in\\{1,\dots,N\\}$ and all $C$ as

$\displaystyle\mathbb{P}(Z=1\mid X=0,Y=0,C)$
$\displaystyle=\frac{p_{x}+\delta^{C}_{X}}{1-p_{x}-\delta^{C}_{X}}\sum_{n=1}^{N}\frac{p_{y_{n}}}{p_{y_{0}}}\mathbb{P}(Z=1\mid
X=1,Y=n)+\frac{p^{E}_{10C}p_{y_{0}}-p^{T}_{11C}(p_{x}+\delta^{C}_{X})}{(1-p_{x}-\delta^{C}_{X})p_{y_{0}}}$
$\displaystyle\mathbb{P}(Z=1\mid X=1,Y=0,C)$
$\displaystyle=\frac{p^{T}_{11C}}{p_{y_{0}}}-\sum_{n=1}^{N}\frac{p_{y_{n}}}{p_{y_{0}}}\mathbb{P}(Z=1\mid
X=1,Y=n,C)$ $\displaystyle\mathbb{P}(Z=1\mid X=0,Y=1,C)$
$\displaystyle=\frac{p^{E}_{11C}}{1-p_{x}-\delta^{C}_{X}}-\frac{p_{x}+\delta^{C}_{X}}{1-p_{x}-\delta^{C}_{X}}\mathbb{P}(Z=1\mid
X=1,Y=1,C)$ $\displaystyle\mathbb{P}(Z=1\mid X=0,Y=2,C)$
$\displaystyle=\frac{p^{E}_{12C}}{1-p_{x}-\delta^{C}_{X}}-\frac{p_{x}+\delta^{C}_{X}}{1-p_{x}-\delta^{C}_{X}}\mathbb{P}(Z=1\mid
X=1,Y=2,C)$ $\displaystyle\vdots$ $\displaystyle\mathbb{P}(Z=1\mid X=0,Y=N,C)$
$\displaystyle=\frac{p^{E}_{1N}}{1-p_{x}-\delta^{C}_{X}}-\frac{p_{x}+\delta^{C}_{X}}{1-p_{x}-\delta^{C}_{X}}\mathbb{P}(Z=1\mid
X=1,Y=N,C)$

###### Proof.

The proof for this Lemma follows two major steps, the first being a proof by
double mathematical induction to represent the Reduced Row Echelon Form (RREF)
for the system of equations for arbitrary dimensionality of $Y$ and $C$. The
second part of the proof applies properties of probability distributions to
express the system of equations in the form given in the Lemma.

We start start with the proof by double mathematical induction, and state the
proposition we prove.

Proposition: Let $P(m,n)$ denote the proposition that given the system of
equations corresponding to the constraints provided in Lemma, represented in
matrix form $A_{m,n}\mathbf{x}_{m,n}=\mathbf{b}_{m,n}$ provided below.

$\displaystyle A_{m,n}$
$\displaystyle=\begin{bmatrix}A_{0,n}&[\mathbf{0}]&\dots&[\mathbf{0}]\\\
[\mathbf{0}]&A_{1,n}&\dots&[\mathbf{0}]\\\
[\mathbf{0}]&[\mathbf{0}]&\ddots&[\mathbf{0}]\\\
[\mathbf{0}]&[\mathbf{0}]&\dots&A_{m,n}\end{bmatrix}\hskip
14.45377ptb_{m,n}=\begin{bmatrix}b_{0,n}\\\ \vdots\\\
b_{m,n}\end{bmatrix}\hskip
14.45377pt\mathbf{x}_{m,n}=\begin{bmatrix}\mathbf{x}_{0,n}\\\ \vdots\\\
\mathbf{x}_{m,n}\end{bmatrix}$ Where $A_{i,n}$, $b_{i,n}$ and
$\mathbf{x}_{i,n}$ will be of the form $\displaystyle A_{i,n}$
$\displaystyle=\begin{bmatrix}p_{y_{0}}&0&p_{y_{1}}&0&\dots&p_{y_{N-1}}&0&p_{y_{N}}&0\\\
0&p_{y_{0}}&0&p_{y_{1}}&\dots&0&p_{y_{N-1}}&0&p_{y_{N}}\\\
1-p_{x}-\delta^{i}_{X}&p_{x}+\delta^{i}_{X}&0&0&\dots&0&0&0&0\\\
0&0&1-p_{x}-\delta^{i}_{X}&p_{x}+\delta^{i}_{X}&\dots&0&0&0&0\\\
\vdots&\ddots&&&&&&&\vdots\\\
0&0&\dots&0&0&1-p_{x}-\delta^{i}_{X}&p_{x}+\delta^{i}_{X}&0&0\\\
0&0&\dots&0&&0&0&1-p_{x}-\delta^{i}_{X}&p_{x}-\delta^{i}_{X}\\\ \end{bmatrix}$
$\displaystyle b_{i,n}$ $\displaystyle=\begin{bmatrix}p^{T}_{10i}\\\
p^{T}_{11i}\\\ p^{E}_{10i}\\\ p^{E}_{11i}\\\ p^{E}_{12i}\vdots\\\
p^{E}_{1(n-1)i}\\\ p^{E}_{1ni}\\\ \end{bmatrix}\hskip
72.26999pt\mathbf{x}_{i,n}=\begin{bmatrix}P(Z=1\mid X=0,Y=0,C=i)\\\ P(Z=1\mid
X=1,Y=0,C=i)\\\ P(Z=1\mid X=0,Y=1,C=i)\\\ P(Z=1\mid X=1,Y=1,C=i)\\\ \vdots\\\
P(Z=1\mid X=0,Y=N,C=i)\\\ P(Z=1\mid X=1,Y=N,C=i)\\\ \end{bmatrix}$

Given this system of equations, the RREF of the augmented matrix ${A\mid
b}_{m,n}$ corresponding to this system of equations will be of the form

$\displaystyle{A\mid
b}^{R}_{m,n}=\begin{bmatrix}A^{R}_{0,n}&[\mathbf{0}]&[\mathbf{0}]&b^{R}_{0,n}\\\
[\mathbf{0}]&A^{R}_{1,n}&[\mathbf{0}]&b^{R}_{1,n}\\\
[\mathbf{0}]&[\mathbf{0}]&\ddots&\vdots\\\
[\mathbf{0}]&\dots&A^{R}_{m,n}&b^{R}_{m,n}\\\
0&\dots&0&\frac{\sum_{j=0}^{N}p^{E}_{1j0}p_{y_{j}}-p^{T}_{10}(1-p_{x}-\delta^{0}_{X})-p^{T}_{110}(p_{x}+\delta^{0}_{X})}{p_{y_{N}}}\\\
\vdots&\dots&0&\vdots\\\
0&\dots&0&\frac{\sum_{j=0}^{N}p^{E}_{1j0}p_{y_{j}}-p^{T}_{10}(1-p_{x}-\delta^{0}_{X})-p^{T}_{110}(p_{x}+\delta^{0}_{X})}{p_{y_{N}}}\\\
\end{bmatrix}$

Where $A^{R}_{i,n}$ and $b^{R}_{i,n}$ will be of the form

$\displaystyle
A^{R}_{i,n}=\begin{bmatrix}1&0&0&-\frac{(p_{x}+\delta^{i}_{X})p_{y_{1}}}{(1-p_{x}-\delta^{i}_{X})p_{y_{0}}}&\dots&0&-\frac{(p_{x}+\delta^{i}_{X})p_{y_{N-1}}}{(1-p_{x}-\delta^{i}_{X})p_{y_{0}}}&0&-\frac{(p_{x}+\delta^{i}_{X})p_{y_{N}}}{(1-p_{x}-\delta^{i}_{X})p_{y_{0}}}\\\
0&1&0&\frac{p_{y_{1}}}{p_{y_{0}}}&\dots&0&\frac{p_{y_{N-1}}}{p_{y_{0}}}&0&\frac{p_{y_{N}}}{p_{y_{0}}}\\\
0&0&1&\frac{p_{x}+\delta^{i}_{X}}{1-p_{x}-\delta^{i}_{X}}&\dots&0&0&0&0\\\
\vdots&&&&&&&\ddots\\\
0&0&0&0&\dots&0&0&1&\frac{p_{x}+\delta^{i}_{X}}{1-p_{x}-\delta^{i}_{X}}\\\
\end{bmatrix}$

And $b^{R}_{i,n}$ will be of the form

$\displaystyle\begin{bmatrix}\frac{p^{E}_{10i}p_{y_{0}}-p^{T}_{11i}p_{x}}{(1-p_{x}-\delta^{i}_{X})p_{y_{0}}}\\\
\frac{p^{T}_{11i}}{p_{y_{0}}}\\\ \vdots\\\
\frac{p^{T}_{10i}(1-p_{x}-\delta^{i}_{X})+p^{T}_{11i}(p_{x}+\delta^{i}_{X})-\sum_{j=0}^{N-1}p^{E}_{1ji}p_{y_{j}}}{p_{y_{N}}(1-p_{x}-\delta^{i}_{X})}\end{bmatrix}$

To prove $P(m,n)$ using double mathematical induction, we first prove the base
case with $m=n=1$.

Base Case: Using a symbolic solver to put the matrix in RREF, we obtain the
RREF form of $A\mid b$ as

$\displaystyle\begin{bmatrix}1&0&0&-\frac{p_{y_{1}}(\delta^{0}_{X}+p_{x})}{p_{y_{0}}(1-p_{x}-\delta^{0}_{X})}&0&0&0&0&\frac{p^{E}_{100}p_{y_{0}}-p^{T}_{110}(\delta^{0}_{X}+p_{x})}{p_{y_{0}}(1-p_{x}-\delta^{0}_{X})}\\\
0&1&0&\frac{p_{y_{1}}}{p_{y_{0}}}&0&0&0&0&\frac{p^{T}_{110}}{p_{y_{0}}}\\\
0&0&1&\frac{\delta^{0}_{X}+x}{1-x-\delta^{0}_{X}}&0&0&0&0&\frac{p^{T}_{100}(1-p_{x}-\delta^{0}_{X})+p^{T}_{110}(\delta^{0}_{X}+x)-p^{E}_{100}p_{y_{0}}}{p_{y_{1}}(1-p_{x}-\delta^{0}_{X})}\\\
0&0&0&0&1&0&0&-\frac{p_{y_{1}}(\delta^{1}_{X}+x)}{p_{y_{0}}(1-p_{x}-\delta^{1}_{X})}&\frac{p^{E}_{101}p_{y_{0}}-p^{T}_{111}(\delta^{1}_{X}+p_{x})}{p_{y_{0}}(1-p_{x}-\delta^{1}_{X})}\\\
0&0&0&0&0&1&0&\frac{p_{y_{1}}}{p_{y_{0}}}&\frac{p^{T}_{111}}{p_{y_{0}}}\\\
0&0&0&0&0&0&1&\frac{p_{x}+\delta^{1}_{X}}{1-x-\delta^{1}_{X}}&\frac{p^{T}_{101}(1-p_{x}-\delta^{1}_{X})+p^{T}_{111}(\delta^{1}_{X}+x)-p^{E}_{101}p_{y_{0}}}{p_{y_{1}}(1-p_{x}-\delta^{1}_{X})}\\\
0&0&0&0&0&0&0&0&\frac{p^{E}_{110}p_{y_{1}}+p^{E}_{100}p_{y_{0}}-p_{100}(1-p_{x}-\delta^{0}_{X})-p_{110}(\delta^{0}_{X}+p_{x})}{p_{y_{1}}}\\\
0&0&0&0&0&0&0&0&\frac{p^{E}_{111}p_{y_{1}}+p^{E}_{101}p_{y_{0}}-p^{T}_{101}(1-p_{x}-\delta^{1}_{X})-p_{111}(\delta^{1}_{X}+p_{x})}{p_{y_{1}}}\end{bmatrix}$

And by comparison, this matches the base case. Hence the base case holds.

Induction Step I: Here we prove that if $P(1,n-1)$ holds, then $P(1,n)$ holds
as well. For $P(1,n)$, $A|b$ will look like

$\displaystyle{A\mid b}_{1,n}$
$\displaystyle=\begin{bmatrix}A_{0,n}&[\mathbf{0}]&b_{0,n}\\\
[\mathbf{0}]&A_{1,n}&b_{1,n}\end{bmatrix}$

Since $A_{0,n}$ and $A_{1,n}$ are non-overlapping diagional sub-matrices of
${A\mid b}_{1,n}$, a similar proof to Lemma X for each level of of $C$ can be
applied, i.e. for each pair $A_{0,n}$, $b_{0,n}$ and $A_{1,n}$, $b_{1,n}$ to
put each of them in their RREF matrix separately. Then the last rows
correspoding to the RREF of $A_{0,n}$, $b_{0,n}$ and $A_{1,n}$, $b_{1,n}$ can
be shuffled as needed to put ${A\mid b}_{1,n}$ in RREF, proving the first step
of the induction.

Induction Step II: Next, we must prove that for all $n\geq 1$, $P(m,n)$ is
true then $P(m+1,n)$ is true. For $P(m+1,n)$, the system of equations is
represented in matrix form as

$\displaystyle A_{m,n}$
$\displaystyle=\begin{bmatrix}A_{0,n}&[\mathbf{0}]&\dots&[\mathbf{0}]&[\mathbf{0}]\\\
[\mathbf{0}]&A_{1,n}&\dots&[\mathbf{0}]&[\mathbf{0}]\\\
[\mathbf{0}]&[\mathbf{0}]&\ddots&[\mathbf{0}]&[\mathbf{0}]\\\
[\mathbf{0}]&[\mathbf{0}]&\dots&A_{m,n}&[\mathbf{0}]\\\
[\mathbf{0}]&[\mathbf{0}]&\dots&[\mathbf{0}]&A_{m+1,n}\end{bmatrix}\hskip
14.45377ptb_{m,n}=\begin{bmatrix}b_{0,n}\\\ \vdots\\\ b_{m,n}\\\
b_{m+1,n}\end{bmatrix}\hskip
14.45377pt\mathbf{x}_{m,n}=\begin{bmatrix}\mathbf{x}_{0,n}\\\ \vdots\\\
\mathbf{x}_{m,n}\\\ \mathbf{x}_{m+1,n}\end{bmatrix}$ Where $A_{i,n}$,
$b_{i,n}$ and $\mathbf{x}_{i,n}$ will be of the form $\displaystyle A_{i,n}$
$\displaystyle=\begin{bmatrix}p_{y_{0}}&0&p_{y_{1}}&0&\dots&p_{y_{N-1}}&0&p_{y_{N}}&0\\\
0&p_{y_{0}}&0&p_{y_{1}}&\dots&0&p_{y_{N-1}}&0&p_{y_{N}}\\\
1-p_{x}-\delta^{i}_{X}&p_{x}+\delta^{i}_{X}&0&0&\dots&0&0&0&0\\\
0&0&1-p_{x}-\delta^{i}_{X}&p_{x}+\delta^{i}_{X}&\dots&0&0&0&0\\\
\vdots&\ddots&&&&&&&\vdots\\\
0&0&\dots&0&0&1-p_{x}-\delta^{i}_{X}&p_{x}+\delta^{i}_{X}&0&0\\\
0&0&\dots&0&&0&0&1-p_{x}-\delta^{i}_{X}&p_{x}-\delta^{i}_{X}\\\ \end{bmatrix}$
$\displaystyle b_{i,n}$ $\displaystyle=\begin{bmatrix}p^{T}_{10i}\\\
p^{T}_{11i}\\\ p^{E}_{10i}\\\ p^{E}_{11i}\\\ p^{E}_{12i}\vdots\\\
p^{E}_{1(n-1)i}\\\ p^{E}_{1ni}\\\ \end{bmatrix}\hskip
72.26999pt\mathbf{x}_{i,n}=\begin{bmatrix}P(Z=1\mid X=0,Y=0,C=i)\\\ P(Z=1\mid
X=1,Y=0,C=i)\\\ P(Z=1\mid X=0,Y=1,C=i)\\\ P(Z=1\mid X=1,Y=1,C=i)\\\ \vdots\\\
P(Z=1\mid X=0,Y=N,C=i)\\\ P(Z=1\mid X=1,Y=N,C=i)\\\ \end{bmatrix}$

Since we assume $P(m,n)$ is true, the submatrix corresponding to ${A\mid
b}_{m,n}$ in ${A\mid b}_{m+1,n}$ is non-overlapping with $A_{m+1,n}$ and
$b_{m+1,n}$, and can be put into RREF form, presented below.

$\displaystyle{A\mid
b}_{m+1,n}=\begin{bmatrix}A^{R}_{0,n}&[\mathbf{0}]&\dots&[\mathbf{0}]&[\mathbf{0}]&b^{R}_{0,n}\\\
[\mathbf{0}]&A^{R}_{1,n}&\dots&[\mathbf{0}]&[\mathbf{0}]&b^{R}_{1,n}\\\
[\mathbf{0}]&[\mathbf{0}]&\ddots&[\mathbf{0}]&[\mathbf{0}]&\vdots\\\
[\mathbf{0}]&[\mathbf{0}]&\dots&A^{R}_{m,n}&[\mathbf{0}]&b^{R}_{m,n}\\\
0&\dots&0&0&[\mathbf{0}]&\frac{\sum_{j=0}^{N}p^{E}_{1j0}p_{y_{j}}-p^{T}_{10}(1-p_{x}-\delta^{0}_{X})-p^{T}_{110}(p_{x}+\delta^{0}_{X})}{p_{y_{N}}}\\\
\vdots&\dots&0&0&[\mathbf{0}]&\vdots\\\
0&\dots&0&0&[\mathbf{0}]&\frac{\sum_{j=0}^{N}p^{E}_{1j0}p_{y_{j}}-p^{T}_{10}(1-p_{x}-\delta^{0}_{X})-p^{T}_{110}(p_{x}+\delta^{0}_{X})}{p_{y_{N}}}\\\
0&\dots&0&0&A_{m+1,n}&b_{m+1,n}\end{bmatrix}$

Now, putting $A_{m+1,n}$ and $b_{m+1,n}$ in RREF form using a similar approach
to Lemma X, and shuffling all the rows that only have $0$ in every entry
except the last gives us the RREF as stated and proved $P(m+1,n)$. This
concludes the proof of the double induction.

Next, to complete this proof, we prove the following identities conclude the
proof.

$\frac{\sum_{j=0}^{N}p^{E}_{1ji}p_{y_{j}}-p^{T}_{10i}(1-p_{x}-\delta^{i}_{X})-p^{T}_{11i}(p_{x}+\delta^{i}_{X})}{p_{y_{N}}}=0$

To prove this, note that

$p^{T}_{10i}(1-p_{x}-\delta^{0}_{X})&=\sum_{j=0}^{N}P(Z=1\mid
X=0,Y=j,C=i)p_{y_{j}}(1-p_{x}-\delta^{0}_{X})$

Along with

$p^{T}_{11i}(p_{x}+\delta^{i}_{X})=\sum_{j=0}^{N}P(Z=1\mid
X=1,Y=j,C=i)p_{y_{j}}(p_{x}+\delta^{i}_{X})$

Consequently,

$\displaystyle
p^{T}_{10i}(1-p_{x}-\delta^{0}_{X})+p^{T}_{11i}(p_{x}+\delta^{i}_{X})=$
$\displaystyle\sum_{j=0}^{N}p_{y_{j}}\Bigg{(}P(Z=1\mid
X=0,Y=j,C=i)(1-p_{x}-\delta^{0}_{X})+P(Z=1\mid
X=1,Y=j,C=i)(p_{x}+\delta^{i}_{X})\Big{)}$
$\displaystyle=\sum_{j=0}^{N}p_{y_{j}}\Bigg{(}P(Z=1\mid
X=0,Y=j,C=i)P^{E}(X=0\mid C=i)$ $\displaystyle\hskip 28.90755pt+P(Z=1\mid
X=1,Y=j,C=i)P^{E}(X=1\mid C=i)\Big{)}$
$\displaystyle=\sum_{j=0}^{N}p_{y_{j}}p^{E}_{1ji}$

And this will be subtracted from $\sum_{j=0}^{N}p_{y_{j}}p^{E}_{1ji}$, hence
it equals $0$.

$\frac{p^{T}_{10i}(1-p_{x}-\delta^{i}_{X})+p^{T}_{11i}(p_{x}+\delta^{i}_{X})-\sum_{j=0}^{N-1}p^{E}_{1ji}p_{y_{j}}}{p_{y_{N}}(1-p_{x}-\delta^{i}_{X})}=$

Following a similar approach as before, this will simplify to

$\displaystyle\frac{\sum_{j=0}^{N}p_{y_{j}}p^{E}_{1ji}-\sum_{j=0}^{N-1}p_{y_{j}}p^{E}_{1ji}}{p_{y_{N}}(1-p_{x}-\delta^{i}_{X})}=\frac{p^{E}_{1Ni}}{1-p_{x}-\delta^{i}_{X}}$

This concludes the proof of the Lemma. ∎

Proof of Lemma 51: Using the result presented in Lemma 27, when $Y$ has
support $\\{0,\dots,N\\}$ and we are given marginals $P(Z,X)$ and $P(Z,Y)$, we
can write the system of equations using free parameters as

$\displaystyle P^{T}(Z=1\mid X=0,Y=0)$
$\displaystyle=\frac{p_{x}+\delta_{X}}{1-p_{x}-\delta_{X}}\sum_{n=1}^{N}\frac{p_{y_{n}}}{p_{y_{0}}}P^{T}(Z=1\mid
X=1,Y=n)+\frac{p^{E}_{10}p_{y_{0}}-p^{T}_{11}(p_{x}+\delta_{X})}{(1-p_{x}-\delta_{X})p_{y_{0}}}$
$\displaystyle P^{T}(Z=1\mid X=1,Y=0)$
$\displaystyle=\frac{p^{T}_{11}}{p_{y_{0}}}-\sum_{n=1}^{N}\frac{p_{y_{n}}}{p_{y_{0}}}P^{T}(Z=1\mid
X=1,Y=n)$ $\displaystyle P^{T}(Z=1\mid X=0,Y=1)$
$\displaystyle=\frac{p^{E}_{11}}{1-p_{x}-\delta_{X}}-\frac{p_{x}+\deltaXX}{1-p_{x}-\delta_{X}}P^{T}(Z=1\mid
X=1,Y=1)$ $\displaystyle P^{T}(Z=1\mid X=0,Y=2)$
$\displaystyle=\frac{p^{E}_{12}}{1-p_{x}-\delta_{X}}-\frac{p_{x}+\delta_{X}}{1-p_{x}-\delta_{X}}P^{T}(Z=1\mid
X=1,Y=2)$ $\displaystyle\vdots$ $\displaystyle P^{T}(Z=1\mid X=0,Y=N)$
$\displaystyle=\frac{p^{E}_{1N}}{1-p_{x}-\delta_{X}}-\frac{p_{x}+\delta_{X}}{1-p_{x}-\delta_{X}}P^{T}(Z=1\mid
X=1,Y=N)$

Now, each of the basic variables in the system above must be a valid
probability, i.e. between zero and one, inclusive. Employing these constraints
on the free parameters gives

$\displaystyle 0$ $\displaystyle\leq P^{T}(Z=1\mid X=0,Y=0)\leq 1$
$\displaystyle\implies
0\leq\frac{p_{x}+\delta_{X}}{1-p_{x}-\delta_{X}}\sum_{n=1}^{N}$
$\displaystyle\frac{p_{y_{n}}}{p_{y_{0}}}P^{T}(Z=1\mid
X=1,Y=n)+\frac{p^{E}_{10}p_{y_{0}}-p^{T}_{11}(p_{x}+\delta_{X})}{(1-p_{x}-\delta_{X})p_{y_{0}}}\leq
1$

Solving for the lower bound first, we have

$\displaystyle 0$
$\displaystyle\leq\frac{p_{x}+\delta_{X}}{1-p_{x}-\delta_{X}}\sum_{n=1}^{N}\frac{p_{y_{n}}}{p_{y_{0}}}P^{T}(Z=1\mid
X=1,Y=n)+\frac{p^{E}_{10}p_{y_{0}}-p^{T}_{11}(p_{x}+\delta_{X})}{(1-p_{x}-\delta_{X})p_{y_{0}}}$
$\displaystyle\implies\frac{p^{T}_{11}(p_{x}+\delta_{X})-p^{E}_{10}p_{y_{0}}}{p_{x}+\delta_{X}}\leq\sum_{n=1}^{N}P(Z=1\mid
X=1,Y=n)$

Next, with the upper bound

$\displaystyle\frac{p_{x}+\delta_{X}}{1-p_{x}-\delta_{X}}\sum_{n=1}^{N}\frac{p_{y_{n}}}{p_{y_{0}}}P^{T}(Z=1\mid
X=1,Y=n)+\frac{p^{E}_{10}p_{y_{0}}-p^{T}_{11}(p_{x}+\delta_{X})}{(1-p_{x}-\delta_{X})p_{y_{0}}}\leq
1$
$\displaystyle\implies(p_{x}+\delta_{X})\sum_{n=1}^{N}p_{y_{n}}P^{T}(Z=1\mid
X=1,Y=n)+p^{E}_{10}p_{y_{0}}-p^{T}_{11}(p_{x}+\delta_{X})\leq(1-p_{x}-\delta_{X})p_{y_{0}}$
$\displaystyle\implies\sum_{n=1}^{N}p_{y_{n}}P^{T}(Z=1\mid
X=1,Y=n)\leq\frac{(1-p_{x}-\delta_{X})p_{y_{0}}+p^{T}_{11}(p_{x}+\delta_{X})-p^{E}_{10}p_{y_{0}}}{p_{x}+\delta_{X}}$

Following a similar calculation for $P^{T}(Z=1\mid X=1,Y=0)$, we have

$\displaystyle p^{T}_{11}-p_{y_{0}}\leq\sum_{n=1}^{N}y_{n}P^{T}(Z=1\mid
X=1,Y=n)\leq p^{T}_{11}$

Since all of the bounds need to hold simultaneously, we can re-express them
using using maximum and minimum operators as

$\displaystyle\max\begin{bmatrix}\frac{p^{T}_{11}(p_{x}+\delta_{X})-p^{E}_{10}p_{y_{0}}}{p_{x}}\\\
p^{T}_{11}-p_{y_{0}}\end{bmatrix}\leq\sum_{n=1}^{N}y_{n}\mathbb{P}(Z=1\mid
X=1,Y=n)\leq\min\begin{bmatrix}\frac{(1-p_{x}-\delta_{X})p_{y_{0}}+p^{T}_{11}(p_{x}+\delta_{X})-p^{E}_{10}p_{y_{0}}}{p_{x}+\delta_{X}}\\\
p^{T}_{11}\end{bmatrix}$

Next, for $i\in\\{1,\dots,N\\}$, we have the following constraints on the
basic variables $P(Z=1\mid X=0,Y=i)$

$\displaystyle 0$ $\displaystyle\leq P^{T}(Z=1\mid X=0,Y=i)\leq 1$
$\displaystyle 0$
$\displaystyle\leq\frac{p^{E}_{11}}{1-p_{x}-\delta_{X}}-\frac{p_{x}+\delta_{X}}{1-p_{x}-\delta_{X}}P^{T}(Z=1\mid
X=1,Y=i)\leq 1$ $\displaystyle\implies$
$\displaystyle\frac{p^{E}_{1i}-(1-p_{x}-\delta_{X})}{p_{x}+\delta_{X}}\leq
P^{T}(Z=1\mid X=1,Y=i)\leq\frac{p^{E}_{1i}}{p_{x}+\delta_{X}}$

And since the free parameters are probabilities themselves, they must be
between $0$ and $1$ and hence the bounds for $P(Z=1\mid X=1,Y=i)$ become

$\displaystyle\max\begin{bmatrix}0\\\
\frac{p^{E}_{1i}-(1-p_{x}-\delta_{X})}{p_{x}+\delta_{X}}\end{bmatrix}\leq
P(Z=1\mid X=1,Y=i)\leq\min\begin{bmatrix}1\\\
\frac{p^{E}_{1i}}{p_{x}+\delta_{X}}\end{bmatrix}$

This concludes the proof for the lemma.

##  Appendix H Bounds For The Causal Marginal Problem

Problem Setup: Random variables $X$, $Y$ and $Z$ are arranged in the following
graph structure

$X$ $Y$ $Z$ RCT 2RCT 1 Figure 1: (a) Causal graph representing independent
clinical trials studying identical outcomes.

From the joint distribution $P(Z,X,Y)$ obeying the causal graph structure
above, we observe marginals $P(Z,X)$ and $P(Z,Y)$, and we are interested in
combining these marginals to tighten our bounds on counterfactual quantities.
Specifically, we can obtain bounds on counterfactuals just using $P(Z,X)$,
however, we would like to tighten these bounds by adding in information from
$P(Z,Y)$ as well. The Probabilities of Causation for $X\rightarrow Z$ will be
used as the running example throughout, which is defined as follows:

$PNS\equiv P(Z_{\mathbf{x}}=1,Z_{\mathbf{x^{\prime}}}=0)$

Where $\mathbf{x}$ and $\mathbf{x^{\prime}}$ denote the setting of $X$ to $1$
and $0$ respectively.
Now, bounds on the PNS under the above causal graph when we have access to the
joint $P(Z,X,Y)$ are given in dawid2017probability as

$\displaystyle\Delta\leq PNS\leq\mathbb{P}(Z=1\mid X=1)-\Gamma$

Where

$\Delta=\sum_{y}\mathbb{P}(Y=y)\max\\{0,\mathbb{P}(Z=1\mid
X=1,Y=y)-\mathbb{P}(Z=1\mid X=0,Y=y)\\}$

And $\Gamma$ is

$\displaystyle\Gamma=\sum_{y}\mathbb{P}(Y=y)\max\\{0,\mathbb{P}(Z=1\mid
X=1,Y=y)-\mathbb{P}(Z=0\mid X=0,Y=y)\\}$

These bounds are a function of $P(Z,X,Y)$, and since we only have marginals
$P(Z,X)$ and $P(Z,Y)$, we must modify these bounds before they can be used.
Specifically, the two marginals $P(Z,X)$ and $P(Z,Y)$ are consistent with many
possible joint distributions over $Z$, $X$ and $Y$, and so we must use the
above bounds in the most conservative way otherwise we risk giving erroneous
answers. We demonstrate this in a toy example where there are two possible
joint distributions over $Z$, $X$ and $Y$ that are compatible with the
observed marginals. These joint are denoted as $P_{1}(Z,X,Y)$ and
$P_{2}(Z,X,Y)$. By the properties of the max operator, the expression for
$\Delta$ can be written as

$\displaystyle\max\begin{bmatrix}0\\\ P(Z=1\mid X=1)-P(Z=1\mid X=0)\\\
\mathbb{P}(Z=1\mid X=1,Y=0)-\mathbb{P}(Z=1\mid X=0,Y=0)\\\ \mathbb{P}(Z=1\mid
X=1,Y=1)-\mathbb{P}(Z=1\mid X=0,Y=1)\end{bmatrix}\leq PNS$

The above bound can be computed for $P_{1}(Z,X,Y)$ and $P_{2}(Z,X,Y)$. Each of
these distributions will imply the following lower bound on PNS:

$\displaystyle\max\begin{bmatrix}0\\\ P(Z=1\mid X=1)-P(Z=1\mid X=0)\\\
P_{1}(Z=1\mid X=1,Y=0)-P_{1}(Z=1\mid X=0,Y=0)\\\ P_{1}(Z=1\mid
X=1,Y=1)-P_{1}(Z=1\mid X=0,Y=1)\end{bmatrix}\leq PNS$

And similarly,

$\displaystyle\max\begin{bmatrix}0\\\ P(Z=1\mid X=1)-P(Z=1\mid X=0)\\\
P_{2}(Z=1\mid X=1,Y=0)-P_{2}(Z=1\mid X=0,Y=0)\\\ P_{2}(Z=1\mid
X=1,Y=1)-P_{2}(Z=1\mid X=0,Y=1)\end{bmatrix}\leq PNS$

Now the subtlety lies in that we don’t know whether the true underlying
distribution that produced the marginals is $P_{1}$ or $P_{2}$, i.e. if we
choose the tighter bound on PNS that arises from using $P_{2}$ but the true
distribution that generated the marginals is $P_{1}$, our lower bound is more
restrictive and consequently incorrect.

To remedy this, we choose the lower bound on PNS arising from $P_{1}$ and
$P_{2}$ that is smaller, since this is guaranteed to be correct, i.e. if
$P_{2}$ produces a greater lower bound that $P_{1}$, the bound obtained from
$P_{1}$ will be correct all the time - when either $P_{1}$ or $P_{2}$ is the
true distribution that generated the data, whereas the bound produced by
$P_{2}$ will only be correct when $P_{2}$ is the true distribution that
generated the observed marginals.

This line of reasoning can be formally expressed as

$\displaystyle\min_{P_{i}}\max\begin{bmatrix}0\\\ P(Z=1\mid X=1)-P(Z=1\mid
X=0)\\\ \mathbb{P}(Y=0)\\{P_{i}(Z=1\mid X=1,Y=0)-P_{i}(Z=1\mid X=0,Y=0)\\}\\\
\mathbb{P}(Y=1)\\{P_{i}(Z=1\mid X=1,Y=1)-P_{i}(Z=1\mid
X=0,Y=1)\\}\end{bmatrix}\leq PNS$

And similarly for the upper bound, since it is expressed in the form of
$P(Z=1\mid X=1)-\Gamma$, we focus on $\Gamma$, which can be equivalently
expressed as

$\displaystyle PNS\leq\min_{P_{i}}\max\begin{bmatrix}0\\\
\sum_{y}\mathbb{P}(Y=y)\\{\mathbb{P}(Z=1\mid X=1,Y=y)-\mathbb{P}(Z=0\mid
X=0,Y=y)\\}\\\ P(Y=0)\\{P_{i}(Z=1\mid X=1,Y=0)-P_{i}(Z=0\mid X=0,Y=0)\\}\\\
P(Y=1)\\{P_{i}(Z=1\mid X=1,Y=1)-P_{i}(Z=0\mid X=0,Y=1)\\}\end{bmatrix}$

Now, to compute the minimum values in the above matrices, we need to enumerate
the set of solutions for $P(Z,X,Y)$ which are compatible with the observed
marginals, which is done in the next section.

##  Appendix I Enumerating The Set Of Joint Distributions Compatible With
Observed Marginals

For the graph shaped in 1, we can enumerate the following equations for the
$P(Z,X,Y)$

$\displaystyle\mathbb{P}(Z=1\mid X=1)$ $\displaystyle=\mathbb{P}(Z=1\mid
X=1,Y=1)\mathbb{P}(Y=1)+\mathbb{P}(Z=1\mid X=1,Y=0)\mathbb{P}(Y=0)$
$\displaystyle\mathbb{P}(Z=1\mid X=0)$ $\displaystyle=\mathbb{P}(Z=1\mid
X=0,Y=1)\mathbb{P}(Y=1)+\mathbb{P}(Z=1\mid X=0,Y=0)\mathbb{P}(Y=0)$
$\displaystyle\mathbb{P}(Z=1\mid Y=1)$ $\displaystyle=\mathbb{P}(Z=1\mid
X=0,Y=1)\mathbb{P}(X=0)+\mathbb{P}(Z=1\mid X=1,Y=1)\mathbb{P}(X=1)$
$\displaystyle\mathbb{P}(Z=1\mid Y=0)$ $\displaystyle=\mathbb{P}(Z=1\mid
X=0,Y=0)\mathbb{P}(X=0)+\mathbb{P}(Z=1\mid X=1,Y=0)\mathbb{P}(X=1)$

This represents a system of equations with $4$ unknowns and $3$ linearly
independent constraints, and for notational ease we introduce the following
notation. $OR(X)=\frac{P(X=1)}{P(X=0)}$ and same for $OR(Y)$. $P(X=1)=x$ and
$P(Y=1)=y$, and $P(Z=1\mid X=1)=p_{11}$, $P(Z=1\mid X=0)=p_{10}$, $P(Z=1\mid
Y=1)=p^{\prime}_{11}$, $P(Z=1\mid Y=0)=p^{\prime}_{10}$.

$\displaystyle\begin{bmatrix}1&0&0&-OR(X)OR(Y)&\frac{p_{10}(1-x)-p^{\prime}_{11}y}{(1-x)(1-y)}\\\
0&1&0&OR(Y)&\frac{p_{11}}{1-y}\\\ 0&0&1&OR(X)&\frac{p^{\prime}_{11}}{1-x}\\\
0&0&0&0&0\end{bmatrix}$

And this can equivalently be represented in equation form using the free
parameter $P(Z=1\mid X=1,Y=1)$

$\displaystyle\mathbb{P}(Z=1\mid X=0,Y=0)-OR(X)OR(Y)\mathbb{P}(Z=1\mid
X=1,Y=1)=\frac{p_{10}(1-x)-p^{\prime}_{11}y}{(1-x)(1-y)}$ (67)
$\mathbb{P}(Z=1\mid X=1,Y=0)+OR(Y)\mathbb{P}(Z=1\mid
X=1,Y=1)=\frac{p_{11}}{1-y}$ (68) $\mathbb{P}(Z=1\mid
X=0,Y=1)+OR(X)\mathbb{P}(Z=1\mid X=1,Y=1)=\frac{p^{\prime}_{11}}{1-x}$ (69)

Now we must find bounds on $\mathbb{P}(Z=1\mid X=1,Y=1)$ such that every term
in this system of equations is a valid probability. This is presented in the
subssections below.

###  I.1 Bounds From $\mathbb{P}(Z=1\mid X=0,Y=1)$

The following requirement can be used to compute bounds

$0\leq\mathbb{P}(Z=1\mid X=0,Y=1)\leq 1$

Expressing this in terms of $\mathbb{P}(Z=1\mid X=1,Y=1)$

$0\leq\frac{p^{\prime}_{11}}{1-x}-OR(X)\mathbb{P}(Z=1\mid X=1,Y=1)\leq 1$

  * •

Starting with the lower bound:

$\displaystyle 0\leq\frac{p^{\prime}_{11}}{1-x}-OR(X)\mathbb{P}(Z=1\mid
X=1,Y=1)$ $\displaystyle\implies x\mathbb{P}(Z=1\mid X=1,Y=1)\leq
p^{\prime}_{11}$ $\displaystyle\implies\mathbb{P}(Z=1\mid
X=1,Y=1)\leq\frac{p^{\prime}_{11}}{x}$

This implies the following bound

$\mathbb{P}(Z=1\mid X=1,Y=1)\leq\frac{p^{\prime}_{11}}{x}$ (70)

  * •

Repeating a similar calculation for the upper bound

$\displaystyle\frac{p^{\prime}_{11}}{1-x}-OR(X)\mathbb{P}(Z=1\mid X=1,Y=1)\leq
1$ $\displaystyle\implies\frac{p^{\prime}_{11}-(1-x)}{x}\leq\mathbb{P}(Z=1\mid
X=1,Y=1)$

This implies the following upper bound

$\frac{p^{\prime}_{11}-(1-x)}{x}\leq\mathbb{P}(Z=1\mid X=1,Y=1)$ (71)

###  I.2 Bounds From $\mathbb{P}(Z=1\mid X=1,Y=0)$

Repeating the above process to obtain bounds on $P(Z=1\mid X=1,Y=0)$

$\frac{p_{11}-(1-y)}{y}\leq\mathbb{P}(Z=1\mid X=1,Y=1)$ (72)

And the upper bound

$\mathbb{P}(Z=1\mid X=1,Y=1)\leq\frac{p_{11}}{y}$ (73)

###  I.3 Bounds Obtained From Equation I

From Equation I

$\displaystyle
0\leq\frac{p_{10}(1-x)-p^{\prime}_{11}y}{(1-x)(1-y)}+OR(X)OR(Y)\mathbb{P}(Z=1\mid
X=1,Y=1)\leq 1$

  * •

Starting with the lower bound, we get

$\displaystyle
0\leq\frac{p_{10}(1-x)-p^{\prime}_{11}y}{(1-x)(1-y)}+OR(X)OR(Y)\mathbb{P}(Z=1\mid
X=1,Y=1)$ $\displaystyle\implies 0\leq
p_{10}(1-x)-p^{\prime}_{11}y+xy\mathbb{P}(Z=1\mid X=1,Y=1)$
$\displaystyle\implies\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\leq\mathbb{P}(Z=1\mid
X=1,Y=1)$

And so, the bound implied is

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\leq\mathbb{P}(Z=1\mid X=1,Y=1)$ (74)

  * •

Repeating for the upper bound:

$\displaystyle\frac{p_{10}(1-x)-p^{\prime}_{11}y}{(1-x)(1-y)}+OR(X)OR(Y)\mathbb{P}(Z=1\mid
X=1,Y=1)\leq 1$ $\displaystyle\implies\mathbb{P}(Z=1\mid
X=1,Y=1)\leq\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$

So, the bound implied is

$\mathbb{P}(Z=1\mid
X=1,Y=1)\leq\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$ (75)

So, the final bounds on $P(Z=1\mid X=1,Y=1)$ are

$\displaystyle\max\begin{bmatrix}0\\\ \frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p_{11}-(1-y)}{y}\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq\mathbb{P}(Z=1\mid
X=1,Y=1)\leq\min\begin{bmatrix}1\\\
\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}\\\ \frac{p_{11}}{y}\\\
\frac{p^{\prime}_{11}}{x}\end{bmatrix}$ (76)

Since there is no implicit ordering on which is the min amongst the options,
as well as which one is the max amongst all the options, this reduces to a
problem of checking all at the relevant extreme points.

###  I.4 Implications Of Upper And Lower Bounds

Each of the choice of the lower bounds being the max or the min has
implications on the system constraints, which we now enumerate.

####  I.4.1 The Lower Bound Is $0$

If $0$ is the lower bound, it means that it must be greater than the rest of
the terms, i.e.

  1. 1.

$0>\frac{p^{\prime}_{11}-(1-x)}{x}\implies p^{\prime}_{11}<1-x$ (77)

  2. 2.

$0>\frac{p_{11}-(1-y)}{y}\implies p_{11}<1-y$ (78)

  3. 3.

$\displaystyle 0>\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\implies
p_{10}(1-x)>p^{\prime}_{11}y$ (79)

####  I.4.2 The Lower Bound Is $\frac{p^{\prime}_{11}-(1-x)}{x}$

This implies the following conditions

  1. 1.

$\frac{p^{\prime}_{11}-(1-x)}{x}>0\implies p^{\prime}_{11}>1-x$ (80)

  2. 2.

$\displaystyle\frac{p^{\prime}_{11}-(1-x)}{x}>\frac{p_{11}-(1-y)}{y}\implies
p^{\prime}_{01}y<p_{01}x$ (81)

  3. 3.

$\displaystyle\frac{p^{\prime}_{11}-(1-x)}{x}>\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\implies
p_{10}>y$ (82)

####  I.4.3 The Lower Bound is $\frac{p_{11}-(1-y)}{y}$

This implies the following conditions

  1. 1.

$\frac{p_{11}-(1-y)}{y}>0\implies p_{11}>(1-y)$ (83)

  2. 2.

$\displaystyle\frac{p_{11}-(1-y)}{y}>\frac{p^{\prime}_{11}-(1-x)}{x}\implies
p_{01}x<p^{\prime}_{01}y$ (84)

  3. 3.

$\displaystyle\frac{p_{11}-(1-y)}{y}>\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\implies
p^{\prime}_{10}>x$ (85)

####  I.4.4 When The Lower Bound Is $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$

The constraints here just amount to inverting the corresponding constrains
above, so we get

$p^{\prime}_{11}y>p_{10}(1-x)$ (86)

$y>p_{10}\\\ $ (87)

$x>p^{\prime}_{10}$ (88)

###  I.5 When The Upper Bound is $1$

  1. 1.

$\displaystyle 1<\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$ It is
worth noting that $xy-(1-x)(1-y)=y-(1-x)$ $\displaystyle\implies
p^{\prime}_{01}y<p_{00}(1-x)$ (89)

  2. 2.

$1<\frac{p_{11}}{y}\implies y<p_{11}$ (90)

  3. 3.

$1<\frac{p^{\prime}_{11}}{x}\implies x<p^{\prime}_{11}$ (91)

###  I.6 The Upper Bound is $\frac{p_{11}}{y}$

  1. 1.

$\frac{p_{11}}{y}<1\implies p_{11}<y$ (92)

  2. 2.

$\displaystyle\frac{p_{11}}{y}<\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}\implies
p^{\prime}_{10}<(1-x)$ (93)

  3. 3.

$\frac{p_{11}}{y}<\frac{p^{\prime}_{11}}{x}\implies p_{11}x<p^{\prime}_{11}y$
(94)

###  I.7 The Upper Bound Is $\frac{p^{\prime}_{11}}{x}$

  1. 1.

$\frac{p^{\prime}_{11}}{x}<1\implies p^{\prime}_{11}<x$ (95)

  2. 2.

$\frac{p^{\prime}_{11}}{x}<\frac{p_{11}}{y}\implies p^{\prime}_{11}y<p_{11}x$
(96)

  3. 3.

$\displaystyle\frac{p^{\prime}_{11}}{x}<\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}\implies
p_{10}<1-y$ (97)

###  I.8 When The Upper Bound is
$\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$

All the relevant constraints here can be obtained by flipping the inequlaities
before, and are enumerated as

$\displaystyle p_{00}(1-x)<p^{\prime}_{01}y$ (98)

$\displaystyle p^{\prime}_{10}>(1-x)$ (99)

$\displaystyle p_{10}>1-y$ (100)

##  Appendix J Deriving Bounds On PNS

Having enumerated the space of solutions for $P(Z,X,Y)$, we can plug them into
our modification of Phil Dawid’s bounds in order to obtain tighter bounds on
PNS. Phil Dawid’s bounds are given as

$\displaystyle\frac{\Delta}{\mathbb{P}(Z=1\mid X=1)}\leq PC\leq
1-\frac{\Gamma}{\mathbb{P}(Z=1\mid X=1)}$

Where

$\Delta=\sum_{y}\mathbb{P}(Y=y)\max\\{0,\mathbb{P}(Z=1\mid
X=1,Y=y)-\mathbb{P}(Z=1\mid X=0,Y=y)\\}$

And $\Gamma$ is

$\displaystyle\Gamma=\sum_{y}\mathbb{P}(Y=y)\max\\{0,\mathbb{P}(Z=1\mid
X=1,Y=y)-\mathbb{P}(Z=0\mid X=0,Y=y)\\}$

For binary $Y$ these can be written as

$\displaystyle\Delta$
$\displaystyle=\mathbb{P}(Y=0)\max\\{0,\mathbb{P}(Z=1\mid
X=1,Y=0)-\mathbb{P}(Z=1\mid X=0,Y=0)\\}$
$\displaystyle+\mathbb{P}(Y=1)\max\\{0,\mathbb{P}(Z=1\mid
X=1,Y=1)-\mathbb{P}(Z=1\mid X=0,Y=1)\\}$ $\displaystyle\Gamma$
$\displaystyle=\mathbb{P}(Y=0)\max\\{0,\mathbb{P}(Z=1\mid
X=1,Y=0)-\mathbb{P}(Z=0\mid X=0,Y=0)\\}$
$\displaystyle+\mathbb{P}(Y=1)\max\\{0,\mathbb{P}(Z=1\mid
X=1,Y=1)-\mathbb{P}(Z=0\mid X=0,Y=1)\\}$

Now as seen in gresele2022causal, we must recover the result that the lower
bound can’t be tightened (since the upper bound of $\lambda_{X}$ can’t be
tightened). So we start with computing $\Delta$, aka the lower bound. To solve
this, we first re-expressing all of the above bounds in terms of the free
parameter $P(Z=1\mid X=1,Y=1)$.

###  J.1 Re-expressing $\Delta$ in terms of $P(Z=1\mid X=1,Y=1)$

Re-expressing all the terms in $\Delta$ in terms of $\mathbb{P}(Z=1\mid
X=1,Y=1)$, using the shortened notation defined previously for ease.

  1. 1.

$P(Y=0)\\{\mathbf{\mathbb{P}(Z=1\mid X=1,Y=0)-\mathbb{P}(Z=1\mid X=0,Y=0)}\\}$

$\displaystyle\mathbf{\mathbb{P}(Z=1\mid X=1,Y=0)-\mathbb{P}(Z=1\mid
X=0,Y=0)}=$
$\displaystyle\frac{p_{11}-p^{\prime}_{10}(1-y)}{(1-x)(1-y)}-\frac{y\mathbb{P}(Z=1\mid
X=1,Y=1)}{(1-x)(1-y)}$

So, this is expressed as

$\frac{p_{11}-p^{\prime}_{10}(1-y)}{(1-x)}-\frac{y\mathbb{P}(Z=1\mid
X=1,Y=1)}{(1-x)}$ (101)

  2. 2.

$P(Y=1)\\{\mathbb{P}(Z=1\mid X=1,Y=1)-\mathbb{P}(Z=1\mid X=0,Y=1)\\}$

$\displaystyle\mathbb{P}(Z=1\mid X=1,Y=1)-\mathbb{P}(Z=1\mid
X=0,Y=1)=\frac{\mathbb{P}(Z=1\mid X=1,Y=1)-p^{\prime}_{11}}{1-x}$

So, this is written as

$\frac{\mathbb{P}(Z=1\mid X=1,Y=1)y-p^{\prime}_{11}y}{1-x}$ (102)

###  J.2 Re-expressing $\Gamma$ in terms of $P(Z=1\mid X=1,Y=1)$

A similar approach to the one used above is utilized, starting with expressing
everything in terms of $P(Z=1\mid X=1,Y=1)$, done below. For notational
convenience, we define the following

$\Phi_{X}=p_{11}+p_{10}-1$

  1. 1.

$P(Y=0)\\{\mathbb{P}(Z=1\mid X=1,Y=0)-\mathbb{P}(Z=0\mid X=0,Y=0)\\}$

$\displaystyle\mathbb{P}(Z=1\mid X=1,Y=0)$
$\displaystyle=\frac{p_{11}}{1-y}-OR(Y)\mathbb{P}(Z=1\mid X=1,Y=1)$
$\displaystyle\mathbb{P}(Z=0\mid X=0,Y=0)$
$\displaystyle=1-\\{\frac{p_{10}(1-x)-p^{\prime}_{11}y}{(1-x)(1-y)}+$
$\displaystyle OR(X)OR(Y)\mathbb{P}(Z=1\mid X=1,Y=1)\\}$

Subtracting the two, we get

$\displaystyle OR(Y)P(Z=1\mid
X=1,Y=1)(OR(X)-1)+\frac{p_{11}}{1-y}+\frac{p_{10}(1-x)-p^{\prime}_{11}y}{(1-x)(1-y)}-1$
$\displaystyle=OR(Y)P(Z=1\mid
X=1,Y=1)(OR(X)-1)+\frac{p_{11}(1-x)+p_{10}(1-x)-p^{\prime}_{11}y-(1-x)(1-y)}{(1-x)(1-y)}$
$\displaystyle=OR(Y)P(Z=1\mid
X=1,Y=1)(OR(X)-1)+\frac{(1-x)\Phi_{X}+(1-x)y-p^{\prime}_{11}y}{(1-x)(1-y)}$
$\displaystyle=OR(Y)P(Z=1\mid
X=1,Y=1)(OR(X)-1)+\frac{\Phi_{X}+y}{1-y}-\frac{p^{\prime}_{11}y}{(1-x)(1-y)}$

Now, this difference is multiplied by $1-y$, so it is finally written as

$\displaystyle=yP(Z=1\mid
X=1,Y=1)(OR(X)-1)+\Phi_{X}+y-\frac{p^{\prime}_{11}y}{1-x}$ (103)

  2. 2.

$P(Y=1)\\{\mathbb{P}(Z=1\mid X=1,Y=1)-\mathbb{P}(Z=0\mid X=0,Y=1)\\}$

$\displaystyle\mathbb{P}(Z=0\mid X=0,Y=1)$
$\displaystyle=1-\Bigg{(}\frac{p^{\prime}_{11}}{1-x}-OR(X)\mathbb{P}(Z=1\mid
X=1,Y=1)\Bigg{)}$
$\displaystyle=1-\frac{p^{\prime}_{11}}{1-x}+OR(X)\mathbb{P}(Z=1\mid X=1,Y=1)$

And so,

$\displaystyle\mathbb{P}(Z=1\mid X=1,Y=1)-\mathbb{P}(Z=0\mid X=0,Y=1)$
$\displaystyle=P(Z=1\mid
X=1,Y=1)-1+\frac{p^{\prime}_{11}}{1-x}-OR(X)\mathbb{P}(Z=1\mid X=1,Y=1)$
$\displaystyle=P(Z=1\mid X=1,Y=1)(1-OR(X))+\frac{p^{\prime}_{11}-(1-x)}{1-x}$

Multiplying by $y$, this will be written as

$P(Z=1\mid X=1,Y=1)(1-OR(X))y+y\frac{p^{\prime}_{11}-(1-x)}{1-x}\\\ $ (104)

###  J.3 Sum of the above two terms

This is a sanity check, since we know these must add up to $p_{11}+p_{10}-1$.

$\displaystyle yP(Z=1\mid
X=1,Y=1)(OR(X)-1)+\Phi_{X}+y-\frac{p^{\prime}_{11}y}{1-x}$
$\displaystyle+P(Z=1\mid
X=1,Y=1)(1-OR(X))y+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$
$\displaystyle=\Phi_{X}+y-\frac{p^{\prime}_{11}y}{1-x}+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$
$\displaystyle=\Phi_{X}$

So, sanity check passed.

##  Appendix K Final Proof: Causal Marginal Problem Lower Bound

Given $P(Z,X)$ and $P(Z,Y)$, and indexing the set of joint distributions
$P(Z,X,Y)$ compatible with these constraints by $P^{(i)}$, the lower bound for
the causal marginal problem is obtained by solving

$\displaystyle\min_{P_{i}}\max\begin{bmatrix}0\\\ P(Z=1\mid X=1)-P(Z=1\mid
X=0)\\\ \mathbb{P}(Y=0)\\{P_{i}(Z=1\mid X=1,Y=0)-P_{i}(Z=1\mid X=0,Y=0)\\}\\\
\mathbb{P}(Y=1)\\{P_{i}(Z=1\mid X=1,Y=1)-P_{i}(Z=1\mid
X=0,Y=1)\\}\end{bmatrix}\leq PNS$

Now, by solving the system of linear equations from the implied consistency
constraints, we see there is a single free parameter $P(Z=1\mid X=1,Y=1)$, and
so the above can be re-expressed as

$\displaystyle\min_{P(Z=1\mid X=1,Y=1)}\max\begin{bmatrix}0\\\ P(Z=1\mid
X=1)-P(Z=1\mid X=0)\\\ \frac{\mathbb{P}(Z=1\mid X=1)-\mathbb{P}(Z=1\mid
Y=0)\mathbb{P}(Y=0)}{\mathbb{P}(X=0)}-\frac{\mathbb{P}(Y=1)\mathbb{P}(Z=1\mid
X=1,Y=1)}{\mathbb{P}(X=0)}\\\ \frac{\mathbb{P}(Z=1\mid
X=1,Y=1)\mathbb{P}(Y=1)-\mathbb{P}(Z=1\mid
Y=1)\mathbb{P}(Y=1)}{\mathbb{P}(X=0)}\end{bmatrix}$

Now, to make proofs easier, we re-write the above as

$\displaystyle\min_{P(Z=1\mid X=1,Y=1)}\max\begin{bmatrix}0\\\ P(Z=1\mid
X=1)-P(Z=1\mid X=0)\\\ \mathbb{P}(Z=1\mid X=1)-\mathbb{P}(Z=1\mid
X=0)+\frac{P(Z=1\mid
Y=1)P(Y=1)}{P(X=0)}-\frac{\mathbb{P}(Y=1)\mathbb{P}(Z=1\mid
X=1,Y=1)}{\mathbb{P}(X=0)}\\\ \frac{\mathbb{P}(Z=1\mid
X=1,Y=1)\mathbb{P}(Y=1)-\mathbb{P}(Z=1\mid
Y=1)\mathbb{P}(Y=1)}{\mathbb{P}(X=0)}\end{bmatrix}$

And the lower bound on $P(Z=1\mid X=1,Y=1)$ will be

$\displaystyle\max\begin{bmatrix}0\\\ \frac{\mathbb{P}(Z=1\mid
Y=1)-\mathbb{P}(X=0)}{\mathbb{P}(X=1)}\\\ \frac{\mathbb{P}(Z=1\mid
X=1)-\mathbb{P}(Y=0)}{\mathbb{P}(Y=1)}\\\ \frac{\mathbb{P}(Z=1\mid
Y=1)\mathbb{P}(Y=1)-\mathbb{P}(Z=1\mid
X=0)\mathbb{P}(X=0)}{\mathbb{P}(X=1)\mathbb{P}(Y=1)}\end{bmatrix}\leq\mathbb{P}(Z=1\mid
X=1,Y=1)$

And the upper bound on $P(Z=1\mid X=1,Y=1)$ will be

$\displaystyle\mathbb{P}(Z=1\mid X=1,Y=1)\leq\min\begin{bmatrix}1\\\
\frac{\mathbb{P}(X=0)\mathbb{P}(Y=0)-\mathbb{P}(Z=1\mid
X=0)\mathbb{P}(X=0)+\mathbb{P}(Z=1\mid
Y=1)\mathbb{P}(Y=1)}{\mathbb{P}(X=1)\mathbb{P}(Y=1)}\\\
\frac{\mathbb{P}(Z=1\mid X=1)}{\mathbb{P}(Y=1)}\\\ \frac{\mathbb{P}(Z=1\mid
Y=1)}{\mathbb{P}(X=1)}\end{bmatrix}$

Now, the choice between $0$ and $P(Z=1\mid X=1)-P(Z=1\mid X=0)$ is a priori
decided based on $P(Z,X)$. So, the max operator must be analyzed in each of
these cases, presented in the subsections below. For convenience, we use the
following notation for each of the terms. $p^{\prime}_{11}=P(Z=1\mid Y=1)$ and
$p_{11}=P(Z=1\mid X=1)$ and $x=P(X=1)$ and same for $y$, with the relevant
extrapolations needed for the rest of the terms.

Claim: $0\geq P(Z=1\mid X=1)-P(Z=1\mid X=0)$ doesn’t restrict the range of
$P(Z=1\mid X=1,Y=1)$ as it doesn’t contradict any of the implied inequalities
in the choice of bounds for $P(Z=1\mid X=1,Y=1)$.

###  K.1 Case I: $0>p_{11}-p_{10}$

We start by examining the behavior of the max operator as the value of
$P(Z=1\mid X=1,Y=1)$ varies. First we notice that line 3 of the max operator
is a decreasing function of $P(Z=1\mid X=1,Y=1)$ and line 4 is an increasing
function, which can be seen below.

$\displaystyle\max\begin{bmatrix}0\\\ p_{11}-p_{10}\\\
p_{11}-p_{10}+\frac{p^{\prime}_{11}y}{1-x}-\frac{y\mathbb{P}(Z=1\mid
X=1,Y=1)}{1-x}\\\ \frac{\mathbb{P}(Z=1\mid
X=1,Y=1)y-p^{\prime}_{11}y}{1-x}\end{bmatrix}$

This is analyzed piece-wise, to see where line 3 is the output, and where line
4 is the output. First we examine where line 3 would be greater than $0$.

$\displaystyle
p_{11}-p_{10}+\frac{p^{\prime}_{11}y}{1-x}-\frac{y\mathbb{P}(Z=1\mid
X=1,Y=1)}{1-x}\geq 0$
$\displaystyle\implies\frac{\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}\geq\mathbb{P}(Z=1\mid
X=1,Y=1)$

Now, repeating for line 4:

$\displaystyle\frac{\mathbb{P}(Z=1\mid X=1,Y=1)y-p^{\prime}_{11}y}{1-x}\geq 0$
$\displaystyle\implies P(Z=1\mid X=1,Y=1)\geq p^{\prime}_{11}$

Since $p_{11}-p_{10}\leq
0\implies\frac{\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}\leq
p^{\prime}_{11}$. So we expect the graph to look like

Figure 2: Identifying the lower bound of PNS (y-axis) for different values of
the free parameter $P(Z=1|X=1,Y=1)$ (x-axis). Whenever the lower bound of the
free parameter is less than $a$ or larger than $b$, then the resulting lower
bound for the PNS results into falsification +++.

The dashed line is meant to convey that the graph may not may not have this
slope, for the following reason. If
$\frac{\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}\geq 0$, then
we will have the dashed line, and if not, we will have the flat line.

As long as the range of $P(Z=1\mid X=1,Y=1)$ allows the Y-axis to include $0$,
we will not obtain any falsification. This can happen in the following
scenarios:

  1. 1.

$\frac{\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}\geq 0$, and
the upper bound is always less than
$\frac{\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}$ (since
otherwise the upper bound can be as low as $0$ and still include the $0$ point
in the max)

  2. 2.

The lower bound for $P(Z=1\mid X=1,Y=1)$ is always $>p^{\prime}_{11}$
Atalanti: isn’t it that max of two things?

Now, we check each of these in the following subsections.

####  K.1.1 Case 1

Assume $\frac{\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}\geq 0$,
and we check whether any of the upper bounds are less than this value.

When the upper bound is $1$, it is impossible for $1$ to be less than this
value since it would require $p^{\prime}_{11}>1$.

Next, we check the upper bound when it equals $\frac{p_{11}}{y}$, and use a
proof by contradiction.

Proof: Assume

$\displaystyle\frac{\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}>\frac{p_{11}}{y}$
$\displaystyle\implies p_{11}(1-x)-p_{10}(1-x)+p^{\prime}_{11}y>p_{11}$
$\displaystyle\implies-p_{11}x-p_{10}(1-x)+p^{\prime}_{11}y>0$ From the
property $p^{\prime}_{11}y-p_{11}x-p_{10}(1-x)=-p^{\prime}_{10}(1-y)$
$\displaystyle\implies-p^{\prime}_{10}(1-y)>0$ $\displaystyle\implies
p^{\prime}_{10}(1-y)<0$

This cannot happen from the properties of probability since $0\leq
p^{\prime}_{10}\leq 1$ and $0\leq y\leq 1$.

Next, we check the upper bound when it equals $\frac{p^{\prime}_{11}}{x}$, and
proceed by a similar proof by contradiction.

Proof: Assume

$\displaystyle\frac{\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}>\frac{p^{\prime}_{11}}{x}$
$\displaystyle\implies
p_{11}(1-x)x-p_{10}(1-x)x+p^{\prime}_{11}yx>p^{\prime}_{11}y$
$\displaystyle\implies p_{11}(1-x)x-p_{10}(1-x)x>p^{\prime}_{11}y(1-x)$
$\displaystyle\implies x\Big{\\{}p_{11}-p_{10}\Big{\\}}>p^{\prime}_{11}y$

Since we assume $p_{11}-p_{10}<0$, this produces a contradiction and concludes
the proof.

Finally, when the upper bound is
$\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$, we proceed by a similar
proof by contradiction strategy.

Proof: Assume

$\displaystyle\frac{\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}>\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$
$\displaystyle\implies\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)+p^{\prime}_{11}y>\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{x}$
$\displaystyle\implies
x\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)>(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y(1-x)$
$\displaystyle\implies
x\Big{\\{}p_{11}-p_{10}\Big{\\}}>(1-y)-p_{10}+p^{\prime}_{11}y$
$\displaystyle\implies p_{11}x+p_{10}(1-x)>(1-y)+p^{\prime}_{11}y$
$\displaystyle\implies p_{11}x+p_{10}(1-x)-p^{\prime}_{11}y>(1-y)$ Using the
inequality $p_{11}x+p_{10}(1-x)-p^{\prime}_{11}y=p^{\prime}_{10}(1-y)$
$\displaystyle\implies p^{\prime}_{10}(1-y)>(1-y)$

And this is a contradiction from the properties of probability.

And so, in Case 1, the upper bound will always be
$\geq\frac{\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}$.

####  K.1.2 Case 2

But, we still need to ensure that the lower bound isn’t so high that we wander
into the right side of the graph. So, we follow a similar proof by
contradiction approach to demonstrate the lower bounds relation to
$p^{\prime}_{11}$.

First, by the properties of probability $p^{\prime}_{11}\geq 0$ will always
hold.

Next, when the lower bound is $\frac{p^{\prime}_{11}-(1-x)}{x}$, applying a
proof by contradiction strategy.

Proof: Assume $\frac{p^{\prime}_{11}-(1-x)}{x}>p^{\prime}_{11}$. This creates
a contradiction since this implies $p^{\prime}_{11}>1$, and so we prove the
opposite.

Next, consider the case where the lower bound is $\frac{p_{11}-(1-y)}{y}$,
following a similar proof by contradiction strategy.

Proof: Assume $\frac{p_{11}-(1-y)}{y}>p^{\prime}_{11}$. This implies

$\displaystyle p_{11}>p^{\prime}_{11}y+(1-y)$ $\displaystyle\implies
1-p_{01}>(p^{\prime}_{11}y-y)+1$ $\displaystyle\implies
p_{01}<p^{\prime}_{01}y$

This is a contradiction because of Equation LABEL:lower-bound-max-constraint,
since an implication of $\frac{p_{11}-(1-y)}{y}>p^{\prime}_{11}$ being the
lower bound is $P(Z=0\mid X=1)P(X=1)\geq P(Z=0\mid Y=1)P(Y=1)$ and this
contradicts that.

Finally, consider the case where $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$,
following a similar proof by contradiction approach.

Proof: Assume $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}>p^{\prime}_{11}$, this
implies

$\displaystyle p^{\prime}_{11}y-p_{10}(1-x)>p^{\prime}_{11}xy$ $\displaystyle
p^{\prime}_{11}y(1-x)>p_{10}(1-x)$ $\displaystyle p^{\prime}_{11}y>p_{10}$

$\displaystyle p^{\prime}_{11}y>p_{10}$ $\displaystyle\ext@arrow
0359\Rightarrowfill@{}$ $\displaystyle
p_{11}x+p_{10}(1-x)-p^{\prime}_{10}(1-y)>p_{10}$ $\displaystyle\implies
p_{11}x>p^{\prime}_{10}(1-y)+p_{10}x$
$\displaystyle\implies(p_{11}-p_{10})x>p^{\prime}_{10}(1-y)$

which is a contradiction because $p_{11}-p_{10}<0$.

###  K.2 Case II: $p_{11}-p_{10}>0$

The turning points in this graph are obtained as follows:

$\displaystyle
p_{11}-p_{10}+\frac{p^{\prime}_{11}y}{1-x}-\frac{y\mathbb{P}(Z=1\mid
X=1,Y=1)}{1-x}\geq p_{11}-p_{10}$
$\displaystyle\implies\frac{p^{\prime}_{11}y}{1-x}-\frac{y\mathbb{P}(Z=1\mid
X=1,Y=1)}{1-x}\geq 0\implies p^{\prime}_{11}\geq P(Z=1\mid X=1,Y=1)$

Now, doing a similar calculation for the other turning point

$\displaystyle\frac{\mathbb{P}(Z=1\mid X=1,Y=1)y-p^{\prime}_{11}y}{1-x}\geq
p_{11}-p_{10}$ $\displaystyle\implies P(Z=1\mid
X=1,Y=1)\geq\frac{\Big{\\{}p^{\prime}_{11}-p^{\prime}_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}$

Now, we do a similar case by case analysis for falsification.

####  K.2.1 Case 1

We want to check whether the upper bounds can be less that $p^{\prime}_{11}$.

When the upper bound is $1$, this can never be less than $p^{\prime}_{11}$
from the properties of probability.

Next, when the upper bound is $\frac{p^{\prime}_{11}}{x}$, this will also hold
since $x\leq 1$.

Next, when the upper bound is $\frac{p_{11}}{y}$, we follow a proof by
contradiction, with the proof presented below.

Proof: Assume $p^{\prime}_{11}>\frac{p_{11}}{y}$

$\displaystyle\implies p^{\prime}_{11}y>p_{11}$ $\displaystyle\implies
p_{11}x+p_{10}(1-x)-p^{\prime}_{10}(1-y)>p_{11}$
$\displaystyle\implies-p^{\prime}_{10}(1-y)>\\{p_{11}-p_{10}\\}(1-x)$

This produces a contradiction since $p_{11}-p_{10}>0$ by assumption.

Next, when the upper bound is
$\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$, following a similar
proof by contradiction strategy.

Proof: Assume
$p^{\prime}_{11}>\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$

$\displaystyle\implies
p^{\prime}_{11}xy>(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y$
$\displaystyle\implies 0<(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y(1-x)$
$\displaystyle\implies 0<(1-y)-p_{10}+p^{\prime}_{11}y$ $\displaystyle\implies
0<p_{00}-p^{\prime}_{01}y$ $\displaystyle\implies p^{\prime}_{01}y<p_{00}$

The above proof is wrong ;) . Please double check and replace with the one I
wrote here:

$\displaystyle\implies
p^{\prime}_{11}xy>(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y$
$\displaystyle\implies 0>(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y(1-x)$
$\displaystyle\implies 0>(1-y)-p_{10}+p^{\prime}_{11}y$ $\displaystyle\implies
p_{10}>(1-y)+p^{\prime}_{11}y$ $\displaystyle\implies
0>(1-y)+p_{11}x-p_{10}x-p^{\prime}_{10}(1-y)$ $\displaystyle\implies
p^{\prime}_{10}(1-y)>(1-y)+x(p_{11}-p_{10})$
$\displaystyle\implies(p^{\prime}_{10}-1)>x\frac{p_{11}-p_{10}}{1-y}$

which is a contradiction because $x\frac{p_{11}-p_{10}}{1-y}$ is a positive
quantity and $p^{\prime}_{10}-1$ is a negative one, since $p^{\prime}_{10}$ is
a probability.

####  K.2.2 Case 2

We check whether the lower bounds can be greater than
$\frac{\Big{\\{}p^{\prime}_{11}-p^{\prime}_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}$.

  * •

Lower Bound Of $0$: This produces a contradiction since $p_{11}-p_{10}$ is
positive, and $p^{\prime}_{11}$ is also positive, and so a sum of two positive
quantities cannot be negative.

  * •

Lower Bound of $\frac{p^{\prime}_{11}-(1-x)}{x}$: This follows a proof by
contradiction.

$\displaystyle\frac{\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}<\frac{p^{\prime}_{11}-(1-x)}{x}]$
$\displaystyle\implies
x\\{p_{11}-p_{10}\\}(1-x)+p^{\prime}_{11}xy<p^{\prime}_{11}y-y(1-x)$
$\displaystyle\implies x\\{p_{11}-p_{10}\\}(1-x)<p^{\prime}_{11}y(1-x)-y(1-x)$
$\displaystyle\implies x\\{p_{11}-p_{10}\\}<p^{\prime}_{11}y-y$
$\displaystyle\implies x\\{p_{11}-p_{10}\\}<-yp^{\prime}_{01}$

This produces a contradiction since $p_{11}-p_{10}>0$

  * •

Lower Bound Of $\frac{p_{11}-(1-y)}{y}$: Following a similar proof by
contradiction:

$\displaystyle\frac{\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}<\frac{p_{11}-(1-y)}{y}$
$\displaystyle\implies\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)+p^{\prime}_{11}y<p_{11}-(1-y)$
$\displaystyle\implies-p_{10}(1-x)-p_{11}x+p^{\prime}_{11}y<-(1-y)$
$\displaystyle\ext@arrow
0359\Rightarrowfill@{}{p_{11}x+p_{10}(1-x)=p^{\prime}_{11}y+p^{\prime}_{10}(1-y)}-p^{\prime}_{10}(1-y)<-(1-y)\implies
p^{\prime}_{10}>1$

This is a contradiction from the properties of probability.

  * •

When the lower bound is $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$. Following a
similar proof by contradiction

$\displaystyle\frac{\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)}{y}+p^{\prime}_{11}<\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$
$\displaystyle\implies\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)+p^{\prime}_{11}y<\frac{p^{\prime}_{11}y-p_{10}(1-x)}{x}$
$\displaystyle\implies
x\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)+p^{\prime}_{11}yx<p^{\prime}_{11}y-p_{10}(1-x)$
$\displaystyle\implies
x\Big{\\{}p_{11}-p_{10}\Big{\\}}(1-x)<p^{\prime}_{11}y(1-x)-p_{10}(1-x)$
$\displaystyle\implies
x\Big{\\{}p_{11}-p_{10}\Big{\\}}<p^{\prime}_{11}y-p_{10}$
$\displaystyle\implies xp_{11}+p_{10}(1-x)-p^{\prime}_{11}y<0$
$\displaystyle\ext@arrow
0359\Rightarrowfill@{}{p_{11}x+p_{10}(1-x)=p^{\prime}_{11}y+p^{\prime}_{10}(1-y)}p^{\prime}_{10}(1-y)<0$

This produces a contradiction as well.

So we have shown in both cases where $0>p_{11}-p_{10}$ and vice versa, that
$0$ or $p_{11}-p_{10}$ will always be included based on the allowed range of
$P(Z=1\mid X=1,Y=1)$. This concludes the proof for the lower bound.

##  Appendix L Final Proof: Causal Marginal Problem Upper Bound

Given $P(Z,X)$ and $P(Z,Y)$, and indexing the set of joint distributions
$P(Z,X,Y)$ compatible with these constraints by $P^{(i)}$, we want to pick the
most upper bound for PNS/PC. To do this, we use the following upper bound

$\displaystyle PNS\leq p_{11}-\Gamma$

Where $\Gamma$ is

$\displaystyle\Gamma=\sum_{y}\mathbb{P}(Y=y)\max\\{0,\mathbb{P}(Z=1\mid
X=1,Y=y)-\mathbb{P}(Z=0\mid X=0,Y=y)\\}$

Since we want to pick the highest upper bound, we want to pick the smallest
value of $\Gamma$ since it is subtracted from $P(Z=1\mid X=1)$. So this is
equivalent to

$\min_{P^{(i)}}\Gamma$

Now, based on Equations 1 and 104, we can rewrite the above as

$\displaystyle\min_{P_{i}}\max\begin{bmatrix}0\\\ p_{11}+p_{10}-1\\\
\mathbb{P}(Y=0)\\{\mathbb{P}(Z=1\mid X=1,Y=0)-\mathbb{P}(Z=0\mid
X=0,Y=0)\\}\\\ \mathbb{P}(Y=1)\\{\mathbb{P}(Z=1\mid
X=1,Y=1)-\mathbb{P}(Z=0\mid X=0,Y=1)\\}\end{bmatrix}$

Which from Equations 1 and 104 is equal to

$\displaystyle\min_{P_{i}}\max\begin{bmatrix}0\\\ p_{10}+p_{11}-1\\\
yP(Z=1\mid X=1,Y=1)(OR(X)-1)+\Phi_{X}+y-\frac{p^{\prime}_{11}y}{1-x}\\\
P(Z=1\mid X=1,Y=1)(1-OR(X))y+y\frac{p^{\prime}_{11}-(1-x)}{1-x}\\\
\end{bmatrix}$

Now, this is analyzed on a case by case basis, and for begin with analyzing
the situation where $p_{11}+p_{10}-1<0$

###  L.1 Case I: $p_{11}+p_{10}-1<0$

As we can see from above, depending on whether $OR(X)-1$ is positive or
negative, line 3 and line 4 in the above matrix will be increasing or
decreasing functions of $P(Z=1\mid X=1,Y=1)$.

Assuming $x>1-x$, line 3 is an increasing function of $P(Z=1\mid X=1,Y=1)$ and
line 4 is a decreasing function of this. Since falsification happens when line
3 and line 4 of the matrix are greater than 0, we first check the values of
$P(Z=1\mid X=1,Y=1)$ for which this happens.

  * •

When $yP(Z=1\mid X=1,Y=1)(OR(X)-1)+\Phi_{X}+y-\frac{p^{\prime}_{11}y}{1-x}\geq
0$:

$\implies P(Z=1\mid
X=1,Y=1)\geq\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}\\\ $ (105)

We denote this value as

$b=\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$ (106)

  * •

When $P(Z=1\mid X=1,Y=1)(1-OR(X))y+y\frac{p^{\prime}_{11}-(1-x)}{1-x}\geq 0$

$\implies\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}\geq P(Z=1\mid X=1,Y=1)$ (107)

We denote this value as

$a=\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$ (108)

.

###  L.2 Case By Case Upper Bound Calculation

Each of these sections considers a different case where the values of
$p^{\prime}_{11}$, $p^{\prime}_{10}$ are picked relative to $x$ and $1-x$.

####  L.2.1 Case I

This case assumes the following

$\displaystyle\Phi_{X}<0$ (109) $\displaystyle 1-x<x$ (110) $\displaystyle
1-x<p^{\prime}_{11}<x$ (111) $\displaystyle p^{\prime}_{10}<1-x<x$ (112)

The first two have been assumed in a larger context, and the last two are made
for this specific case for this section.

These conditions, based on the implications in subsection I.4, limit the
choice of bounds on $P(Z=1\mid X=1,Y=1)$ from (76) to

$\max\begin{bmatrix}\frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq\min\begin{bmatrix}\frac{p_{11}}{y}\\\
\frac{p^{\prime}_{11}}{x}\end{bmatrix}$

Now, we know falsification will happen when either lower bound is greater than
106, or the upper bound is less than 108. So we examine each choice of upper
and lower bound, and see whether they can be less than 108 or 106.

  1. 1.

$\frac{p^{\prime}_{11}-(1-x)}{x}$: This will be less than 108, i.e.
$\frac{p^{\prime}_{11}-(1-x)}{x}<\frac{p^{\prime}_{11}-(1-x)}{2x-1}$ since the
denominator is smaller on the right hand side. Since this is less than $(a)$,
it will also be less than 106.

  2. 2.

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$: This will always be less than 106,
which we prove below. Assume

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}>\frac{p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)}{y(x-(1-x))}$

Checking the right hand side of the implication

$\displaystyle\implies
p^{\prime}_{11}y(x-(1-x))-p_{10}(1-x)(x-(1-x))>p^{\prime}_{11}xy-x(1-x)(p_{11}+p_{10}-1)-xy(1-x)$
$\displaystyle\implies-p^{\prime}_{11}y-p_{10}(x-(1-x))>-x(p_{11}+p_{10}-1)-xy$
$\displaystyle\implies-p^{\prime}_{11}y+p_{10}(1-x)-p_{10}x>-xp_{11}-xp_{10}+x-xy$
$\displaystyle\implies-p^{\prime}_{11}y+p_{10}(1-x)+xp_{11}>x(1-y)$
$\displaystyle\implies p^{\prime}_{10}(1-y)>x(1-y)\implies p^{\prime}_{10}>x$

This is a contradiction based on the case we are in.

And so, the lower bounds will always be less than 106, so any falsification
will come from the upper bounds being less than 108, which we check below.

  1. 1.

When the upper bound is $\frac{p_{11}}{y}$: Checking this against 106 using a
proof by contradiction, we assume

$\displaystyle\frac{p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)}{y(x-(1-x))}<\frac{p_{11}}{y}$
$\displaystyle\implies
p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)<p_{11}x-p_{11}(1-x)$
$\displaystyle\implies
p_{11}x+p_{10}(1-x)-p^{\prime}_{10}(1-y)-(1-x)(p_{11}+p_{10}-1)-y(1-x)<p_{11}x-p_{11}(1-x)$
$\displaystyle\implies
p_{10}(1-x)+p_{11}(1-x)-p^{\prime}_{10}(1-y)-(1-x)(p_{11}+p_{10}-1)-y(1-x)<0$
$\displaystyle\implies-p^{\prime}_{10}(1-y)+(1-x)-y(1-x)<0$
$\displaystyle\implies-p^{\prime}_{10}(1-y)+(1-x)(1-y)<0$
$\displaystyle\implies-p^{\prime}_{10}+(1-x)<0$

This is a contradiction, and so
$\frac{p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)}{y(x-(1-x))}\geq\frac{p_{11}}{y}$.
Similarly, checking $\frac{p_{11}}{y}$ against 108

$\displaystyle\frac{p_{11}}{y}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}\implies
p_{11}x-p_{11}(1-x)<p^{\prime}_{11}y-y(1-x)$ $\displaystyle\implies
p_{11}x-p_{11}(1-x)<p_{11}x+p_{10}(1-x)-p^{\prime}_{10}(1-y)-y(1-x)$
$\displaystyle\implies-p_{11}(1-x)<p_{10}(1-x)-p^{\prime}_{10}(1-y)-y(1-x)$
$\displaystyle\implies p^{\prime}_{10}(1-y)<p_{10}(1-x)+p_{11}(1-x)-y(1-x)$
$\displaystyle\implies p^{\prime}_{10}(1-y)<(1-x)(p_{10}+p_{11}-(1-(1-y)))$
$\displaystyle\implies p^{\prime}_{10}(1-y)<(1-x)(p_{10}+p_{11}-1)+(1-x)(1-y)$
$\displaystyle\implies\frac{(1-y)(p^{\prime}_{10}-(1-x))}{1-x}<\Phi_{X}$

So, when $\frac{(1-y)(p^{\prime}_{10}-(1-x))}{1-x}<\Phi_{X}$,
$\frac{p_{11}}{y}$ will be less than 108 and we can get falsification.

  2. 2.

When the upper bound is $\frac{p^{\prime}_{11}}{x}$: Checking this against the
first turning point, since the graphical argument above is reliant on this.

$\displaystyle\frac{p^{\prime}_{11}}{x}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}\implies
p^{\prime}_{11}x-p^{\prime}_{11}(1-x)<p^{\prime}_{11}x-x(1-x)$
$\displaystyle\implies p^{\prime}_{11}>x$

This is a contradiction because of the case we are in, and so
$\frac{p^{\prime}_{11}}{x}\geq\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$

Putting this all together, any falsification will come from the upper bounds
being less than 108. The only upper bound capable of being less than 108 is
$\frac{p_{11}}{y}$. This gives us the first needed condition for falsification

$\frac{p_{11}}{y}>\frac{p^{\prime}_{11}}{x}\implies p_{11}x>p^{\prime}_{11}y$

Next, we examine the pairing with each of the lower bounds.

  1. 1.

When the lower bound is $\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$: As proved
before, this will always be less than 108, so in this case we only check when
$\frac{p_{11}}{y}$ is less than (a). We have solved this before and the
condition is $\frac{(1-y)(p^{\prime}_{10}-(1-x))}{1-x}<\Phi_{X}$. Then
$\frac{p_{11}}{y}$ is substituted into $P(Z=1\mid
X=1,Y=1)(1-OR(X))y+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$. And this equals

$\displaystyle p_{11}(1-OR(X))+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$
$\displaystyle=p_{11}-\frac{p_{11}x}{1-x}+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$
$\displaystyle=p_{11}+\frac{p^{\prime}_{11}y-y(1-x)-p_{11}x}{1-x}$
$\displaystyle=p_{11}+\frac{p_{10}(1-x)-p^{\prime}_{10}(1-y)-y(1-x)}{1-x}$
$\displaystyle=\frac{p_{11}(1-x)+p_{10}(1-x)-p^{\prime}_{10}(1-y)-(1-(1-y))(1-x)}{1-x}$
$\displaystyle=\Phi_{X}+\frac{-p^{\prime}_{10}(1-y)+(1-y)(1-x)}{1-x}$
$\displaystyle=\Phi_{X}-\frac{(1-y)(p^{\prime}_{10}(1-y)-(1-x)}{1-x}$

This will be the output of $\Gamma$.

  2. 2.

When the lower bound is $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$: Since the
upper bound is the same, the falsification calculation is the same. Since the
lower bound is always less than the upper bound, we have the same calculation
and same condition for falsification.

Putting this all together, in Case I, when the following conditions hold:

$\displaystyle p_{11}x>p^{\prime}_{11}y$
$\displaystyle\frac{(1-y)(p^{\prime}_{10}-(1-x))}{1-x}<\Phi_{X}$

We will get falsification, where $\Gamma$ is calculated as

$\Gamma=\Phi_{X}-\frac{(1-y)(p^{\prime}_{10}-(1-x))}{1-x}$

Now, subtracting this from $p_{11}$, we have

$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{11}-(p_{11}+p_{10}-1)+\frac{(1-y)(p^{\prime}_{10}-(1-x))}{1-x}$
$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{00}-\frac{(1-y)((1-x)-p^{\prime}_{10})}{1-x}$

####  L.2.2 Case II

This case assumes the following

$\displaystyle\Phi_{X}<0$ (113) $\displaystyle 1-x<x$ (114) $\displaystyle
1-x<p^{\prime}_{11}<x$ (115) $\displaystyle 1-x<p^{\prime}_{10}<x$ (116)

These conditions limit the choice of bounds on $P(Z=1\mid X=1,Y=1)$ from (76)
to

$max\begin{bmatrix}\frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq min\begin{bmatrix}\frac{p^{\prime}_{11}}{x}\\\
\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}\end{bmatrix}$

Now, we check whether the lower bounds are greater than (b) and if the upper
bounds are less than (a).

  1. 1.

First, $\frac{p^{\prime}_{11}-(1-x)}{x}<\frac{p^{\prime}_{11}-(1-x)}{2x-1}$
since the denominator is smaller on the right hand side. This lower bound is
less than (a), and will consequently be less than (b).

  2. 2.

Checking the lower bound of $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$ against
(b), using a proof by contradiction. Assume,

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}>\frac{p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)}{y(x-(1-x))}$

Checking the right hand side of the implication

$\displaystyle\implies
p^{\prime}_{11}y(x-(1-x))-p_{10}(1-x)(x-(1-x))>p^{\prime}_{11}xy-x(1-x)(p_{11}+p_{10}-1)-xy(1-x)$
$\displaystyle\implies-p^{\prime}_{11}y-p_{10}(x-(1-x))>-x(p_{11}+p_{10}-1)-xy$
$\displaystyle\implies-p^{\prime}_{11}y+p_{10}(1-x)-p_{10}x>-xp_{11}-xp_{10}+x-xy$
$\displaystyle\implies-p^{\prime}_{11}y+p_{10}(1-x)+xp_{11}>x(1-y)$
$\displaystyle\implies p^{\prime}_{10}(1-y)>x(1-y)\implies p^{\prime}_{10}>x$

This is a contradiction based on the case we are in. So,

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\leq\frac{p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)}{y(x-(1-x))}$

Checking the upper bounds now.

  1. 1.

When the upper bound is $\frac{p^{\prime}_{11}}{x}$: Checking this against the
(a).

$\displaystyle\frac{p^{\prime}_{11}}{x}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}\implies
p^{\prime}_{11}x-p^{\prime}_{11}(1-x)<p^{\prime}_{11}x-x(1-x)$
$\displaystyle\implies p^{\prime}_{11}>x$

This is a contradiction because of the case we are in. So

$\frac{p^{\prime}_{11}}{x}\geq\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$

  2. 2.

When the upper bound is $\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$:
Assume

$\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$
$\displaystyle\implies(1-x)(1-y)(x-(1-x))-p_{10}(1-x)(x-(1-x))+p^{\prime}_{11}y(x-(1-x))<p^{\prime}_{11}xy-
xy(1-x)$
$\displaystyle\implies(1-x)(1-y)(x-(1-x))-p_{10}(1-x)(x-(1-x))-p^{\prime}_{11}y(1-x)<-xy(1-x)$
$\displaystyle\implies(1-y)(x-(1-x))-p_{10}(x-(1-x))-p^{\prime}_{11}y<-xy$
$\displaystyle\implies x-xy-(1-y)(1-x)-p_{10}(x-(1-x))-p^{\prime}_{11}y<-xy$
$\displaystyle\implies x-(1-y)(1-x)-p_{10}(x-(1-x))-p^{\prime}_{11}y<0$
$\displaystyle\implies x-p_{10}x-(1-y)(1-x)+p_{10}(1-x)-p^{\prime}_{11}y<0$
$\displaystyle\implies x-p_{10}x-(1-y)(1-x)+p^{\prime}_{10}(1-y)-p_{11}x<0$
$\displaystyle\implies x(1-p_{10}-p_{11})-(1-y)(1-x)+p^{\prime}_{10}(1-y)<0$
$\displaystyle\implies(1-y)(p^{\prime}_{10}-(1-x))<x(p_{11}+p_{10}-1)$

This creates a contradiction since the LHS is positive but the RHS is
negative.

So, no falsification in this case. So,

$\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}\geq\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$

Since the lower bounds are less than (b), and upper bounds are greater than
(a), so $\Gamma=0$.

####  L.2.3 Case III

This case assumes the following conditions

$\displaystyle\Phi_{X}<0$ (117) $\displaystyle 1-x<x$ (118) $\displaystyle
1-x<p^{\prime}_{11}<x$ (119) $\displaystyle 1-x<x<p^{\prime}_{10}$ (120)

These conditions limit the choice of bounds on $P(Z=1\mid X=1,Y=1)$ from (76)
to

$\max\begin{bmatrix}\frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p_{11}-(1-y)}{y}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq\min\begin{bmatrix}\frac{p^{\prime}_{11}}{x}\\\
\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}\end{bmatrix}$

Following the usual approach, we start with checking the lower bounds against
106.

  1. 1.

The lower bound of $\frac{p^{\prime}_{11}-(1-x)}{x}$ will be less than 108,
and so less than 106.

  2. 2.

Checking $\frac{p_{11}-(1-y)}{y}$: Checking whether
$\frac{p_{11}-(1-y)}{y}\geq\frac{p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)}{y(x-(1-x))}$.
Assume

$\displaystyle\frac{p_{11}-(1-y)}{y}<\frac{p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)}{y(x-(1-x))}$
$\displaystyle\implies
p_{11}-(1-y)<\frac{p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)}{(x-(1-x))}$
$\displaystyle\implies
p_{11}x-p_{11}(1-x)-(1-y)(x-(1-x))<p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)$
$\displaystyle\implies
p^{\prime}_{10}(1-y)-p_{10}(1-x)-p_{11}(1-x)-(1-y)x+(1-y)(1-x)<-(1-x)(p_{11}+p_{10}-1)-y(1-x)$
$\displaystyle\implies
p^{\prime}_{10}(1-y)-(1-x)(p_{10}+p_{11}-1)-(1-y)x-y(1-x)<-(1-x)(p_{11}+p_{10}-1)-y(1-x)$
$\displaystyle\implies p^{\prime}_{10}<x$

This contradicts the assumptions of this case, so we expect
$\frac{p_{11}-(1-y)}{y}\geq\frac{p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)}{y(x-(1-x))}$
to hold. And we will come back to this for falsification calculations.

Now, we check the upper bounds against 108.

  1. 1.

Upper bound is $\frac{p^{\prime}_{11}}{x}$:

$\displaystyle\frac{p^{\prime}_{11}}{x}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}\implies
p^{\prime}_{11}x-p^{\prime}_{11}(1-x)<p^{\prime}_{11}x-x(1-x)$
$\displaystyle\implies p^{\prime}_{11}>x$

This is a contradiction because of the case we are in. So

$\frac{p^{\prime}_{11}}{x}\geq\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$

  2. 2.

Upper bound is $\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$. Assume

$\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$
$\displaystyle\implies(1-y)(p^{\prime}_{10}-(1-x))<x\Phi_{X}$

This is a contradiction since the RHS is negative, but the LHS in positive.

Now, the first condition that must be satisfied is $\frac{p_{11}-(1-y)}{y}$ is
the lower bound, which requires the following condition to hold

$p_{01}x<p^{\prime}_{01}y$

Then, we substitute the lower bound into the increasing function of $P(Z=1\mid
X=1,Y=1)$

$\displaystyle\frac{(1-x)(p_{10}+p_{11}-1)-p^{\prime}_{11}y+P(Z=1\mid
X=1,Y=1)y(x-(1-x))+(1-x)y}{1-x}$
$\displaystyle=\frac{(1-x)(p_{10}+p_{11}-1)-p^{\prime}_{11}y+(p_{11}-(1-y))(x-(1-x))+(1-x)y}{1-x}$
$\displaystyle=\frac{(1-x)(p_{10}+p_{11}-1)+p^{\prime}_{10}(1-y)-p_{10}(1-x)-(p_{11}-(1-y))(1-x)+(1-x)y}{1-x}$
$\displaystyle=\frac{(1-x)(p_{10}+p_{11}-1)+p^{\prime}_{10}(1-y)-(1-x)(p_{10}+p_{11}-1)-x(1-y)}{1-x}$
$\displaystyle=\frac{p^{\prime}_{10}(1-y)-x(1-y)}{1-x}=\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

So in this case, when $p_{01}x<p^{\prime}_{01}y$ holds, we get falsification
with $\Gamma$ evaluating to

$\Gamma=\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

And so the upper bound for PNS will be

$p_{11}-\Gamma=p_{11}-\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

####  L.2.4 Case IV

We assume the following conditions

$\displaystyle\Phi_{X}<0$ (121) $\displaystyle 1-x<x$ (122) $\displaystyle
p^{\prime}_{11}<1-x<x$ (123) $\displaystyle p^{\prime}_{10}<1-x<x$ (124)

These constraints imply the following lower and upper bounds on $P(Z=1\mid
X=1,Y=1)$ from (76) to

$\displaystyle\max\begin{bmatrix}0\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq\min\begin{bmatrix}\frac{p_{11}}{y}\\\
\frac{p^{\prime}_{11}}{x}\end{bmatrix}$

Here, since 108 will always be negative, we only need to check whether the
lower bounds will be greater than 106.

  1. 1.

Lower bound is $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$: Assume

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}>\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$
$\displaystyle\implies
p^{\prime}_{11}yx-p^{\prime}_{11}y(1-x)-p_{10}x(1-x)+p_{10}(1-x)^{2}>p^{\prime}_{11}xy-x(1-x)\Phi_{X}-xy(1-x)$
$\displaystyle\implies-p^{\prime}_{11}y-p_{10}x+p_{10}(1-x)>-x\Phi_{X}-xy$
$\displaystyle\implies p^{\prime}_{10}(1-y)-p_{10}x-p_{11}x+xy>-x\Phi_{X}$
$\displaystyle\implies
p^{\prime}_{10}(1-y)-x(p_{10}+p_{11}-1)-(1-y)x>-x\Phi_{X}$
$\displaystyle\implies p^{\prime}_{10}(1-y)-(1-y)x>0$

This is a contradiction since this implies $p^{\prime}_{10}>x$, and so by
contradiction

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\leq\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$

  2. 2.

Lower bound is $0$: It requires this condition to hold

$\frac{y(p^{\prime}_{11}-(1-x))}{1-x}<\Phi_{X}$

The lower bound $0$ will be greater than (b).

Consequently, when $0$ is the lower bound, which requires the following
condition to hold

$p_{10}(1-x)>p^{\prime}_{11}y$

Then $\Gamma$ will evaluate to

$\Phi_{X}-\frac{y(p^{\prime}_{11}-(1-x))}{1-x}$

And the upper bound on PNS will be

$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{11}-\Phi_{X}+\frac{y(p^{\prime}_{11}-(1-x))}{1-x}$
$\displaystyle=p_{00}-\frac{y((1-x)-p^{\prime}_{11})}{1-x}$

####  L.2.5 Case V

We assume the following conditions

$\displaystyle\Phi_{X}<0$ (125) $\displaystyle 1-x<x$ (126) $\displaystyle
p^{\prime}_{11}<1-x<x$ (127) $\displaystyle 1-x<p^{\prime}_{10}<x$ (128)

These constraints imply the following lower bounds on $P(Z=1\mid X=1,Y=1)$
from (76). Only the lower bound is checked against 106 since 108 will be
negative in this case.

$\displaystyle\max\begin{bmatrix}0\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq P(Z=1\mid X=1,Y=1)$

The same argument as before will apply for
$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$ being less than 106.

The same condition holds on $0$ as well, so we get the requirement

$p_{10}(1-x)>p^{\prime}_{11}y$

Then, $\Gamma$ will evaluate to

$\Phi_{X}-\frac{y(p^{\prime}_{11}-(1-x))}{1-x}$

And so, the upper bound on the PNS

$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{11}-\Phi_{X}+\frac{y(p^{\prime}_{11}-(1-x))}{1-x}$
$\displaystyle=p_{00}-\frac{y((1-x)-p^{\prime}_{11})}{1-x}$

####  L.2.6 Case VI

We assume the following conditions

$\displaystyle\Phi_{X}<0$ (129) $\displaystyle 1-x<x$ (130) $\displaystyle
p^{\prime}_{11}<1-x<x$ (131) $\displaystyle 1-x<x<p^{\prime}_{10}$ (132)

Similar to the previous case, only the lower bound needs to be checked. These
constraints imply the following lower bounds on $P(Z=1\mid X=1,Y=1)$ from (76)
as

$\displaystyle max\begin{bmatrix}0\\\ \frac{p_{11}-(1-y)}{y}\end{bmatrix}\leq
P(Z=1\mid X=1,Y=1)$

Now, we check whether each of these lower bounds can be greater than 106.

  1. 1.

When lower bound is $\frac{p_{11}-(1-y)}{y}$: We provide a proof by
contradiction. Assume

$\frac{p_{11}-(1-y)}{y}<\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$
$\displaystyle\implies
p_{11}x-p_{11}(1-x)-(1-y)x+(1-y)(1-x)<p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)$
$\displaystyle\implies
p^{\prime}_{10}(1-y)-p_{10}(1-x)-p_{11}(1-x)-(1-y)x+(1-x)-y(1-x)<-(1-x)\Phi_{X}-(1-x)y$
$\displaystyle\implies
p^{\prime}_{10}(1-y)-(1-x)\Phi_{X}-(1-y)x<-(1-x)\Phi_{X}$
$\displaystyle\implies p^{\prime}_{10}(1-y)-(1-y)x<0$ $\displaystyle\implies
p^{\prime}_{10}-x<0$

This is a contradiction since we assume $p^{\prime}_{10}>x$. So, this lower
bound is capable of falsifying.

  2. 2.

When the lower bound is $0$: If the following condition holds as before, we
get falsification:

$\frac{y(p^{\prime}_{11}-(1-x))}{1-x}<\Phi_{X}$

So, the falsification will be as follows.

When $p_{11}>1-y$, $\frac{p_{11}-(1-y)}{y}$ will be the lower bound, and
$\Gamma$ will be evaluating by inserting this lower bound into $yP(Z=1\mid
X=1,Y=1)(OR(X)-1)+\Phi_{X}+y-\frac{p^{\prime}_{11}y}{1-x}$. And $\Gamma$ will
evaluate to

$\Gamma=\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

And therefore the upper bound on PNS will be

$\displaystyle p_{11}-\Gamma=p_{11}-\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

Next, when $p_{11}<1-y$, $0$ will be the lower bound, and $\Gamma$ will
evaluate to

$\Gamma=\Phi_{X}-\frac{y(p^{\prime}_{11}-(1-x))}{1-x}$

And the upper bound on PNS will be

$\displaystyle p_{11}-\Gamma=p_{00}-\frac{y((1-x)-p^{\prime}_{11})}{1-x}$

####  L.2.7 Case VII: Atalanti

We assume the following conditions

$\displaystyle\Phi_{X}<0$ (133) $\displaystyle 1-x<x$ (134) $\displaystyle
1-x<x<p^{\prime}_{11}$ (135) $\displaystyle p^{\prime}_{10}<1-x<x$ (136)

These constraints imply based on (76) and the implications on Subsection I.4,
the following lower and upper bounds on $P(Z=1\mid X=1,Y=1)$:

$\displaystyle\max\begin{bmatrix}\frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq\min\begin{bmatrix}1\\\ \frac{p_{11}}{y}\end{bmatrix}$

Now we will check if each of the possible lower bounds are greater than
$b=\frac{(1-x)(p^{\prime}_{11}y-\Phi_{X}-y)}{y(x-(1-x))}$ (106).

  1. 1.

$\frac{p^{\prime}_{11}-(1-x)}{x}>\frac{(1-x)(p^{\prime}_{11}y-\Phi_{X}-y)}{y(x-(1-x))}$

. This is a contradiction because we see that
$\frac{p^{\prime}_{11}-(1-x)}{x}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}=a$.
Which implies that $\frac{p^{\prime}_{11}-(1-x)}{x}<b$.

  2. 2.

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}>\frac{(1-x)(p^{\prime}_{11}y-\Phi_{X}-y)}{y(x-(1-x))}$

. Checking the right hand side of the implication

$\displaystyle\implies
p^{\prime}_{11}y(x-(1-x))-p_{10}(1-x)(x-(1-x))>p^{\prime}_{11}xy-x(1-x)(p_{11}+p_{10}-1)-xy(1-x)$
$\displaystyle\implies-p^{\prime}_{11}y-p_{10}(x-(1-x))>-x(p_{11}+p_{10}-1)-xy$
$\displaystyle\implies-p^{\prime}_{11}y+p_{10}(1-x)-p_{10}x>-xp_{11}-xp_{10}+x-xy$
$\displaystyle\implies-p^{\prime}_{11}y+p_{10}(1-x)+xp_{11}>x(1-y)$
$\displaystyle\implies p^{\prime}_{10}(1-y)>x(1-y)\implies p^{\prime}_{10}>x$

This leads to a contradiction of our case assumption $p^{\prime}_{10}<x$.

Now we will check if each of the possible upper bounds are less than
$a=\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$ (108). NOTE: Unsure whether the
argument for $1$ holds since 108 doesn’t need to be a probability.

  1. 1.

$\displaystyle 1>\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$ $\displaystyle\implies
x-(1-x)>p^{\prime}_{11}-(1-x)$ $\displaystyle\implies x>p^{\prime}_{11}$

This is a contradiction based on our assumption, so $1$ is capable of
falsifying.

  2. 2.

Checking if $\frac{p_{11}}{y}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$.

$\displaystyle\frac{p_{11}}{y}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}\implies
p_{11}x-p_{11}(1-x)<p^{\prime}_{11}y-y(1-x)$ $\displaystyle\implies
p_{11}x-p_{11}(1-x)<p_{11}x+p_{10}(1-x)-p^{\prime}_{10}(1-y)-y(1-x)$
$\displaystyle\implies-p_{11}(1-x)<p_{10}(1-x)-p^{\prime}_{10}(1-y)-y(1-x)$
$\displaystyle\implies p^{\prime}_{10}(1-y)<p_{10}(1-x)+p_{11}(1-x)-y(1-x)$
$\displaystyle\implies p^{\prime}_{10}(1-y)<(1-x)(p_{10}+p_{11}-(1-(1-y)))$
$\displaystyle\implies p^{\prime}_{10}(1-y)<(1-x)(p_{10}+p_{11}-1)+(1-x)(1-y)$
$\displaystyle\implies\frac{(1-y)(p^{\prime}_{10}-(1-x))}{1-x}<\Phi_{X}$

So, when $\frac{(1-y)(p^{\prime}_{10}-(1-x))}{1-x}<\Phi_{X}$,
$\frac{p_{11}}{y}$ will be less than $a$ and we can get falsification.

Now, for each of these settings, we evaluate $\Gamma$ and PNS.

  1. 1.

Upper bound is $1$:

$\displaystyle\Gamma$
$\displaystyle=(1-OR(X))y+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$
$\displaystyle=y-\frac{xy}{1-x}+\frac{yp^{\prime}_{11}-y(1-x)}{1-x}$
$\displaystyle=\frac{y(p^{\prime}_{11}-x)}{1-x}$

And the PNS upper bound will be

$\displaystyle p_{11}-\Gamma$ $\displaystyle=p_{11}-\Gamma$
$\displaystyle\implies$
$\displaystyle=p_{11}-\frac{y(p^{\prime}_{11}-x)}{1-x}$

  2. 2.

Upper bound is $\frac{p_{11}}{y}$: Then

$\displaystyle\Gamma$
$\displaystyle=p_{11}(1-OR(X))+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$
$\displaystyle=p_{11}+\frac{p^{\prime}_{11}y-p_{11}x-y(1-x)}{1-x}$
$\displaystyle=p_{11}+\frac{p_{10}(1-x)-p^{\prime}_{10}(1-y)-y(1-x)}{1-x}$
$\displaystyle=p_{11}+\frac{p_{10}(1-x)-p^{\prime}_{10}(1-y)-(1-(1-y))(1-x)}{1-x}$
$\displaystyle=p_{11}+p_{10}-1+\frac{-p^{\prime}_{10}(1-y)+(1-y)(1-x)}{1-x}$
$\displaystyle=\Phi_{X}-\frac{(1-y)(p^{\prime}_{10}-(1-x))}{1-x}$

And the upper bound on PNS will evaluate to

$\displaystyle p_{11}-\Gamma=p_{00}-\frac{(1-y)((1-x)-p^{\prime}_{10})}{1-x}$

####  L.2.8 Case VIII: Atalanti

We assume the following conditions

$\displaystyle p_{11}+p_{10}-1<0$ (137) $\displaystyle 1-x<x$ (138)
$\displaystyle 1-x<x<p^{\prime}_{11}$ (139) $\displaystyle
1-x<p^{\prime}_{10}<x$ (140)

These constraints imply based on (76) and the implications on Subsection I.4,
the following lower and upper bounds on $P(Z=1\mid X=1,Y=1)$:

$\displaystyle\max\begin{bmatrix}\frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq\min\begin{bmatrix}1\\\
\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}\end{bmatrix}$

Now we will check if each of the possible lower bounds are greater than
$b=\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$ (106).

  1. 1.

$\frac{p^{\prime}_{11}-(1-x)}{x}>\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$

. This is a contradiction because we see that
$\frac{p^{\prime}_{11}-(1-x)}{x}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}=a$.
Which implies that $\frac{p^{\prime}_{11}-(1-x)}{x}<b$.

  2. 2.

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}>\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$

. Checking the right hand side of the implication

$\displaystyle\implies
p^{\prime}_{11}y(x-(1-x))-p_{10}(1-x)(x-(1-x))>p^{\prime}_{11}xy-x(1-x)(p_{11}+p_{10}-1)-xy(1-x)$
$\displaystyle\implies-p^{\prime}_{11}y-p_{10}(x-(1-x))>-x(p_{11}+p_{10}-1)-xy$
$\displaystyle\implies-p^{\prime}_{11}y+p_{10}(1-x)-p_{10}x>-xp_{11}-xp_{10}+x-xy$
$\displaystyle\implies-p^{\prime}_{11}y+p_{10}(1-x)+xp_{11}>x(1-y)$
$\displaystyle\implies p^{\prime}_{10}(1-y)>x(1-y)\implies p^{\prime}_{10}>x$

This leads to a contradiction of our case assumption $p^{\prime}_{10}<x$.

Next, we check if each of the possible upper bounds are less than
$a=\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$ (108).

  1. 1.

$\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}<1$ $\displaystyle\implies
p^{\prime}_{11}<x$

This is a contradiction, and so $1\leq(a)$.

  2. 2.

Checking

$\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$

$\displaystyle\implies(1-y)(p^{\prime}_{10}-(1-x))<x\Phi_{X}$

This is a contradiction since the RHS is negative, but the LHS in positive.

So, when $1<\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$, then $\Gamma$
will be evaluated by subsititung $1$ as the value for $P(Z=1\mid X=1,Y=1)$
into $P(Z=1\mid X=1,Y=1)(1-OR(X))y+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$, to get

$\displaystyle\Gamma$
$\displaystyle=(1-OR(X))y+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$
$\displaystyle=\frac{y(p^{\prime}_{11}-x)}{1-x}$

And the upper bound on PNS will be

$\displaystyle p_{11}-\Gamma=p_{11}-\frac{y(p^{\prime}_{11}-x)}{1-x}$

####  L.2.9 Case IX: Atalanti

We assume the following conditions

$\displaystyle\Phi_{X}<0$ (141) $\displaystyle 1-x<x$ (142) $\displaystyle
1-x<x<p^{\prime}_{11}$ (143) $\displaystyle 1-x<x<p^{\prime}_{10}$ (144)

These constraints imply based on (76) and the implications on Subsection I.4,
the following lower and upper bounds on $P(Z=1\mid X=1,Y=1)$:

$\displaystyle\max\begin{bmatrix}\frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p_{11}-(1-y)}{y}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq\min\begin{bmatrix}1\\\
\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}\end{bmatrix}$

Now we will check if each of the possible lower bounds are greater than
$b=\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$ (106).

  1. 1.

$\frac{p^{\prime}_{11}-(1-x)}{x}>\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$

. This is a contradiction because we see that
$\frac{p^{\prime}_{11}-(1-x)}{x}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}=a$.
Which implies that $\frac{p^{\prime}_{11}-(1-x)}{x}<b$.

  2. 2.

$\displaystyle\frac{p_{11}-(1-y)}{y}>\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$
$\displaystyle\implies(p_{11}-(1-y))(x-(1-x))>p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)$
$\displaystyle\implies
p_{11}x-p_{11}(1-x)-(1-y)x+(1-y)(1-x)>p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)$
$\displaystyle\implies
p^{\prime}_{10}(1-y)-p_{10}(1-x)-p_{11}(1-x)-(1-y)x+(1-x)-y(1-x)>-(1-x)\Phi_{X}-(1-x)y$
$\displaystyle\implies
p^{\prime}_{10}(1-y)-(1-x)\Phi_{X}-(1-y)x>-(1-x)\Phi_{X}$
$\displaystyle\implies p^{\prime}_{10}(1-y)-(1-y)x>0$ $\displaystyle\implies
p^{\prime}_{10}-x>0$

Which agrees with our assumption $p^{\prime}_{10}>x$. So, this lower bound is
capable of falsifying, since the opposite of this inequality would create a
contradiction. So,

$\frac{p_{11}-(1-y)}{y}\geq\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$

Now we will check if each of the possible upper bounds are less than
$a=\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$ (108).

  1. 1.

$\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}<1$ $\displaystyle\implies
p^{\prime}_{11}<x$

This is a contradiction, and so $1\leq(a)$.

  2. 2.

$\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$
$\displaystyle\implies(1-y)(p^{\prime}_{10}-(1-x))<x\Phi_{X}$

This is a contradiction since the RHS is negative, but the LHS in positive.

So, when $1$ is the upper bound, the upper bound on PNS will be

$\displaystyle p_{11}-\Gamma=p_{11}-\frac{y(p^{\prime}_{11}-x)}{1-x}$

And when $\frac{p_{11}-(1-y)}{y}$ is the lower bound, $\Gamma$ will evaluate
to

$\displaystyle\Gamma$
$\displaystyle=(p_{11}-(1-y))(OR(X)-1)+\Phi_{X}+y-\frac{p^{\prime}_{11}y}{1-x}$
$\displaystyle=\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

And the upper bound on PNS will be

$p_{11}-\Gamma=p_{11}-\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

###  L.3 When $1-x>x$

Here, since the $OR(X)$ will be less than $1$, we need to recompute the
turning points, which we do now.

$\displaystyle yP(Z=1\mid
X=1,Y=1)(OR(X)-1)+\Phi_{X}+y-\frac{p^{\prime}_{11}y}{1-x}\geq 0$
$\displaystyle\implies\Phi_{X}+y-\frac{p^{\prime}_{11}y}{1-x}\geq yP(Z=1\mid
X=1,Y=1)(OR(X)-1)$
$\displaystyle\implies\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{1-x}\geq\frac{yP(Z=1\mid
X=1,Y=1)((1-x)-x)}{1-x}$
$\displaystyle\implies\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{y((1-x)-x)}\geq
P(Z=1\mid X=1,Y=1)$

We denote as (a). Next,

$\displaystyle P(Z=1\mid
X=1,Y=1)(1-OR(X))y+y\frac{p^{\prime}_{11}-(1-x)}{1-x}\geq 0$
$\displaystyle\implies P(Z=1\mid
X=1,Y=1)(1-OR(X))+\frac{p^{\prime}_{11}-(1-x)}{1-x}\geq 0$
$\displaystyle\implies P(Z=1\mid
X=1,Y=1)(1-OR(X))\geq\frac{(1-x)-p^{\prime}_{11}}{1-x}$ $\displaystyle\implies
P(Z=1\mid X=1,Y=1)\geq\frac{(1-x)-p^{\prime}_{11}}{(1-x)-x}$

And we denote this as (b). Now, we enumerate cases in a similar fashion as
before.

####  L.3.1 Case I

This case assumes the following

$\displaystyle\Phi_{X}<0$ (145) $\displaystyle x<1-x$ (146) $\displaystyle
x<p^{\prime}_{11}<1-x$ (147) $\displaystyle p^{\prime}_{10}<x<1-x$ (148)

Here, the chosen bounds will be

$\displaystyle\max\begin{bmatrix}0\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq
PNS\leq\min\begin{bmatrix}1\\\ \frac{p_{11}}{y}\end{bmatrix}$

Now, we check the upper and lower bounds against (a) and (b).

  1. 1.

Lower bound is $0$: $0$ will be less than (b) since we assume
$p^{\prime}_{11}<1-x$.

  2. 2.

Lower bound is $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$: This will be less
than (b) when the following condition holds:

$\Phi_{X}<\frac{(1-y)(p^{\prime}_{10}-x)}{x}$

Next, checking the upper bounds

  1. 1.

Upper bound of $1$: Checking this against (a), using a proof by contradiction.
Assume

$\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{y((1-x)-x)}>1$
$\displaystyle(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y>y((1-x)-x)$
$\displaystyle\implies(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y>y(1-x)-xy$
$\displaystyle\implies(1-x)y+(1-x)\Phi_{X}-p^{\prime}_{11}y>y(1-x)-xy$
$\displaystyle\implies(1-x)\Phi_{X}>y(p^{\prime}_{11}-x)$

This is a contradiction since LHS is negative and RHS is positive.

  2. 2.

Upper bound is $\frac{p_{11}}{y}$: Following proof by contradiction, assume

$\frac{p_{11}}{y}<\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{y((1-x)-x)}$
$\displaystyle\implies
p_{11}(1-x)-p_{11}x<(1-x)\Phi_{X}+y(1-x)-p^{\prime}_{11}y$
$\displaystyle\implies(1-x)\Phi_{X}-p^{\prime}_{10}(1-y)<(1-x)\Phi_{X}-(1-y)(1-x)$
$\displaystyle\implies 0<(1-y)(p^{\prime}_{10}-(1-x))$

This is a contradiction based on our assumptions.

So, when $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$ is the lower bound, i.e.
$p^{\prime}_{11}y>p_{10}(1-x)$ and
$\Phi_{X}>\frac{(1-y)(p^{\prime}_{10}-x)}{x}$, we compute $\Gamma$ by
inserting this lower bound into increasing function $P(Z=1\mid
X=1,Y=1)(1-OR(X))y+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$.

$\displaystyle\Gamma$
$\displaystyle=\frac{p^{\prime}_{11}y-p_{10}(1-x)}{x}(1-OR(X))+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$
$\displaystyle\Gamma$
$\displaystyle=\Phi_{X}-\frac{(1-y)(p^{\prime}_{10}-x)}{x}$

And the upper bound on PNS is computed as

$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{11}-\Phi_{X}+\frac{(1-y)(p^{\prime}_{10}-x)}{x}$
$\displaystyle=p_{00}-\frac{(1-y)(x-p^{\prime}_{10})}{x}$

####  L.3.2 Case II

$\displaystyle p_{10}+p_{10}-1<0$ (149) $\displaystyle x<1-x$ (150)
$\displaystyle x<p^{\prime}_{11}<1-x$ (151) $\displaystyle
x<p^{\prime}_{10}<1-x$ (152)

Here, the chosen bounds will be

$\displaystyle\max\begin{bmatrix}0\\\ \frac{p_{11}-(1-y)}{y}\end{bmatrix}\leq
PNS\leq\min\begin{bmatrix}1\\\ \frac{p_{11}}{y}\end{bmatrix}$

Checking whether the lower bounds will be greater than (b) first.

  1. 1.

Lower bound of $0$: $0$ will be less than (b) since $p^{\prime}_{11}<1-x$.

  2. 2.

Lower bound of $\frac{p_{11}-(1-y)}{y}$: Following a proof by contradiction,
assume

$\frac{p_{11}-(1-y)}{y}>\frac{(1-x)-p^{\prime}_{11}}{(1-x)-x}$
$\displaystyle\implies(1-x)\Phi_{X}>(1-y)(p^{\prime}_{10}-x)$

Contradiction because LHS is negative and RHS is positive

Now, for the upper bound $1$ will not be less than (a) using the same argument
as above. Similarly, $\frac{p_{11}}{y}$ will not be less than (a) either.

So no falsification will be observed in this case.

####  L.3.3 Case III

$\displaystyle\Phi_{X}<0$ (153) $\displaystyle x<1-x$ (154) $\displaystyle
x<p^{\prime}_{11}<1-x$ (155) $\displaystyle x<1-x<p^{\prime}_{10}$ (156)

Here, the chosen bounds will be

$\displaystyle\max\begin{bmatrix}0\\\ \frac{p_{11}-(1-y)}{y}\end{bmatrix}\leq
PNS\leq\min\begin{bmatrix}1\\\
\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}\end{bmatrix}$

Checking whether the lower bounds will be greater than (b).

  1. 1.

Lower bound on $0$: Using arguments before, this will be less than (b).

  2. 2.

Lower bound of $\frac{p_{11}-(1-y)}{y}$: This generates a contradiction as
before.

So, neither of the lower bounds are capable of falsifying. Next, we move on to
the upper bounds.

  1. 1.

Upper bound of $1$: Using the argument from Case I, we know this will be
greater than (a).

  2. 2.

: Upper bound of $\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$:
Following a proof by contradiction: Assume

$\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{y((1-x)-x)}<\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$
$\displaystyle\implies\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{(1-x)-x}<\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{x}$
$\displaystyle\implies(1-y)(p^{\prime}_{10}-(1-x))<0$

This is a contradiction, so

$\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{y((1-x)-x)}\geq\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$

Now, this will create a contradiction. So when this is the upper bound, i.e.
this condition holds

$\displaystyle p^{\prime}_{01}y>p_{00}(1-x)$

Then we can insert $\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$ into
$P(Z=1\mid X=1,Y=1)y(OR(X)-1)+\Phi_{X}+y-\frac{p^{\prime}_{11}y}{1-x}$. to
evaluate $\Gamma$ as

$\Gamma=\frac{(1-y)(p^{\prime}_{10}-(1-x))}{x}$

And this gives the following upper bound on PNS as

$\displaystyle p_{11}-\Gamma=p_{11}-\frac{(1-y)(p^{\prime}_{10}-(1-x))}{x}$

####  L.3.4 Case IV: Atalanti

$\displaystyle\Phi_{X}<0$ (157) $\displaystyle x<1-x$ (158) $\displaystyle
p^{\prime}_{11}<x<1-x$ (159) $\displaystyle x<1-x<p^{\prime}_{10}$ (160)

These constraints imply based on (76) and the implications on Subsection I.4,
the following lower and upper bounds on $P(Z=1\mid X=1,Y=1)$:

$\displaystyle max\begin{bmatrix}0\\\ \frac{p_{11}-(1-y)}{y}\end{bmatrix}\leq
P(Z=1\mid X=1,Y=1)\leq min\begin{bmatrix}\frac{p^{\prime}_{11}}{x}\\\
\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}\end{bmatrix}$

Now we will check if each of the possible lower bounds are greater than
$b=\frac{(1-x)-p^{\prime}_{11}}{(1-x)-x}$.

  1. 1.

$0<b$ holds since $1-x>p^{\prime}_{11}$ by assumption.

  2. 2.

$\displaystyle\frac{p_{11}-(1-y)}{y}>\frac{(1-x)-p^{\prime}_{11}}{(1-x)-x}$
$\displaystyle\implies\Phi_{X}>\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

This is contradiction since LHS is negative but RHS is positive.

Now we will check if each of the possible upper bounds are less than
$a=\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$.

  1. 1.

Upper bound is $\frac{p^{\prime}_{11}}{x}$:

$\displaystyle\frac{p^{\prime}_{11}}{x}<\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{y((1-x)-x)}$
$\displaystyle\implies\frac{y(p^{\prime}_{11}-x)}{x}<\Phi_{X}$

So, when the above holds, the upper bound will be capable of falsifying.

  2. 2.

$\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}<\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{y((1-x)-x)}$.

$\displaystyle\implies 0<(1-y)(p^{\prime}_{10}-(1-x))$

This bound is capable of falsifying as well.

So now, $\Gamma$ will be evaluated for different values of upper bounds. When
the upper bound is $\frac{p^{\prime}_{11}}{x}$, $\Gamma$ is evaluated as

$\displaystyle\Gamma$
$\displaystyle=\frac{p^{\prime}_{11}y}{x}(OR(X)-1)+\Phi_{X}+y-\frac{p^{\prime}_{11}y}{1-x}$
$\displaystyle=\Phi_{X}-\frac{y(p^{\prime}_{11}-x)}{x}$

And the upper bound on PNS is evaluated as

$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{11}-\Phi_{X}+\frac{y(p^{\prime}_{11}-x)}{x}$ $\displaystyle
p_{11}-\Gamma$ $\displaystyle=p_{00}-\frac{y(p^{\prime}_{11}-x)}{x}$

When the upper bound is $\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$,
$\Gamma$ is evaluated as

$\Gamma=\frac{(1-y)(p^{\prime}_{10}-(1-x))}{x}$

And this gives the following upper bound on PNS as

$\displaystyle p_{11}-\Gamma=p_{11}-\frac{(1-y)(p^{\prime}_{10}-(1-x))}{x}$

####  L.3.5 Case V: Atalanti

$\displaystyle\Phi_{X}<0$ (161) $\displaystyle x<1-x$ (162) $\displaystyle
p^{\prime}_{11}<x<1-x$ (163) $\displaystyle x<p^{\prime}_{10}<1-x$ (164)

These constraints imply based on (76) and the implications on Subsection I.4,
the following lower and upper bounds on $P(Z=1\mid X=1,Y=1)$:

$\displaystyle\max\begin{bmatrix}0\\\ \frac{p_{11}-(1-y)}{y}\end{bmatrix}\leq
P(Z=1\mid X=1,Y=1)\leq\min\begin{bmatrix}\frac{p_{11}}{y}\\\
\frac{p^{\prime}_{11}}{x}\end{bmatrix}$

Now we will check if each of the possible lower bounds are greater than $b$.

  1. 1.

$0<b$ holds since $1-x>p^{\prime}_{11}$ by assumption.

  2. 2.

Lower bound is $\frac{p_{11}-(1-y)}{y}$:

$\displaystyle\frac{p_{11}-(1-y)}{y}>\frac{(1-x)-p^{\prime}_{11}}{(1-x)-x}$
$\displaystyle\implies\Phi_{X}>\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

This is contradiction since LHS is negative but RHS is positive.

Now we will check if each of the possible upper bounds are less than $a$.

  1. 1.

Upper bound is $\frac{p_{11}}{y}$: Following proof by contradiction, assume

$\frac{p_{11}}{y}<\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{y((1-x)-x)}$
$\displaystyle\implies
p_{11}(1-x)-p_{11}x<(1-x)\Phi_{X}+y(1-x)-p^{\prime}_{11}y$
$\displaystyle\implies(1-x)\Phi_{X}-p^{\prime}_{10}(1-y)<(1-x)\Phi_{X}-(1-y)(1-x)$
$\displaystyle\implies 0<(1-y)(p^{\prime}_{10}-(1-x))$

This is a contradiction.

  2. 2.

Upper bound is $\frac{p^{\prime}_{11}}{x}$:

$\displaystyle\frac{p^{\prime}_{11}}{x}<\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{y((1-x)-x)}$
$\displaystyle\implies\frac{y(p^{\prime}_{11}-x)}{x}<\Phi_{X}$

So, when the above holds, the upper bound will be capable of falsifying.

When the upper bound is $\frac{p^{\prime}_{11}}{x}$, $\Gamma$ will evaluate to

$\Gamma=\Phi_{X}-\frac{y(p^{\prime}_{11}-x)}{x}$

And the upper bound on PNS is evaluated as

$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{11}-\Phi_{X}+\frac{y(p^{\prime}_{11}-x)}{x}$ $\displaystyle
p_{11}-\Gamma$ $\displaystyle=p_{00}-\frac{y(p^{\prime}_{11}-x)}{x}$

####  L.3.6 Case VI: Atalanti

$\displaystyle\Phi_{X}<0$ (165) $\displaystyle x<1-x$ (166) $\displaystyle
p^{\prime}_{11}<x<1-x$ (167) $\displaystyle p^{\prime}_{10}<x<1-x$ (168)

These constraints imply based on (76) and the implications on Subsection I.4,
the following lower and upper bounds on $P(Z=1\mid X=1,Y=1)$:

$\displaystyle max\begin{bmatrix}0\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq min\begin{bmatrix}\frac{p_{11}}{y}\\\
\frac{p^{\prime}_{11}}{x}\end{bmatrix}$

Now we will check if each of the possible lower bounds are greater than $b$.

  1. 1.

$0<b$ holds since $1-x>p^{\prime}_{11}$ by assumption.

  2. 2.

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}>\frac{(1-x)-p^{\prime}_{11}}{(1-x)-x}$
$\displaystyle\implies x\Phi_{X}>(1-y)(p^{\prime}_{10}-x)$

When this condition holds, this lower bound can falsify.

Now we will check if each of the possible upper bounds are less than $a$.

  1. 1.

Upper bound is $\frac{p_{11}}{y}$: Following proof by contradiction, assume

$\frac{p_{11}}{y}<\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{y((1-x)-x)}$
$\displaystyle\implies
p_{11}(1-x)-p_{11}x<(1-x)\Phi_{X}+y(1-x)-p^{\prime}_{11}y$
$\displaystyle\implies(1-x)\Phi_{X}-p^{\prime}_{10}(1-y)<(1-x)\Phi_{X}-(1-y)(1-x)$
$\displaystyle\implies 0<(1-y)(p^{\prime}_{10}-(1-x))$

This is a contradiction.

  2. 2.

Upper bound is $\frac{p^{\prime}_{11}}{x}$:

$\displaystyle\frac{p^{\prime}_{11}}{x}<\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{y((1-x)-x)}$
$\displaystyle\implies\frac{y(p^{\prime}_{11}-x)}{x}<\Phi_{X}$

So, when the above holds, the upper bound will be capable of falsifying.

So, when $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$ is the lower bound, i.e.
$p^{\prime}_{11}y>p_{10}(1-x)$ and
$\Phi_{X}>\frac{(1-y)(p^{\prime}_{10}-x)}{x}$, we compute $\Gamma$ by
inserting this lower bound into increasing function $P(Z=1\mid
X=1,Y=1)(1-OR(X))y+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$.

$\displaystyle\Gamma$
$\displaystyle=\frac{p^{\prime}_{11}y-p_{10}(1-x)}{x}(1-OR(X))+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$
$\displaystyle\Gamma$
$\displaystyle=\Phi_{X}-\frac{(1-y)(p^{\prime}_{10}-x)}{x}$

And the upper bound on PNS is computed as

$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{11}-\Phi_{X}+\frac{(1-y)(p^{\prime}_{10}-x)}{x}$
$\displaystyle=p_{00}-\frac{(1-y)(x-p^{\prime}_{10})}{x}$

And when $\frac{p^{\prime}_{11}}{x}$ is the upper bound, $\Gamma$ will
evaluate to

$\Gamma=\Phi_{X}-\frac{y(p^{\prime}_{11}-x)}{x}$

And the upper bound on PNS is evaluated as

$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{11}-\Phi_{X}+\frac{y(p^{\prime}_{11}-x)}{x}$ $\displaystyle
p_{11}-\Gamma$ $\displaystyle=p_{00}-\frac{y(p^{\prime}_{11}-x)}{x}$

####  L.3.7 Case VII

$\displaystyle\Phi_{X}<0$ (169) $\displaystyle x<1-x$ (170) $\displaystyle
x<1-x<p^{\prime}_{11}$ (171) $\displaystyle p^{\prime}_{10}<x<1-x$ (172)

These constraints imply based on (76) and the implications on Subsection I.4,
the following lower and upper bounds on $P(Z=1\mid X=1,Y=1)$:

$\displaystyle\max\begin{bmatrix}\frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq min\begin{bmatrix}1\\\ \frac{p_{11}}{y}\end{bmatrix}$

Now we will check if each of the possible lower bounds are greater than $b$.

  1. 1.

$\frac{p^{\prime}_{11}-(1-x)}{x}<\frac{(1-x)-p^{\prime}_{11}}{(1-x)-x}$. This
is a contradiction since the LHS is positive based on our assumptions while
the RHS is negative.

  2. 2.

Assume

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}>\frac{(1-x)-p^{\prime}_{11}}{(1-x)-x}$

.

$\displaystyle\implies\Phi_{X}>\frac{(1-y)(p^{\prime}_{10}-x)}{x}$

So, when this holds, this lower bound is capable of falsification.

Now we will check if each of the possible upper bounds are less than $a$.

  1. 1.

Following proof by contradiction, assume

$\frac{p_{11}}{y}<\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{y((1-x)-x)}$
$\displaystyle\implies
p_{11}(1-x)-p_{11}x<(1-x)\Phi_{X}+y(1-x)-p^{\prime}_{11}y$
$\displaystyle\implies(1-x)\Phi_{X}-p^{\prime}_{10}(1-y)<(1-x)\Phi_{X}-(1-y)(1-x)$
$\displaystyle\implies 0<(1-y)(p^{\prime}_{10}-(1-x))$

This is a contradiction.

  2. 2.

Upper bound is $1$: Assume

$1<\frac{(1-x)(\Phi_{X}+y)-p^{\prime}_{11}y}{y((1-x)-x)}$
$\displaystyle\implies(1-x)(\Phi_{X}+y)>(1-x)y-xy+p^{\prime}_{11}y$
$\displaystyle\implies(1-x)\Phi_{X}>y(p^{\prime}_{11}-x)$

This is a contradiction since LHS is negative but RHS is positive.

So, when the lower bound is $\frac{p^{\prime}_{11}-(1-x)}{x}$, $\Gamma$ will
evaluate to

$\displaystyle\Gamma$
$\displaystyle=\frac{p^{\prime}_{11}y-y(1-x)}{x}(1-OR(X))+\frac{p^{\prime}_{11}y-y(1-x)}{1-x}$
$\displaystyle=\frac{y(p^{\prime}_{11}-(1-x))}{x}$

And so the upper bound on PNS will be

$\displaystyle p_{11}-\Gamma=p_{11}-\frac{y(p^{\prime}_{11}-(1-x))}{x}$

Next, when the lower bound is $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$,
$\Gamma$ is computed by inserting this lower bound into increasing function
$P(Z=1\mid X=1,Y=1)(1-OR(X))y+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$.

$\displaystyle\Gamma$
$\displaystyle=\frac{p^{\prime}_{11}y-p_{10}(1-x)}{x}(1-OR(X))+y\frac{p^{\prime}_{11}-(1-x)}{1-x}$
$\displaystyle\Gamma$
$\displaystyle=\Phi_{X}-\frac{(1-y)(p^{\prime}_{10}-x)}{x}$

And the upper bound on PNS is computed as

$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{11}-\Phi_{X}+\frac{(1-y)(p^{\prime}_{10}-x)}{x}$
$\displaystyle=p_{00}-\frac{(1-y)(x-p^{\prime}_{10})}{x}$

####  L.3.8 Case VIII

$\displaystyle\Phi_{X}<0$ (173) $\displaystyle x<1-x$ (174) $\displaystyle
x<1-x<p^{\prime}_{11}$ (175) $\displaystyle x<p^{\prime}_{10}<1-x$ (176)

These constraints imply based on (76) and the implications on Subsection I.4,
the following lower and upper bounds on $P(Z=1\mid X=1,Y=1)$:

$\displaystyle max\begin{bmatrix}\frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p_{11}-(1-y)}{y}\end{bmatrix}\leq P(Z=1\mid X=1,Y=1)\leq
min\begin{bmatrix}1\\\ \frac{p_{11}}{y}\end{bmatrix}$

The same arguments as the previous case are utilized for the upper bounds, and
consequently they cannot falsify. The same argument for
$\frac{p^{\prime}_{11}-(1-x)}{x}$ is used, and this can falsify.

When the lower bound is $\frac{p_{11}-(1-y)}{y}$:

$\displaystyle\frac{p_{11}-(1-y)}{y}>\frac{(1-x)-p^{\prime}_{11}}{(1-x)-x}$
$\displaystyle\implies\Phi_{X}>\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

This is contradiction since LHS is negative but RHS is positive.

So, when the lower bound is $\frac{p^{\prime}_{11}-(1-x)}{x}$, $\Gamma$ will
evaluate to

$\displaystyle\Gamma$
$\displaystyle=\frac{p^{\prime}_{11}y-y(1-x)}{x}(1-OR(X))+\frac{p^{\prime}_{11}y-y(1-x)}{1-x}$
$\displaystyle=\frac{y(p^{\prime}_{11}-(1-x))}{x}$

And so the upper bound on PNS will be

$\displaystyle p_{11}-\Gamma=p_{11}-\frac{y(p^{\prime}_{11}-(1-x))}{x}$

####  L.3.9 Case IX

$\displaystyle\Phi_{X}<0$ (177) $\displaystyle x<1-x$ (178) $\displaystyle
x<1-x<p^{\prime}_{11}$ (179) $\displaystyle x<1-x<p^{\prime}_{10}$ (180)

These constraints imply based on (76) and the implications on Subsection I.4,
the following lower and upper bounds on $P(Z=1\mid X=1,Y=1)$:

$\displaystyle\max\begin{bmatrix}\frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p_{11}-(1-y)}{y}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq\min\begin{bmatrix}1\\\
\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}\end{bmatrix}$

The same arguments from the above case are used for the lower bounds, with the
conclusion that $\frac{p^{\prime}_{11}-(1-x)}{x}$ is capable of inducing
falsification.

Following the same argument as before for $1$, it is incapable of falsifying.
$\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$ is capable of falsifying.

So, when the lower bound is $\frac{p^{\prime}_{11}-(1-x)}{x}$, $\Gamma$ will
evaluate to

$\displaystyle\Gamma$
$\displaystyle=\frac{p^{\prime}_{11}y-y(1-x)}{x}(1-OR(X))+\frac{p^{\prime}_{11}y-y(1-x)}{1-x}$
$\displaystyle=\frac{y(p^{\prime}_{11}-(1-x))}{x}$

And so the upper bound on PNS will be

$\displaystyle p_{11}-\Gamma=p_{11}-\frac{y(p^{\prime}_{11}-(1-x))}{x}$

When the upper bound is $\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$,
$\Gamma$ is evaluated as

$\Gamma=\frac{(1-y)(p^{\prime}_{10}-(1-x))}{x}$

And this gives the following upper bound on PNS as

$\displaystyle p_{11}-\Gamma=p_{11}-\frac{(1-y)(p^{\prime}_{10}-(1-x))}{x}$

###  L.4 Case II: $\Phi_{X}\geq 0$

The usual breakdown around $x>1-x$ or not will happen. Assuming $x>1-x$,
falsification will happen when line 3 and line 4 of the matrix are greater
than 0, we first check the values of $P(Z=1\mid X=1,Y=1)$ for which this
happens.

  * •

When $yP(Z=1\mid
X=1,Y=1)(OR(X)-1)+\Phi_{X}+y-\frac{p^{\prime}_{11}y}{1-x}\geq\Phi_{X}$:

$\implies P(Z=1\mid X=1,Y=1)\geq\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}\\\ $
(181)

We denote this value as

$\ b=\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$ (182)

  * •

When $P(Z=1\mid
X=1,Y=1)(1-OR(X))y+y\frac{p^{\prime}_{11}-(1-x)}{1-x}\geq\Phi_{X}$

$\implies\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}\geq P(Z=1\mid
X=1,Y=1)$ (183)

We denote this value as

$a=\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$ (184)

.

###  L.5 Case By Case Upper Bound Calculation

Each of these sections considers a different case where the values of
$p^{\prime}_{11}$, $p^{\prime}_{10}$ are picked relative to $x$ and $1-x$.

####  L.5.1 Case I

This case assumes the following

$\displaystyle\Phi_{X}\geq 0$ (185) $\displaystyle 1-x<x$ (186) $\displaystyle
1-x<p^{\prime}_{11}<x$ (187) $\displaystyle p^{\prime}_{10}<1-x<x$ (188)

The first two have been assumed in a larger context, and the last two are made
for this specific case for this section.

These conditions, based on the implications in subsection I.4, limit the
choice of bounds on $P(Z=1\mid X=1,Y=1)$ from (76) to

$\max\begin{bmatrix}\frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq\min\begin{bmatrix}\frac{p_{11}}{y}\\\
\frac{p^{\prime}_{11}}{x}\end{bmatrix}$

Now, we know falsification will happen when either lower bound is greater than
(b), or the upper bound is less than 108. So we examine each choice of upper
and lower bound, and see whether they can be less than 108 or 106.

  1. 1.

$\frac{p^{\prime}_{11}-(1-x)}{x}$: This will be less than (b), i.e.
$\frac{p^{\prime}_{11}-(1-x)}{x}<\frac{p^{\prime}_{11}-(1-x)}{2x-1}$ since the
denominator is smaller on the right hand side.

  2. 2.

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$: Assume

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}>\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$

$\displaystyle\implies\Phi_{X}<\frac{(1-y)(p^{\prime}_{10}-x)}{x}$

This is a contradiction since RHS is negative and LHS in positive.

And so, the lower bounds will always be less than (b), so any falsification
will come from the upper bounds being less than (a), which we check below.

  1. 1.

When the upper bound is $\frac{p_{11}}{y}$: Checking this against (a) using a
proof by contradiction, we assume

$\displaystyle\frac{p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)}{y(x-(1-x))}<\frac{p_{11}}{y}$
$\displaystyle\implies
p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)<p_{11}x-p_{11}(1-x)$
$\displaystyle\implies
p_{11}x+p_{10}(1-x)-p^{\prime}_{10}(1-y)-(1-x)(p_{11}+p_{10}-1)-y(1-x)<p_{11}x-p_{11}(1-x)$
$\displaystyle\implies
p_{10}(1-x)+p_{11}(1-x)-p^{\prime}_{10}(1-y)-(1-x)(p_{11}+p_{10}-1)-y(1-x)<0$
$\displaystyle\implies-p^{\prime}_{10}(1-y)+(1-x)-y(1-x)<0$
$\displaystyle\implies-p^{\prime}_{10}(1-y)+(1-x)(1-y)<0$
$\displaystyle\implies-p^{\prime}_{10}+(1-x)<0$

This is a contradiction, and so
$\frac{p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)}{y(x-(1-x))}\geq\frac{p_{11}}{y}$
and this bound is capable of falsifying.

  2. 2.

Upper bound is $\frac{p^{\prime}_{11}}{x}$:

$\displaystyle\frac{p^{\prime}_{11}}{x}<\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$
$\displaystyle\implies\Phi_{X}<\frac{y(p^{\prime}_{11}-x)}{x}$

This is a contradiction, so this bound cannot falsify.

Falsification will happen when $\frac{p_{11}}{y}$ is the upper bound, which
requires the following condition to hold

$\frac{p_{11}}{y}<\frac{p^{\prime}_{11}}{x}\implies p_{11}x<p^{\prime}_{11}y$

Putting this all together, in Case I, when the following condition holds:

$\displaystyle p_{11}x<p^{\prime}_{11}y$

We will get falsification, where $\Gamma$ is calculated as

$\Gamma=\Phi_{X}-\frac{(1-y)(p^{\prime}_{10}-(1-x))}{1-x}$

Now, subtracting this from $p_{11}$, we have

$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{11}-(p_{11}+p_{10}-1)+\frac{(1-y)(p^{\prime}_{10}-(1-x))}{1-x}$
$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{00}-\frac{(1-y)((1-x)-p^{\prime}_{10})}{1-x}$

####  L.5.2 Case II

This case assumes the following

$\displaystyle\Phi_{X}<0$ (189) $\displaystyle 1-x<x$ (190) $\displaystyle
1-x<p^{\prime}_{11}<x$ (191) $\displaystyle 1-x<p^{\prime}_{10}<x$ (192)

These conditions limit the choice of bounds on $P(Z=1\mid X=1,Y=1)$ from (76)
to

$\max\begin{bmatrix}\frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq\min\begin{bmatrix}\frac{p^{\prime}_{11}}{x}\\\
\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}\end{bmatrix}$

Now, we check whether the lower bounds are greater than (b) and if the upper
bounds are less than (a).

  1. 1.

First, $\frac{p^{\prime}_{11}-(1-x)}{x}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$
since the denominator is smaller on the right hand side. This lower bound is
less than (b).

  2. 2.

Checking the lower bound of $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$ against
(b), using a proof by contradiction. Assume,

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}>\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$
$\displaystyle(1-y)(p^{\prime}_{10}-x)>x\Phi_{X}$

This is a contradiction since the RHS is positive but he LHS is negative.

Checking the upper bounds now.

  1. 1.

When the upper bound is $\frac{p^{\prime}_{11}}{x}$: Checking this against the
(a).

$\displaystyle\frac{p^{\prime}_{11}}{x}<\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$
$\displaystyle\implies\Phi_{X}<\frac{y(p^{\prime}_{11}-x)}{x}$

This is a contradiction because of the case we are in.

  2. 2.

When the upper bound is $\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$:
Assume

$\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}<\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}\\\
$ $\displaystyle\implies(1-y)(p^{\prime}_{10}-(1-x))<0$

This creates a contradiction since the LHS is positive but the RHS is 0.

Since the lower bounds are less than (b), and upper bounds are greater than
(a), so $\Gamma=\Phi_{X}$.

####  L.5.3 Case III

This case assumes the following conditions

$\displaystyle\Phi_{X}\geq 0$ (193) $\displaystyle 1-x<x$ (194) $\displaystyle
1-x<p^{\prime}_{11}<x$ (195) $\displaystyle 1-x<x<p^{\prime}_{10}$ (196)

These conditions limit the choice of bounds on $P(Z=1\mid X=1,Y=1)$ from (76)
to

$\max\begin{bmatrix}\frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p_{11}-(1-y)}{y}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq\min\begin{bmatrix}\frac{p^{\prime}_{11}}{x}\\\
\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}\end{bmatrix}$

Following the usual approach, we start with checking the lower bounds against
106.

  1. 1.

First, $\frac{p^{\prime}_{11}-(1-x)}{x}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$
since the denominator is smaller on the right hand side. This lower bound is
less than (b).

  2. 2.

Checking $\frac{p_{11}-(1-y)}{y}$: Assume

$\frac{p_{11}-(1-y)}{y}>\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$
$\displaystyle\implies\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}<\Phi_{X}$

So, when this condition holds this lower bound will be greater than (b).

Now, we check the upper bounds against (a).

  1. 1.

When the upper bound is $\frac{p^{\prime}_{11}}{x}$: Checking this against the
(a).

$\displaystyle\frac{p^{\prime}_{11}}{x}<\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}$
$\displaystyle\implies\Phi_{X}<\frac{y(p^{\prime}_{11}-x)}{x}$

This is a contradiction because of the case we are in.

  2. 2.

When the upper bound is $\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}$:
Assume

$\frac{(1-x)(1-y)-p_{10}(1-x)+p^{\prime}_{11}y}{xy}<\frac{p^{\prime}_{11}y-(1-x)(\Phi_{X}+y)}{y(x-(1-x))}\\\
$ $\displaystyle\implies(1-y)(p^{\prime}_{10}-(1-x))<0$

This creates a contradiction since the LHS is positive but the RHS is 0.

Now, the first condition that must be satisfied is $\frac{p_{11}-(1-y)}{y}$ is
the lower bound, which requires the following condition to hold

$p_{01}x<p^{\prime}_{01}y$

Then, we substitute the lower bound into the increasing function of $P(Z=1\mid
X=1,Y=1)$

$\displaystyle\frac{(1-x)(p_{10}+p_{11}-1)-p^{\prime}_{11}y+P(Z=1\mid
X=1,Y=1)y(x-(1-x))+(1-x)y}{1-x}$
$\displaystyle=\frac{(1-x)(p_{10}+p_{11}-1)-p^{\prime}_{11}y+(p_{11}-(1-y))(x-(1-x))+(1-x)y}{1-x}$
$\displaystyle=\frac{(1-x)(p_{10}+p_{11}-1)+p^{\prime}_{10}(1-y)-p_{10}(1-x)-(p_{11}-(1-y))(1-x)+(1-x)y}{1-x}$
$\displaystyle=\frac{(1-x)(p_{10}+p_{11}-1)+p^{\prime}_{10}(1-y)-(1-x)(p_{10}+p_{11}-1)-x(1-y)}{1-x}$
$\displaystyle=\frac{p^{\prime}_{10}(1-y)-x(1-y)}{1-x}=\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

So in this case, when $p_{01}x<p^{\prime}_{01}y$ holds, we get falsification
with $\Gamma$ evaluating to

$\Gamma=\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

And so the upper bound for PNS will be

$p_{11}-\Gamma=p_{11}-\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

####  L.5.4 Case IV

We assume the following conditions

$\displaystyle\Phi_{X}\geq 0$ (197) $\displaystyle 1-x<x$ (198) $\displaystyle
p^{\prime}_{11}<1-x<x$ (199) $\displaystyle p^{\prime}_{10}<1-x<x$ (200)

These constraints imply the following lower and upper bounds on $P(Z=1\mid
X=1,Y=1)$ from (76) to

$\displaystyle\max\begin{bmatrix}0\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq\min\begin{bmatrix}\frac{p_{11}}{y}\\\
\frac{p^{\prime}_{11}}{x}\end{bmatrix}$

Here, since (b) will always be negative, we expect both lower bounds to
falsify.

  1. 1.

Lower bound is $0$: Since the numerator in (b) is negative, $0$ will be
greater.

  2. 2.

Lower bound is $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$: Assume

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}>\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$
$\displaystyle\implies(1-y)(p^{\prime}_{10}-x)>x\Phi_{X}$

This is a contradiction.

Consequently, when $0$ is the lower bound, which requires the following
condition to hold

$p_{10}(1-x)>p^{\prime}_{11}y$

Then $\Gamma$ will evaluate to

$\Phi_{X}-\frac{y(p^{\prime}_{11}-(1-x))}{1-x}$

And the upper bound on PNS will be

$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{11}-\Phi_{X}+\frac{y(p^{\prime}_{11}-(1-x))}{1-x}$
$\displaystyle=p_{00}-\frac{y((1-x)-p^{\prime}_{11})}{1-x}$

####  L.5.5 Case V

We assume the following conditions

$\displaystyle\Phi_{X}\geq 0$ (201) $\displaystyle 1-x<x$ (202) $\displaystyle
p^{\prime}_{11}<1-x<x$ (203) $\displaystyle 1-x<p^{\prime}_{10}<x$ (204)

These constraints imply the following lower bounds on $P(Z=1\mid X=1,Y=1)$
from (76). Only the lower bound is checked against 106 since 108 will be
negative in this case.

$\displaystyle\max\begin{bmatrix}0\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq P(Z=1\mid X=1,Y=1)$

$0$ will be greater than (b) since (b) is negative, and the same argument as
above will apply to $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$.

Then, $\Gamma$ will evaluate to Then $\Gamma$ will evaluate to

$\Phi_{X}-\frac{y(p^{\prime}_{11}-(1-x))}{1-x}$

And the upper bound on PNS will be

$\displaystyle p_{11}-\Gamma$
$\displaystyle=p_{11}-\Phi_{X}+\frac{y(p^{\prime}_{11}-(1-x))}{1-x}$
$\displaystyle=p_{00}-\frac{y((1-x)-p^{\prime}_{11})}{1-x}$

####  L.5.6 Case VI

We assume the following conditions

$\displaystyle\Phi_{X}\geq 0$ (205) $\displaystyle 1-x<x$ (206) $\displaystyle
p^{\prime}_{11}<1-x<x$ (207) $\displaystyle 1-x<x<p^{\prime}_{10}$ (208)

Similar to the previous case, only the lower bound needs to be checked. These
constraints imply the following lower bounds on $P(Z=1\mid X=1,Y=1)$ from (76)
as

$\displaystyle max\begin{bmatrix}0\\\ \frac{p_{11}-(1-y)}{y}\end{bmatrix}\leq
P(Z=1\mid X=1,Y=1)$

Now, we check whether each of these lower bounds can be greater than 106.

  1. 1.

Lower bound is $0$ \- the same argument as the previous case holds.

  2. 2.

Lower bound is $\frac{p_{11}-(1-y)}{y}$: Assume

$\frac{p_{11}-(1-y)}{y}>\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$
$\displaystyle\implies\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}<\Phi_{X}$

So, when this condition holds this lower bound will be greater than (b).

So, the falsification will be as follows.

When $p_{11}>1-y$, $\frac{p_{11}-(1-y)}{y}$ will be the lower bound, and
$\Gamma$ will be evaluating by inserting this lower bound into $yP(Z=1\mid
X=1,Y=1)(OR(X)-1)+\Phi_{X}+y-\frac{p^{\prime}_{11}y}{1-x}$. And $\Gamma$ will
evaluate to

$\Gamma=\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

And therefore the upper bound on PNS will be

$\displaystyle p_{11}-\Gamma=p_{11}-\frac{(1-y)(p^{\prime}_{10}-x)}{1-x}$

Next, when $p_{11}<1-y$, $0$ will be the lower bound, and $\Gamma$ will
evaluate to

$\Gamma=\Phi_{X}-\frac{y(p^{\prime}_{11}-(1-x))}{1-x}$

And the upper bound on PNS will be

$\displaystyle p_{11}-\Gamma=p_{00}-\frac{y((1-x)-p^{\prime}_{11})}{1-x}$

####  L.5.7 Case VII

We assume the following conditions

$\displaystyle\Phi_{X}\geq 0$ (209) $\displaystyle 1-x<x$ (210) $\displaystyle
1-x<x<p^{\prime}_{11}$ (211) $\displaystyle p^{\prime}_{10}<1-x<x$ (212)

These constraints imply based on (76) and the implications on Subsection I.4,
the following lower and upper bounds on $P(Z=1\mid X=1,Y=1)$:

$\displaystyle\max\begin{bmatrix}\frac{p^{\prime}_{11}-(1-x)}{x}\\\
\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}\end{bmatrix}\leq P(Z=1\mid
X=1,Y=1)\leq\min\begin{bmatrix}1\\\ \frac{p_{11}}{y}\end{bmatrix}$

Now we will check if each of the possible lower bounds are greater than (b).

  1. 1.

First, $\frac{p^{\prime}_{11}-(1-x)}{x}<\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$
since the denominator is smaller on the right hand side. This lower bound is
less than (b).

  2. 2.

Lower bound is $\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}$: Assume

$\frac{p^{\prime}_{11}y-p_{10}(1-x)}{xy}>\frac{p^{\prime}_{11}-(1-x)}{x-(1-x)}$
$\displaystyle\implies(1-y)(p^{\prime}_{10}-x)>x\Phi_{X}$

This is a contradiction.

Now we will check if each of the possible upper bounds are less than (a).

  1. 1.

When the upper bound is $\frac{p_{11}}{y}$: Checking this against (a) using a
proof by contradiction, we assume

$\displaystyle\frac{p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)}{y(x-(1-x))}<\frac{p_{11}}{y}$
$\displaystyle\implies
p^{\prime}_{11}y-(1-x)(p_{11}+p_{10}-1)-y(1-x)<p_{11}x-p_{11}(1-x)$
$\displaystyle\implies
p_{11}x+p_{10}(1-x)-p^{\prime}_{10}(1-y)-(1-x)(p_{11}+p_{10}-1)-y(1-x)<p_{11}x-p_{11}(1-x)$
$\displaystyle\implies
p_{10}(1-x)+p_{11}(1-x)-p^{\prime}_{10}(1-y)-(1-x)(p_{11}+p_{10}-1)-y(1-x)<0$
$\displaystyle\implies-p^{\prime}_{10}(1-y)+(1-x)-y(1-x)<0$
$\displaystyle\implies-p^{\prime}_{10}(1-y)+(1-x)(1-y)<0$