$\int_{-3/4}^{0}\omega(y,\alpha^{\star})\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{2\sqrt{y+1}}dy<0$

Thus, $G(\alpha)=G_{0}(\alpha)+G_{1}(\alpha)<0$.

      3. iii.

$\hat{\tau_{d}}^{2}(\alpha)-1<-3/4\Longleftrightarrow\alpha<\alpha^{\star}$

$\displaystyle G_{1}(\alpha)$
$\displaystyle=\int_{\hat{\tau_{d}}^{2}(\alpha)-1}^{\hat{\tau_{d}}^{2}(\alpha^{\star})-1}\omega(y,\alpha)\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{2\sqrt{y+1}}dy$
$\displaystyle+\int_{\hat{\tau_{d}}^{2}(\alpha^{\star})-1}^{0}\omega(y,\alpha)\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{2\sqrt{y+1}}dy$
$\displaystyle\leq\int_{\hat{\tau_{d}}^{2}(\alpha)-1}^{\hat{\tau_{d}}^{2}(\alpha^{\star})-1}\omega(y,\alpha^{\star})\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{2\sqrt{y+1}}dy$
$\displaystyle+\int_{\hat{\tau_{d}}^{2}(\alpha^{\star})-1}^{0}\omega(y,\alpha^{\star})\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{2\sqrt{y+1}}dy$
$\displaystyle=\int_{\hat{\tau_{d}}^{2}(\alpha)-1}^{0}\omega(y,\alpha^{\star})\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{2\sqrt{y+1}}dy$
$\displaystyle\leq\int_{-1}^{0}\omega(y,\alpha^{\star})\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{2\sqrt{y+1}}dy$

We can numerically find

$\int_{-1}^{0}\omega(y,\alpha^{\star})\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{2\sqrt{y+1}}dy<0.817$

Thus, $G(\alpha)=G_{0}(\alpha)+G_{1}(\alpha)<-0.96+0.817<0$.

We conclude that $\forall\alpha\in[0,1/2],\ G(\alpha)<0$.

Hence, $Var(\theta_{A}(\sigma_{q}^{2\star}))$ strictly decreases in $\gamma$.

  2. 2.

When $\gamma=0$, we know $\forall\theta,\ w^{\star}(\theta)=0,\
\theta_{A}^{\star}=\theta$. Thus,

$\lim_{\gamma\to
0}Var(\theta^{\star})=\int_{-\infty}^{\infty}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta=Var(\theta)=\sigma_{\theta}^{2}$

When $\gamma\to\infty$, by definition, for any $\theta$, $l\to\infty$ if
$\sigma_{q}$ is finite, so the optimal decision is
$\sigma_{q}^{\star}=+\infty$ with $l^{\star}=(\theta-\mu_{\theta})^{2}$. Thus,
by Equation 23,

$\lim_{\gamma\to\infty}Var(\theta^{\star})=2\int_{\mu_{\theta}+\tau_{a}}^{\infty}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

And by Proposition 4.1, for any $\Gamma>0$, we must have $\tau_{a}>0$, so

$\lim_{\gamma\to\infty}Var(\theta^{\star})=2\int_{\mu_{\theta}+\tau_{a}}^{\infty}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta<2\int_{\mu_{\theta}}^{\infty}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta=Var(\theta)$

Hence, $Var(\theta^{\star})<Var(\theta)$.

  3. 3.

(see Section 11) Next, we want to show $Var(\theta^{\star})<Var(\theta)$ if
$\gamma\geq\sigma_{q}^{2}/2$ or $\Gamma\leq\hat{\Gamma}$ or
$\Gamma\geq\tilde{\Gamma}$ for some $\hat{\Gamma}>0,\tilde{\Gamma}>0$. Let
$D\triangleq Var(\theta)-Var(\theta^{\star})$

First, when $\tau_{d}>\tau_{a}$, Equation 23 yields

$D=\int_{d(\theta)>0}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta-\int_{d(\theta)>\tau_{a}}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

which is positive since $\tau_{a}$ is positive.

Second, when $\tau_{d}\leq\tau_{a}$, Equation 24 yields

$D=\int_{\mu_{\theta}}^{\mu_{\theta}+\tau_{a}}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta-\int_{\mu_{\theta}+\tau_{d}}^{\mu_{\theta}+\tau_{a}}[(1-w^{\star}(\theta))w^{\star}(\theta)\sigma_{\theta}^{2}+(1-w^{\star}(\theta))^{2}(\mu_{\theta}-\theta)^{2}]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

We can do the same change of variables as the above steps. In particular, let
$y=((\theta-\mu_{\theta})/\sigma_{\theta})^{2}-1$, then we have

$D=\dfrac{\sigma_{\theta}}{\sqrt{2\pi}}\left[\int_{-1}^{\hat{\tau_{a}}^{2}-1}(1+y)\dfrac{\exp(-(y+1)/2)}{\sqrt{y+1}}d\theta-\int_{\hat{\tau_{d}}^{2}-1}^{\hat{\tau_{a}}^{2}-1}(1-\hat{w})(1+(1-\hat{w})y)\dfrac{\exp(-(y+1)/2)}{\sqrt{y+1}}d\theta\right]$

where
$\hat{\tau_{a}}=\tau_{a}/\sigma_{\theta},\hat{\tau_{d}}=\tau_{d}/\sigma_{\theta}$,
$\hat{w}=(-1+\sqrt{1+4\alpha y})/(4y)$ and
$\alpha=\gamma/\sigma_{\theta}^{2}$.

    1. (a)

When $\gamma\geq\sigma_{\theta}^{2}/2$, by Lemma 7.2,
$\tau_{d}\geq\sqrt{\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma)}$, so
$\hat{\tau_{d}}\geq 1/\sqrt{2}$.

Let

$f(w)\triangleq\int_{\hat{\tau_{d}}^{2}-1}^{\hat{\tau_{a}}^{2}-1}\omega(w,y)\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{\sqrt{y+1}}d\theta$

where $\omega(w,y)\triangleq(1-w)(1+(1-w)y)$.

Notice that $\dfrac{\partial\omega}{\partial w}=-1-2(1-w)y$, which is non-
positive if and only if $(1-w)y\geq-1/2$. Because $y\geq\hat{\tau_{d}}-1>-1/2$
and $\hat{w}\in[0,1]$, this implies that $(1-\hat{w})y\geq-1/2$ and
$\dfrac{\partial\omega}{\partial\hat{w}}\leq 0$.

Thus,

$\max_{w\in[0,1]}f(w)=\int_{\hat{\tau_{d}}^{2}-1}^{\hat{\tau_{a}}^{2}-1}(1+y)\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{\sqrt{y+1}}d\theta$

So we get a lower bound of $D$:

$D\geq\dfrac{\sigma_{\theta}}{\sqrt{2\pi}}\int_{-1}^{\hat{\tau_{d}}^{2}-1}(y+1)\dfrac{\exp(-(y+1)/2)}{\sqrt{y+1}}d\theta$

And by Lemma 7.2, we know $\forall\gamma>0$, we must have $\tau_{d}>0$. Thus,
$D>0$.

    2. (b)

Let $\hat{\Gamma}\triangleq
l^{\star}(\theta)\big{|}_{\theta=\mu_{\theta}+\tau_{d}}>0$

When $\Gamma\leq\hat{\Gamma}$, this means $\tau_{a}\leq\tau_{d}$, by Equation
23,
$Var(\theta^{\star})=\int_{d(\theta)>\tau_{a}}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$,
which is less than $Var(\theta)$, since $\tau_{a}>0$ whenever $\Gamma>0$.

    3. (c)

Let $\acute{\Gamma}\triangleq
l^{\star}(\theta)\big{|}_{\theta=\mu_{\theta}+\sigma_{\theta}/\sqrt{2}}>0$

When $\Gamma\geq\acute{\Gamma}$, then
$\tau_{a}\geq\sigma_{\theta}/\sqrt{2}\implies\hat{\tau_{a}}\geq
1/\sqrt{2}\implies\hat{\tau_{d}}^{2}-1\geq-1/2$.

Also, in part 3 (a), we have seen that if $y\geq-1/2$,
$\dfrac{\partial\omega}{\partial w}(\hat{w},y)\leq 0$ (since
$\hat{w}\in[0,1]$). This implies that if $y\geq-1/2$,
$\omega(\hat{w},y)\leq\omega(0,y)=(1+y)$.

And if $\hat{\tau_{a}}^{2}-1$ increases to $\hat{\tau_{a}}^{2}-1+\xi$ for any
$\xi>0$, then the change of $D$ is

$\delta_{D}=\dfrac{\sigma_{\theta}}{\sqrt{2\pi}}\left[\int_{\hat{\tau_{a}}^{2}-1}^{\hat{\tau_{a}}^{2}-1+\xi}[(1+y)-(1-\hat{w})(1+(1-\hat{w})y)]\dfrac{\exp(-(y+1)/2)}{\sqrt{y+1}}d\theta\right]\geq
0$

This means $D$ monotonically increases in $\tau_{a}$ for any
$\tau_{a}\geq\acute{\Gamma}$.

In part 1, we have proved that $D>0$ when $\Gamma\to\infty$, meaning that
$D>0$ when $\tau_{a}\to\infty$. Because $D$ is continuous in $\tau_{a}$, we
either have $D>0$ whenever $\Gamma\geq\acute{\Gamma}$ (so
$\tilde{\Gamma}=\acute{\Gamma}$) or there exists another threshold
$\tilde{\Gamma}>\acute{\Gamma}$ such that $D>0$ whenever
$\Gamma\geq\tilde{\Gamma}$.

###  8.2 Proof of the Results in Section 5.

####  8.2.1 Auxiliary lemmas

###### Lemma 8.12

For any $\theta,\sigma_{q}^{2}$,

$e(\theta,\sigma_{q})=\dfrac{\sigma_{q}^{2}(\sigma_{A}^{4}+\sigma_{q}^{2}(\mu_{A}-\theta)^{2})}{(\sigma_{A}^{2}+\sigma_{q}^{2})^{2}}$
(25)

In addition,

  * •

Both $l(\theta,\sigma_{q}^{2})$ and $e(\theta,\sigma_{q})$ strictly increase
in $(\mu_{A}-\theta)^{2}$.

  * •

Both $l(\theta,\sigma_{q}^{2})$ and $e(\theta,\sigma_{q})$ strictly decrease
in $\sigma_{A}^{2}$ for $\sigma_{A}^{2}<(\mu_{A}-\theta)^{2}$ and increase in
$\sigma_{A}^{2}$ for $\sigma_{A}^{2}\geq(\mu_{A}-\theta)^{2}$.

###### Proof 8.13

Proof of Lemma 8.12. By Equation 1,
$\theta_{A}=\dfrac{\sigma_{A}^{2}}{\sigma_{A}^{2}+\sigma_{q}^{2}}q+\dfrac{\sigma_{q}^{2}}{\sigma_{A}^{2}+\sigma_{q}^{2}}\mu_{A}$.
Then,

$\displaystyle e(\theta,\sigma_{q})$
$\displaystyle=E\left[\left(\dfrac{\sigma_{A}^{2}}{\sigma_{A}^{2}+\sigma_{q}^{2}}(\theta+\epsilon_{q})+\dfrac{\sigma_{q}^{2}}{\sigma_{A}^{2}+\sigma_{q}^{2}}\mu_{A}-\theta\right)^{2}|\theta\right]$
$\displaystyle=E\left[\left(\dfrac{\sigma_{A}^{2}}{\sigma_{A}^{2}+\sigma_{q}^{2}}\epsilon_{q}+\dfrac{\sigma_{q}^{2}}{\sigma_{A}^{2}+\sigma_{q}^{2}}(\mu_{A}-\theta)\right)^{2}|\theta\right]$
$\displaystyle=\left(\dfrac{\sigma_{A}^{2}}{\sigma_{A}^{2}+\sigma_{q}^{2}}\right)^{2}E[\epsilon_{q}^{2}]+\left(\dfrac{\sigma_{q}^{2}}{\sigma_{A}^{2}+\sigma_{q}^{2}}(\mu_{A}-\theta)\right)^{2}$
$\displaystyle=\left(\dfrac{\sigma_{A}^{2}}{\sigma_{A}^{2}+\sigma_{q}^{2}}\right)^{2}\sigma_{q}^{2}+\left(\dfrac{\sigma_{q}^{2}}{\sigma_{A}^{2}+\sigma_{q}^{2}}(\mu_{A}-\theta)\right)^{2}$
$\displaystyle=\dfrac{\sigma_{q}^{2}(\sigma_{A}^{4}+\sigma_{q}^{2}(\mu_{A}-\theta)^{2})}{(\sigma_{A}^{2}+\sigma_{q}^{2})^{2}}$

It is clear that $e(\theta,\sigma_{q})$ strictly increases in
$(\mu_{A}-\theta)^{2}$, and $l(\theta,\sigma_{q}^{2})$ strictly increases in
$(\mu_{A}-\theta)^{2}$ (Note that $I(\sigma_{q}^{2})$ does not depend on
either $\mu_{A}$ or $\sigma_{A}$).

Take the derivative of $e(\theta,\sigma_{q})$ with respect to
$\sigma_{A}^{2}$:

$\displaystyle\dfrac{\partial e(\theta,\sigma_{q})}{\partial\sigma_{A}^{2}}$
$\displaystyle=\dfrac{2\sigma_{q}^{2}\sigma_{A}^{2}(\sigma_{A}^{2}+\sigma_{q}^{2})^{2}-2(\sigma_{A}^{2}+\sigma_{q}^{2})\sigma_{q}^{2}(\sigma_{A}^{4}+\sigma_{q}^{2}(\mu_{A}-\theta)^{2})}{(\sigma_{A}^{2}+\sigma_{q}^{2})^{4}}$
$\displaystyle=\dfrac{2\sigma_{q}^{2}\sigma_{A}^{2}(\sigma_{A}^{2}+\sigma_{q}^{2})-2\sigma_{q}^{2}(\sigma_{A}^{4}+\sigma_{q}^{2}(\mu_{A}-\theta)^{2})}{(\sigma_{A}^{2}+\sigma_{q}^{2})^{3}}$
$\displaystyle=\dfrac{2\sigma_{q}^{4}(\sigma_{A}^{2}-(\mu_{A}-\theta)^{2})}{(\sigma_{A}^{2}+\sigma_{q}^{2})^{3}}$

Thus, $\dfrac{\partial e(\theta,\sigma_{q})}{\partial\sigma_{A}^{2}}<0$ if
$\sigma_{A}^{2}<(\mu_{A}-\theta)^{2}$, and $\dfrac{\partial
e(\theta,\sigma_{q})}{\partial\sigma_{A}^{2}}\geq 0$ if
$\sigma_{A}^{2}\geq(\mu_{A}-\theta)^{2}$. This implies that both
$l(\theta,\sigma_{q}^{2})$ and $e(\theta,\sigma_{q})$ strictly decrease in
$\sigma_{A}^{2}$ for $\sigma_{A}^{2}<(\mu_{A}-\theta)^{2}$ and increase in
$\sigma_{A}^{2}$ for $\sigma_{A}^{2}\geq(\mu_{A}-\theta)^{2}$.

###### Lemma 8.14

Let $w^{\star}(\theta,\gamma)=\dfrac{\sigma_{q}^{\star
2}(\theta,\gamma)}{\sigma_{A}^{2}+\sigma_{q}^{\star 2}(\theta,\gamma)}$.
$\forall\theta,\mu_{A},\sigma_{A},\gamma_{1}>\gamma_{2}$,
$w^{\star}(\theta,\gamma_{1})\geq w^{\star}(\theta,\gamma_{2})$.

###### Proof 8.15

Proof of Lemma 8.14. Firstly, by Equation 25, we can write the objective
function as:

$l(\theta,w,\gamma)=w(1-w)\sigma_{A}^{2}+w^{2}(\mu_{A}-\theta)^{2}-\dfrac{\gamma}{2}\ln
w$

where $w=\sigma_{q}^{2}/(\sigma_{A}^{2}+\sigma_{q}^{2})\in[0,1]$.

For the sake of contradiction, assume
$w^{\star}(\theta,\gamma_{1})<w^{\star}(\theta,\gamma_{2})$ for some $\theta$.
Since $I(w)=-\ln(w)/2$ strictly decreases in $w$, we have
$\delta_{I}\triangleq
I(w^{\star}(\theta,\gamma_{1}))-I(w^{\star}(\theta,\gamma_{2}))>0$. Let
$\delta_{e}\triangleq
e(\theta,w^{\star}(\theta,\gamma_{1}))-e(\theta,w^{\star}(\theta,\gamma_{2}))$.
Because $w^{\star}(\theta,\gamma_{1})$ is optimal when $\gamma=\gamma_{1}$, we
must have
$l(\theta,w^{\star}(\theta,\gamma_{1}),\gamma_{1})-l(\theta,w^{\star}(\theta,\gamma_{2}),\gamma_{1})<0$.
This implies $\delta_{e}<0$ and $\delta_{e}+\gamma_{1}\delta_{I}<0$. However,
because $\gamma_{1}>\gamma_{2}$ and $\delta_{I}>0$, we must have
$\delta_{e}+\gamma_{2}\delta_{I}<\delta_{e}+\gamma_{1}\delta_{I}<0$, meaning
that
$l(\theta,w^{\star}(\theta,\gamma_{1}),\gamma_{2})-l(\theta,w^{\star}(\theta,\gamma_{2}),\gamma_{2})<0$.
This contradicts the assumption that $w^{\star}(\theta,\gamma_{2})$ is optimal
when $\gamma=\gamma_{2}$. Therefore $\forall\theta$,
$w^{\star}(\theta,\gamma_{1})\geq w^{\star}(\theta,\gamma_{2})$ whenever
$\gamma_{1}>\gamma_{2}$.

####  8.2.2 Proofs of the results

###### Proof 8.16

Proof of Proposition 5.1.

  * •

Suppose $|\mu_{A_{1}}-\theta|>|\mu_{A_{2}}-\theta|$ for some
$\mu_{A_{1}},\mu_{A_{2}},\theta$. Let $\sigma_{q_{1}}^{\star}$ and
$\sigma_{q_{2}}^{\star}$ denote the optimal decision for user $\theta$ in
Problem (4) when $\mu_{A}=\mu_{A_{1}}$ and $\mu_{A}=\mu_{A_{2}}$,
respectively. By definition of $l$ in Equation 3, let
$l^{\star}_{1}=l(\theta,\sigma_{q_{1}}^{\star},\mu_{A_{1}})$ and
$l^{\star}_{2}=l(\theta,\sigma_{q_{2}}^{\star},\mu_{A_{2}})$.

We want to show $l^{\star}_{1}>l^{\star}_{2}$. For the sake of contradiction,
suppose $l^{\star}_{1}\leq l^{\star}_{2}$. By Lemma 8.12,
$l^{\star}_{1}=l(\theta,\sigma_{q_{1}}^{\star},\mu_{A_{1}})>l(\theta,\sigma_{q_{1}}^{\star},\mu_{A_{2}})$.
This implies
$l(\theta,\sigma_{q_{1}}^{\star},\mu_{A_{2}})<l^{\star}_{2}=l(\theta,\sigma_{q_{2}}^{\star},\mu_{A_{2}})$.
This contradicts the assumption that $\sigma_{q_{2}}^{\star}$ minimizes
$l(\theta,\sigma_{q},\mu_{A_{2}})$. Therefore, $l^{\star}_{1}>l^{\star}_{2}$.
We conclude that $l^{\star}$ strictly increases in $|\mu_{A}-\theta|$.

  * •

Suppose $\sigma_{A_{1}}<\sigma_{A_{2}}<|\mu_{A}-\theta|$ for some
$\sigma_{A_{1}},\sigma_{A_{2}},\mu_{A},\theta$. Let $\sigma_{q_{1}}^{\star}$
and $\sigma_{q_{2}}^{\star}$ denote the optimal decision for user $\theta$ in
Problem (4) when $\sigma_{A}=\sigma_{A_{1}}$ and $\sigma_{A}=\sigma_{A_{2}}$,
respectively. By definition of $l$ in Equation 3, let
$l^{\star}_{1}=l(\theta,\sigma_{q_{1}}^{\star},\sigma_{A_{1}})$ and
$l^{\star}_{2}=l(\theta,\sigma_{q_{2}}^{\star},\sigma_{A_{2}})$.

We want to show $l^{\star}_{1}>l^{\star}_{2}$. For the sake of contradiction,
suppose $l^{\star}_{1}\leq l^{\star}_{2}$. By Lemma 8.12,
$l^{\star}_{1}=l(\theta,\sigma_{q_{1}}^{\star},\sigma_{A_{1}})>l(\theta,\sigma_{q_{1}}^{\star},\sigma_{A_{2}})$.
This implies
$l(\theta,\sigma_{q_{1}}^{\star},\sigma_{A_{2}})<l^{\star}_{2}=l(\theta,\sigma_{q_{2}}^{\star},\sigma_{A_{2}})$.
This contradicts the assumption that $\sigma_{q_{2}}^{\star}$ minimizes
$l(\theta,\sigma_{q},\sigma_{A_{2}})$. Therefore,
$l^{\star}_{1}>l^{\star}_{2}$. We conclude that $l^{\star}$ strictly decreases
in $\sigma_{A}$ when $\sigma_{A}<|\mu_{A}-\theta|$.

Similarly, when $|\mu_{A}-\theta|\leq\sigma_{A_{1}}<\sigma_{A_{2}}$, we want
to show $l^{\star}_{1}\leq l^{\star}_{2}$. For the sake of contradiction,
suppose $l^{\star}_{1}>l^{\star}_{2}$. By Lemma 8.12,
$l^{\star}_{2}=l(\theta,\sigma_{q_{2}}^{\star},\sigma_{A_{2}})>l(\theta,\sigma_{q_{2}}^{\star},\sigma_{A_{1}})$.
This implies
$l(\theta,\sigma_{q_{2}}^{\star},\sigma_{A_{1}})<l^{\star}_{1}=l(\theta,\sigma_{q_{1}}^{\star},\sigma_{A_{1}})$.
This contradicts the assumption that $\sigma_{q_{1}}^{\star}$ minimizes
$l(\theta,\sigma_{q},\sigma_{A_{1}})$. Therefore, $l^{\star}_{1}\leq
l^{\star}_{2}$. We conclude that $l^{\star}$ strictly increases in
$\sigma_{A}$ when $\sigma_{A}\geq|\mu_{A}-\theta|$.

###### Proof 8.17

Proof of Proposition 5.2. Let $\phi(\cdot)$ denote the probability density
function of $N(0,1)$. And let
$w=\sigma_{q}^{2}/(\sigma_{A}^{2}+\sigma_{q}^{2})$.

  * •

Let us first show $E[l^{\star}(\theta,\mu_{A})]$ is minimized at
$\mu_{A}=\mu_{\theta}$. That is, $\forall\mu_{A1}\neq\mu_{\theta}$, we want to
show $E[l^{\star}(\theta,\mu_{A1})]>E[l^{\star}(\theta,\mu_{\theta})]$.
Without loss of generality, suppose $\mu_{A1}>\mu_{\theta}$.

By definition,

$E[l^{\star}(\theta,\mu_{A})]=\displaystyle\int_{-\infty}^{\infty}l^{\star}(\theta,\mu_{A})\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

So we want to show

$\int_{-\infty}^{\infty}[l^{\star}(\theta,\mu_{A1})-l^{\star}(\theta,\mu_{\theta})]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta>0$

By Equation 25, $\forall\sigma_{q},\theta_{1},\theta_{2}$,
$\theta_{1}-\mu_{A}=\mu_{A}-\theta_{2}\implies
e(\theta_{1},\sigma_{q})=e(\theta_{2},\sigma_{q})$, so
$w^{\star}(\theta_{1})=w^{\star}(\theta_{2})$, meaning that
$w^{\star}(\theta)$ and $l^{\star}(\theta,\mu_{A})$ are axisymmetric with
respect to $\theta=\mu_{A}$. Also, $\forall\theta,\mu_{A}$,
$w^{\star}(\theta)$ and $l^{\star}(\theta,\mu_{A})$ are constant as long as
$|\mu_{A}-\theta|$ is constant. This implies
$[l^{\star}(\theta,\mu_{A1})-l^{\star}(\theta,\mu_{\theta})]$ is
centrosymmetric with respect to the point
$\left((\mu_{A1}+\mu_{\theta})/2,0\right)$. That is,
$\forall\theta_{1}>\theta_{2}$,
$\theta_{1}-(\mu_{A1}+\mu_{\theta})/2=(\mu_{A1}+\mu_{\theta})/2-\theta_{2}\implies[l^{\star}(\theta_{1},\mu_{A1})-l^{\star}(\theta_{1},\mu_{\theta})]=-[l^{\star}(\theta_{2},\mu_{A1})-l^{\star}(\theta_{2},\mu_{\theta})]>0$,
which is positive because $l^{\star}(\theta,\mu_{A})$ strictly increases in
$|\mu_{A}-\theta|$ by Proposition 5.1.

Let $\bar{\mu}$ denote $(\mu_{A1}+\mu_{\theta})/2$. Because
$\mu_{A}>\mu_{\theta}\implies\bar{\mu}>\mu_{\theta}$, we have
$Pr(\theta\leq\bar{\mu})>Pr(\theta>\bar{\mu})$, and
$\forall\theta_{1}>\theta_{2}$,
$\theta_{1}-\bar{\mu}=\bar{\mu}-\theta_{2}\implies\phi\left((\theta_{1}-\mu_{\theta})/\sigma_{\theta}\right)<\phi\left((\theta_{2}-\mu_{\theta})/\sigma_{\theta}\right)$.
Because $[l^{\star}(\theta,\mu_{A1})-l^{\star}(\theta,\mu_{\theta})]$ is
centrosymmetric with respect to the point $(\bar{\mu},0)$, these imply
$0<[l^{\star}(\theta_{1},\mu_{A1})-l^{\star}(\theta_{1},\mu_{\theta})]\phi\left((\theta_{1}-\mu_{\theta})/\sigma_{\theta}\right)<-[l^{\star}(\theta_{2},\mu_{A1})-l^{\star}(\theta_{2},\mu_{\theta})]\phi((\theta_{2}-\mu_{\theta})/\sigma_{\theta}$.

This means that $\forall\theta_{1}>\theta_{2}$,
$\theta_{1}-\bar{\mu}=\bar{\mu}-\theta_{2}$, we have

$[l^{\star}(\theta_{1},\mu_{A1})-l^{\star}(\theta_{1},\mu_{\theta})]\phi\left((\theta_{1}-\mu_{\theta})/\sigma_{\theta}\right)+[l^{\star}(\theta_{2},\mu_{A1})-l^{\star}(\theta_{2},\mu_{\theta})]\phi((\theta_{2}-\mu_{\theta})/\sigma_{\theta}>0$

Hence,

$\displaystyle\int_{-\infty}^{\infty}[l^{\star}(\theta,\mu_{A1})-l^{\star}(\theta,\mu_{\theta})]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\bar{\mu}}[l^{\star}(\theta,\mu_{A1})-l^{\star}(\theta,\mu_{\theta})]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{\bar{\mu}}^{\infty}[l^{\star}(\theta,\mu_{A1})-l^{\star}(\theta,\mu_{\theta})]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta>0$

This implies $E[l^{\star}(\theta,\mu_{A})]$ is minimized at
$\mu_{A}=\mu_{\theta}$.

And because $\dfrac{\partial l^{\star}(\theta,\mu_{A})}{\partial\mu_{A}}$ is
continuous at $\mu_{A}=\mu_{\theta}$ and $\sigma_{A}=\sigma_{\theta}$,
$E[l^{\star}(\theta,\mu_{A})]$ is differentiable at $\mu_{A}=\mu_{\theta}$ and
$\sigma_{A}=\sigma_{\theta}$. Thus, we have $\dfrac{\partial
E[l^{\star}]}{\partial\mu_{A}}\Big{\lvert}_{\mu_{A}=\mu_{\theta},\sigma_{A}=\sigma_{\theta}}=0$.

  * •

By Equation 3 and Equation 25,

$l^{\star}(\theta)=\dfrac{\sigma_{q}^{\star
2}(\theta)(\sigma_{A}^{4}+\sigma_{q}^{\star
2}(\theta)(\mu_{A}-\theta)^{2})}{(\sigma_{A}^{2}+\sigma_{q}^{\star
2}(\theta))^{2}}-\dfrac{\gamma}{2}\ln\left(\dfrac{\sigma_{q}^{\star
2}(\theta)}{\sigma_{q}^{\star 2}(\theta)+\sigma_{\theta}^{2}}\right)$

By the chain rule, $\dfrac{\partial
l^{\star}}{\partial\sigma_{A}^{2}}=\dfrac{dl^{\star}}{d\sigma_{q}^{\star
2}}\cdot\dfrac{d\sigma_{q}^{\star
2}}{d\sigma_{A}^{2}}+\dfrac{dl^{\star}}{d\sigma_{A}^{2}}$. Because
$\sigma_{q}^{\star 2}$ is optimal, $\dfrac{dl^{\star}}{d\sigma_{q}^{\star
2}}=0$. This implies $\dfrac{\partial
l^{\star}}{\partial\sigma_{A}^{2}}=\dfrac{dl^{\star}}{d\sigma_{A}^{2}}$. With
some algebra, we can get

$\dfrac{dl^{\star}}{d\sigma_{A}^{2}}\Big{\lvert}_{\mu_{A}=\mu_{\theta},\sigma_{A}=\sigma_{\theta}}=\dfrac{2\sigma_{q}^{\star
4}(\sigma_{\theta}^{2}-(\mu_{\theta}-\theta)^{2})}{(\sigma_{q}^{\star
2}+\sigma_{\theta}^{2})^{3}}$

Since
$w(\theta)=\sigma_{q}^{2}(\theta)/[\sigma_{A}^{2}+\sigma_{q}^{2}(\theta)]$, we
can rewrite it as

$\dfrac{dl^{\star}}{d\sigma_{A}^{2}}\Big{\lvert}_{\mu_{A}=\mu_{\theta},\sigma_{A}=\sigma_{\theta}}=\dfrac{dl^{\star}(\theta)}{d\sigma_{A}^{2}}\Big{\lvert}_{\mu_{A}=\mu_{\theta},\sigma_{A}=\sigma_{\theta}}=\dfrac{2}{\sigma_{\theta}^{2}}w^{\star}(\theta)^{2}(1-w^{\star}(\theta))(\sigma_{A}^{2}-(\mu_{\theta}-\theta)^{2})$

where
$w^{\star}(\theta)=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\Delta}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}$
and
$\Delta=\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$
by Lemma 7.2.

And, by definition,

$\displaystyle
E[l^{\star}]\Big{\lvert}_{\mu_{A}=\mu_{\theta},\sigma_{A}=\sigma_{\theta}}$
$\displaystyle=\int_{-\infty}^{\infty}l^{\star}(\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=\int_{|\mu_{\theta}-\theta|\geq\tau_{d}}l^{\star}(\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{|\mu_{\theta}-\theta|<\tau_{d}}l^{\star}(\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

where $\tau_{d}$ is defined in Lemma 7.2.

When $\mu_{A}=\mu_{\theta}$, $l(\theta)$ is symmetric with respect to
$\theta=\mu_{\theta}$, so

$E[l^{\star}]\Big{\lvert}_{\mu_{A}=\mu_{\theta},\sigma_{A}=\sigma_{\theta}}=2\left[\displaystyle\int_{\mu_{\theta}+\tau_{d}}^{\infty}l^{\star}(\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{0}^{\mu_{\theta}+\tau_{d}}l^{\star}(\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right]$

And when $w=1$ we know $l=(\mu_{\theta}-\theta)^{2}$, so

$E[l^{\star}]\Big{\lvert}_{\mu_{A}=\mu_{\theta},\sigma_{A}=\sigma_{\theta}}=2\left[\displaystyle\int_{\mu_{\theta}+\tau_{d}}^{\infty}l^{\star}(\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{0}^{\mu_{\theta}+\tau_{d}}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right]$

Thus, by the Leibniz integral rule,

$\displaystyle\dfrac{\partial
E[l^{\star}]}{\partial\sigma_{A}^{2}}\Big{\lvert}_{\mu_{A}=\mu_{\theta},\sigma_{A}=\sigma_{\theta}}$
$\displaystyle=2\left[\int_{\mu_{\theta}+\tau_{d}}^{\infty}\dfrac{\partial
l^{\star}(\theta)}{\partial\sigma_{A}^{2}}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta-(\mu_{\theta}-\tau_{d})^{2}\cdot\dfrac{\partial\tau_{d}}{\sigma_{A}^{2}}+(\mu_{\theta}-\tau_{d})^{2}\cdot\dfrac{\partial\tau_{d}}{\sigma_{A}^{2}}\right]$
$\displaystyle=2\left[\int_{\mu_{\theta}+\tau_{d}}^{\infty}\dfrac{\partial
l^{\star}(\theta)}{\partial\sigma_{A}^{2}}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right]$
$\displaystyle=2\left[\int_{\mu_{\theta}+\tau_{d}}^{\infty}\dfrac{2}{\sigma_{\theta}^{2}}w^{\star}(\theta)^{2}(1-w^{\star}(\theta))(\sigma_{A}^{2}-(\mu_{\theta}-\theta)^{2})\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right]$
$\displaystyle=\dfrac{1}{\sigma_{\theta}^{2}}\left[\int_{\mu_{\theta}+\tau_{d}}^{\infty}w^{\star}(\theta)^{2}(1-w^{\star}(\theta))(\sigma_{A}^{2}-(\mu_{\theta}-\theta)^{2})\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right]$

Let $g(\theta)\triangleq
w^{\star}(\theta)^{2}(1-w^{\star}(\theta))(\sigma_{\theta}^{2}-(\mu_{\theta}-\theta)^{2})$

When $\gamma\geq 2\sigma_{\theta}^{2}$, by the proof of Lemma 7.2 (see the
summary at the end of the proof),
$\gamma>2\sigma_{\theta}^{2}\geq\sigma_{\theta}^{2}\implies\tau_{d}=\sqrt{\sigma_{\theta}^{2}/2+\gamma/4}>\sqrt{\sigma_{\theta}^{2}/2+2\sigma_{\theta}^{2}/4}=\sigma_{\theta}$.
So $g(\theta)$ is always negative for any $\theta>\mu_{\theta}+\tau_{d}$.
Thus,

$\int_{\mu_{\theta}+\tau_{d}}^{\infty}g(\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta<0$

###### Proof 8.18

Proof of Theorem 5.3. Let $w=\sigma_{q}^{2}/(\sigma_{A}^{2}+\sigma_{q}^{2})$.
By Equation 1, $\theta_{A}=(1-w)q+w\mu_{A}$, where $q=\theta+\epsilon_{q}$,
$\epsilon_{q}\sim N(0,\sigma_{q}^{2})$ and $\theta\sim
N(\mu_{\theta},\sigma_{\theta}^{2})$. We further define
$w^{\star}(\theta)=\sigma_{q}^{\star
2}(\theta)/[\sigma_{A}^{2}+\sigma_{q}^{\star 2}(\theta)]$. Let $\phi(\cdot)$
denote the probability density function of $N(0,1)$.

$\displaystyle E[\theta^{\star}]$
$\displaystyle=\int_{|\mu_{A}-\theta|\leq\tau_{a}}\int_{-\infty}^{\infty}\theta_{A}^{\star}\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{|\mu_{A}-\theta|>\tau_{a}}\theta\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=\int_{|\mu_{A}-\theta|\leq\tau_{a}}\int_{-\infty}^{\infty}[(1-w^{\star}(\theta))q+w^{\star}(\theta)\mu_{A}]\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle+\int_{|\mu_{A}-\theta|>\tau_{a}}\theta\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=\int_{|\mu_{A}-\theta|\leq\tau_{a}}[(1-w^{\star}(\theta))\theta+w^{\star}(\theta)\mu_{A}]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{|\mu_{A}-\theta|>\tau_{a}}\theta\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=\int_{|\mu_{A}-\theta|\leq\tau_{a}}[(1-w^{\star}(\theta))(\theta-\mu_{A})+\mu_{A}]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{|\mu_{A}-\theta|>\tau_{a}}\theta\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=\int_{|\mu_{A}-\theta|\leq\tau_{a}}[(1-w^{\star}(\theta))(\theta-\mu_{A})+\mu_{A}]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\mu_{\theta}-\int_{|\mu_{A}-\theta|\leq\tau_{a}}\theta\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=\int_{|\mu_{A}-\theta|\leq\tau_{a}}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\mu_{\theta}$

This implies that

$|E[\theta^{\star}]-\mu_{\theta}|=\displaystyle\left\lvert\int_{|\mu_{A}-\theta|\leq\tau_{a}}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right\rvert$
(26)

  1. 1.

    1. (a)

First, we want to show

$|E[\theta^{\star}]-\mu_{\theta}|\leq\left\lvert\int_{-\infty}^{\infty}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right\rvert$

Without loss of generality, suppose $\mu_{A}\geq\mu_{\theta}$. Then,
$Pr(\theta\leq\mu_{A})\geq Pr(\theta>\mu_{A})$, and
$\forall\theta_{1}>\theta_{2}$,
$\theta_{1}-\mu_{A}=\mu_{A}-\theta_{2}\implies\phi\left((\theta_{1}-\mu_{\theta})/\sigma_{\theta}\right)<\phi\left((\theta_{2}-\mu_{\theta})/\sigma_{\theta}\right)$.
Because $w^{\star}(\theta)$ is symmetric with respect to $\theta=\mu_{A}$, we
have $w^{\star}(\theta_{1})=w^{\star}(\theta_{2})$. These imply

$0<-w^{\star}(\theta_{1})(\mu_{A}-\theta_{1})\phi\left((\theta_{1}-\mu_{\theta})/\sigma_{\theta}\right)<w^{\star}(\theta_{2})(\mu_{A}-\theta_{2})\phi\left((\theta_{2}-\mu_{\theta})/\sigma_{\theta}\right)$

which means that $\forall\theta_{1}>\theta_{2}$, if
$\theta_{1}-\mu_{A}=\mu_{A}-\theta_{2}$, then

$w^{\star}(\theta_{2})(\mu_{A}-\theta_{2})\phi\left((\theta_{2}-\mu_{\theta})/\sigma_{\theta}\right)+w^{\star}(\theta_{1})(\mu_{A}-\theta_{1})\phi\left((\theta_{1}-\mu_{\theta})/\sigma_{\theta}\right)>0$

Since $\tau_{a}>0$, we can get

$\int_{\mu_{A}-\tau_{a}}^{\mu_{A}+\tau_{a}}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta>0$

and

$\int_{\mu_{A}+\tau_{a}}^{\infty}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{-\infty}^{\mu_{A}-\tau_{a}}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\geq
0$

Thus,

$\displaystyle\int_{-\infty}^{\infty}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=\int_{\mu_{A}-\tau_{a}}^{\mu_{A}+\tau_{a}}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle+\int_{\mu_{A}+\tau_{a}}^{\infty}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle+\int_{-\infty}^{\mu_{A}-\tau_{a}}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle>0$

and

$|E[\theta^{\star}]-\mu_{\theta}|=\left\lvert\int_{|\mu_{A}-\theta|\leq\tau_{a}}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right\rvert\leq\left\lvert\int_{-\infty}^{\infty}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right\rvert$

Let $\gamma_{1}>\gamma_{2}$. By Lemma 8.14, $\forall\theta$,
$w^{\star}(\theta,\gamma_{1})\geq w^{\star}(\theta,\gamma_{2})$. Because
$w^{\star}(\theta)$ is symmetric with respect to $\theta=\mu_{A}$,
$\forall\theta_{1}>\theta_{2}$, $\theta_{1}-\mu_{A}=\mu_{A}-\theta_{2}$, then

$\displaystyle(w^{\star}(\theta_{2},\gamma_{1})-w^{\star}(\theta_{2},\gamma_{2}))(\mu_{A}-\theta_{2})\phi\left((\theta_{2}-\mu_{\theta})/\sigma_{\theta}\right)$
$\displaystyle\geq$
$\displaystyle-(w^{\star}(\theta_{1},\gamma_{1})-w^{\star}(\theta_{1},\gamma_{2}))(\mu_{A}-\theta_{1})\phi\left((\theta_{1}-\mu_{\theta})/\sigma_{\theta}\right)\geq
0$

This implies

$\displaystyle\int_{\theta\leq\mu_{A}}w^{\star}(\theta,\gamma_{1})(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta-\int_{\theta\leq\mu_{A}}w^{\star}(\theta,\gamma_{2})(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle\geq$
$\displaystyle-\left[\int_{\theta>\mu_{A}}w^{\star}(\theta,\gamma_{1})(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta-\int_{\theta>\mu_{A}}w^{\star}(\theta,\gamma_{2})(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right]\geq
0$

Rearrange the inequality, we can get

$\int_{-\infty}^{\infty}w^{\star}(\theta,\gamma_{1})(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\geq\displaystyle\int_{-\infty}^{\infty}w^{\star}(\theta,\gamma_{2})(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

Thus,

$\int_{-\infty}^{\infty}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

increases in $\gamma$.

And because $w^{\star}(\theta,\gamma)\to 1$ as $\gamma\to\infty$, by the
monotone convergence theorem (Pugh 2015), we get the upper bound

$\int_{-\infty}^{\infty}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\leq\mu_{A}-\mu_{\theta}$

Hence, $|E[\theta^{\star}]-\mu_{\theta}|\leq|\mu_{A}-\mu_{\theta}|$.

  2. 2.

When $\gamma=0$, for any $\theta$, $w^{\star}(\theta)=0$, by Equation 26, we
have $|E[\theta^{\star}]-\mu_{\theta}|=0$. And when $\Gamma=0$, $\tau_{a}=0$,
$|E[\theta^{\star}]-\mu_{\theta}|=\displaystyle\left\lvert\int_{|\mu_{A}-\theta|=0}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right\rvert=0$

  3. 3.

When $\Gamma\to\infty$, by Equation 26,

$|E[\theta^{\star}]-\mu_{\theta}|=\displaystyle\left\lvert\int_{-\infty}^{\infty}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right\rvert$

And when $\gamma\to\infty$, $\forall\theta$, $w^{\star}(\theta)\to 1$.

Without loss of generality, suppose $\mu_{A}\geq\mu_{\theta}$. In part 1, we
have shown that

$\displaystyle\int_{-\infty}^{\infty}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

is non-negative and increases in $\gamma$. By the monotone convergence theorem
(Pugh 2015), we have

$\lim_{\gamma\to\infty}\displaystyle\left\lvert\int_{-\infty}^{\infty}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right\rvert=\displaystyle\left\lvert\int_{-\infty}^{\infty}(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right\rvert=|\mu_{A}-\mu_{\theta}|$

Thus, when $\Gamma\to\infty$ and $\gamma\to\infty$,
$|E[\theta^{\star}]-\mu_{\theta}|=|\mu_{A}-\mu_{\theta}|$.

  4. 4.

When $\Gamma\to\infty$, by Equation 26,

$|E[\theta^{\star}]-\mu_{\theta}|=\displaystyle\left\lvert\int_{-\infty}^{\infty}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\right\rvert$

Without loss of generality, suppose $\mu_{A}\geq\mu_{\theta}$. In part 1, we
have shown

$\int_{-\infty}^{\infty}w^{\star}(\theta)(\mu_{A}-\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

is non-negative and increases in $\gamma$. Hence, when $\Gamma\to\infty$,
$|E[\theta_{A}^{\star}]-\mu_{\theta}|$ increases in $\gamma$.

##  9 The description of the simulation for the self-training loop.

In this section, we explain the simulation process of the self-training loop
depicted in Section 4.3. The detailed pseudo code is provided in Algorithm 1,
Algorithm 2, Algorithm 3, and Algorithm 4. Let $\phi(\cdot)$ denote the
probability density function of $N(0,1)$. To construct Figure 4, we use
$\mu_{\theta}=0,\sigma_{\theta}=1,N=M=1000,n=400$.

Algorithm 1 is the main algorithm that runs the simulation. There are two
points we want to highlight. First, for computational tractability, we
discretize the population distribution of $\theta$. To this end, we evenly
select $M$ points from the range
$[\mu_{\theta}-3\sigma_{\theta},\mu_{\theta}+3\sigma_{\theta}]$ which covers
$99.7\%$ of the sample space of $N(\mu_{\theta},\sigma_{\theta}^{2})$. These
selected points constitute the support for the population distribution of
$\theta$, denoted by $S$. And the probability mass function is given by
$\pi_{\theta}(\theta)=\phi(\theta)/\sum_{x\in S}\phi(x),\ \forall\theta\in S$.
Second, at the end of each iteration, the AI’s prior is updated by the AI
outputs. That is, the AI’s prior is replaced by the distribution of
$\theta_{A}$. This corresponds to the self-training loop in which the AI
perfectly learns the AI content generated in the previous iteration, so that
the distribution of the AI outputs overrides the AI’s prior.

Algorithm 1 The steps of the simulation model

1:Input: $\mu_{\theta},\sigma_{\theta}$, sample size $N$, iteration number
$n$, support size $M$.

2:Output: $\theta_{A,j}^{\star,i}$, $\forall i\in\\{1,2,\dots,n\\}$, $\forall
j\in\\{1,2,\dots,N\\}$.

3:Discretize Normal distribution: Evenly select $M$ points from
$[\mu_{\theta}-3\sigma_{\theta},\mu_{\theta}+3\sigma_{\theta}]$ as the support
of $\theta$, denoted by $S$. And compute
$\pi_{\theta}(\theta)=\phi(\theta)/\sum_{x\in S}\phi(x),\ \forall\theta\in S$
as the population distribution of $\theta$.

4:Generate users’ preferences: Randomly sample
$\\{\theta_{1},\dots,\theta_{N}\\}$ from $S$ with replacement as the set of
users’ preferences.

5:Initialize the AI’s prior: $\pi_{A}^{1}(\theta)=\pi_{\theta}(\theta),\
\forall\theta\in S$

6:for $i=1,2,\dots,n$ do

7: for $j=1,2,\dots,N$ do

8: Find the optimal
$\sigma^{\star,i}_{q,j}=\argmin_{\sigma_{q}}l(\sigma_{q}|\theta_{j},\pi_{A}^{i},\pi_{\theta},S,\gamma)$
(Algorithm 4)

9: Sample $q_{j}^{i}$ from $N(\theta_{j},\sigma^{\star,i}_{q,j})$

10: Output
$\theta_{A,j}^{\star,i}=\theta_{A}(\pi_{A}^{i},q_{j}^{i},\sigma^{\star,i}_{q,j},S)$
(Algorithm 2)

11: end for

12: Compute the distribution of $\theta_{A,j}^{\star,i}$ and use it as the new
AI prior to the next iteration.

13: for $m=1,2,\dots,M$ do

14: $\forall\theta\in
S,\pi_{A}^{i+1}(\theta)=\sum_{j=1}^{N}\boldsymbol{1}_{\theta_{A,j}^{\star,i}=\theta}/N$.

15: end for

16:end for

Algorithm 2 is used to produce the AI output given the information sent by a
user, as depicted in Section 3.

Algorithm 2 Output $\theta_{A}$

1:Input: $\pi_{A}$, $q$, $\sigma_{q}$, $S$

2:Output: $\theta_{A}$

3:Compute the likelihood: $\forall\theta\in
S,p(q|\theta)=\phi((q-\theta)/\sigma_{q})$

4:Compute the posterior with respect to $\pi_{A}$: $\forall\theta\in
S,p(\theta|q)=\dfrac{p(q|\theta)\pi_{A}(\theta)}{\sum_{x\in
S}p(q|x)\pi_{A}(x)}$.

5:Compute the posterior mean: $\theta_{A}=\sum_{\theta\in S}\theta\cdot
p(\theta|q)$

Algorithm 3 is used to compute the posterior distribution with respect to the
population distribution, $\pi_{\theta}$, given $q$. It helps us to compute the
mutual information $e$ given $\sigma_{q}$ in Algorithm 4.

Algorithm 3 Posterior with respect to $\pi_{\theta}$

1:Input: $q$, $\pi_{\theta}$, $\sigma_{q}$, $S$

2:Output: $p_{\theta}(\cdot|q)$

3:Compute the likelihood: $\forall\theta\in
S,p(q|\theta)=\phi((q-\theta)/\sigma_{q})$

4:Compute the posterior with respect to $\pi_{\theta}$: $\forall\theta\in
S,p_{\theta}(\theta|q)=\dfrac{p(q|\theta)\pi_{\theta}(\theta)}{\sum_{x\in
S}p(q|x)\pi_{\theta}(x)}$.

Algorithm 4 is used to compute the utility loss given $\sigma_{q}$ and
$\theta$. Note that we compute $e$ by its definition $e\triangleq
H(\theta)-E[H(\theta|q)]$.

Algorithm 4 Compute the utility loss $l$

1:Input: $\sigma_{q}$, $\theta$, $\pi_{A}$, $\pi_{\theta}$, $S$, $\gamma$

2:Output: $l$

3:Compute the fidelity error
$e=\int[\theta_{A}(\pi_{A},q,\sigma_{q},S)-\theta]^{2}\phi((q-\theta)/\sigma_{q})dq$.

4:Compute the mutual information where $p_{\theta}(\cdot|q)$ is given by
Algorithm 3 $I=-\sum_{\theta\in
S}\pi_{\theta}(\theta)\log(\pi_{\theta}(\theta))+\int\sum_{\theta\in
S}p_{\theta}(\theta|q)\log(p_{\theta}(\theta|q))\phi((q-\theta)/\sigma_{q})dq$

5:Compute $l=e+\gamma I$

##  10 Extensive explanation of Proposition 4.1: Decomposition of the fidelity
error

$-5$$-4$$-3$$-2$$-1$$0$$1$$2$$3$$4$$5$$0$$0.2$$0.4$$0.6$$0.8$user preference
$\theta$fidelity error $e^{\star}$fidelity bias error $e_{b}^{\star}$fidelity
uncertainty error $e_{u}^{\star}$

Figure 9: The black dashed vertical lines are at $d(\theta)=\tau_{d}$, and the
black dotted vertical lines are at $d(\theta)=\tau_{a}$. The white region
indicates the users who simply accept the default output; the yellow region
indicates the users interacting with the AI by sending information; the red
region indicates the users without using AI. We use
$\mu_{\theta}=0,\sigma_{\theta}=1,\gamma=1,\Gamma=1.4$.

To further understand the variation of fidelity error shown in Proposition 4.1
for the users with $d(\theta)<\tau_{a}$, we decompose their the fidelity error
into a bias and a variance term,
$e^{\star}=Var(\theta^{\star}|\theta)+[E(\theta^{\star}|\theta)-\theta]^{2}$,
as introduced in Section 3. Again, we call $Var(\theta^{\star}|\theta)$ the
fidelity uncertainty error denoted by $e_{u}^{\star}$, and
$[E(\theta^{\star}|\theta)-\theta]^{2}$ the fidelity bias error denoted by
$e_{b}^{\star}$. This decomposition is depicted in Figure 9. We can see that
for the users with $d(\theta)<\tau_{d}$, the fidelity bias error $e_{b}$
largely contributes to the fidelity error since they accept the AI’s default
output without sending any informative signal. At the point of
$d(\theta)=\tau_{d}$, the user starts providing information, leading to a
decrease in $e_{b}$ but an increase in $e_{u}$. As the uniqueness further
grows, they share more information, resulting in lower fidelity errors. This
reduction is primarily driven by the decrease in $e_{u}$, since providing more
information effectively reduces the noise of the communication but hardly
eliminates the inherent difference between the mean and their actual
preferences. We formalize this observation in the following Proposition 10.1.

###### Proposition 10.1

For users with $d(\theta)<\tau_{a}$,

  1. 1.

The fidelity uncertainty error $e_{u}^{\star}$ is zero when
$d(\theta)\leq\tau_{d}$, then increases, and finally decreases in $d(\theta)$.

  2. 2.

The fidelity bias error $e_{b}^{\star}$ increases when
$d(\theta)\leq\tau_{d}$, then decreases and finally increases in $d(\theta)$.

###### Proof 10.2

Proof of Proposition 10.1. For users with $d(\theta)<\tau_{a}$, by definition,
$e_{u}(\theta,\sigma_{q})=Var(\theta_{A}|\theta)$ and
$e_{b}(\theta,\sigma_{q}^{2})=[E(\theta_{A}|\theta)-\theta]^{2}$. As what we
did in the proof of Proposition 7.1 and Lemma 8.3, we can show
$e_{u}(\theta,\sigma_{q})=w(1-w)\sigma_{\theta}^{2}$ and
$e_{b}(\theta,\sigma_{q}^{2})=w^{2}(\mu_{\theta}-\theta)^{2}$, where
$w=\dfrac{\sigma_{q}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}$. Thus,
$e_{u}(\theta,\sigma_{q}^{\star}(\theta))=w^{\star}(\theta)(1-w^{\star}(\theta))\sigma_{\theta}^{2}$
and $e_{b}(\theta,\sigma_{q}^{\star}(\theta))=w^{\star
2}(\theta)(\mu_{\theta}-\theta)^{2}$, where
$w^{\star}(\theta)=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}$
given by Lemma 7.2.

Then,

  1. 1.

Fidelity uncertainty error:

We know that $w^{\star}(\theta)=1$ for
$|\mu_{\theta}-\theta|<\tau_{d}(\gamma,\sigma_{\theta})$, and
$w^{\star}(\theta)<1$ for
$|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma,\sigma_{\theta})$. Thus,
$e_{u}(\theta,\sigma_{q}^{\star}(\theta))=0$ for
$|\mu_{\theta}-\theta|<\tau_{d}(\gamma,\sigma_{\theta})$, and
$e_{u}(\theta,\sigma_{q}^{\star}(\theta))>0$ for
$|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma,\sigma_{\theta})$.

When $|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma,\sigma_{\theta})$,

$\displaystyle\dfrac{\partial
e_{u}(\theta,\sigma_{q}^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}$
$\displaystyle=\dfrac{\partial[w^{\star}(\theta)(1-w^{\star}(\theta))\sigma_{\theta}^{2}]}{\partial(\mu_{\theta}-\theta)^{2}}=\sigma_{\theta}^{2}(1-2w^{\star}(\theta))\dfrac{\partial
w^{\star}(\theta)}{\partial(\mu_{\theta}-\theta)^{2}}$

We know $\dfrac{\partial I^{\star}}{\partial(\mu_{\theta}-\theta)^{2}}\geq 0$
by Proposition 4.1 and $I^{\star}=-\dfrac{\gamma}{2}\ln w^{\star}(\theta)$ by
Lemma 8.5. These imply $\dfrac{\partial
w^{\star}(\theta)}{\partial(\mu_{\theta}-\theta)^{2}}\leq 0$. Thus, when
$|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma,\sigma_{\theta})$, the sign of
$\dfrac{\partial
e_{u}(\theta,\sigma_{q}^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}$
depends on $(1-2w^{\star}(\theta))$. If $(1-2w^{\star}(\theta))<0$ for small
$|\mu_{\theta}-\theta|$, then $e_{u}(\theta,\sigma_{q}^{\star}(\theta))$ first
increases and then decreases in $|\mu_{\theta}-\theta|$; if
$(1-2w^{\star}(\theta))\geq 0$ for any $|\mu_{\theta}-\theta|$, monotonically
decreases in $|\mu_{\theta}-\theta|$. (Notice that $(1-2w^{\star}(\theta))$ is
always positive for sufficiently large $|\mu_{\theta}-\theta|$, because
$w^{\star}(\theta)\to 0$ as $|\mu_{\theta}-\theta|\to\infty$.)

Hence, we either have $e_{u}(\theta,\sigma_{q}^{\star}(\theta))=0$ for
$|\mu_{\theta}-\theta|<\tau_{d}(\gamma,\sigma_{\theta})$, first increases and
then decreases in $|\mu_{\theta}-\theta|$ for
$|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma,\sigma_{\theta})$; or
$e_{u}(\theta,\sigma_{q}^{\star}(\theta))=0$ for
$|\mu_{\theta}-\theta|<\tau_{d}(\gamma,\sigma_{\theta})$, then there is a jump
at $|\mu_{\theta}-\theta|=\tau_{d}(\gamma,\sigma_{\theta})$
($e_{u}(\theta,\sigma_{q}^{\star}(\theta))$ jumps to a positive value), and
then $e_{u}(\theta,\sigma_{q}^{\star}(\theta))$ monotonically decreases in
$|\mu_{\theta}-\theta|$.

  2. 2.

Fidelity bias error:

We know that $w^{\star}(\theta)=1$ for
$|\mu_{\theta}-\theta|<\tau_{d}(\gamma,\sigma_{\theta})$, so
$e_{b}(\theta,\sigma_{q}^{\star}(\theta))=(\mu_{\theta}-\theta)^{2}$ for
$|\mu_{\theta}-\theta|<\tau_{d}(\gamma,\sigma_{\theta})$, which is increasing
in $|\mu_{\theta}-\theta|$.

At $|\mu_{\theta}-\theta|=\tau_{d}(\gamma,\sigma_{\theta})$, since
$w^{\star}(\theta)<1$ is optimal, we have
$e_{b}(\theta,\sigma_{q}^{\star}(\theta))=e(\theta,\sigma_{q}^{\star
2}(\theta))-e_{u}(\theta,\sigma_{q}^{\star}(\theta))<e(\theta,\sigma_{q}^{\star
2}(\theta))<(\mu_{\theta}-\theta)^{2}$. Thus,
$e_{b}(\theta,\sigma_{q}^{\star}(\theta))$ decreases at
$|\mu_{\theta}-\theta|=\tau_{d}(\gamma,\sigma_{\theta})$.

When $|\mu_{\theta}-\theta|>\tau_{d}(\gamma,\sigma_{\theta})$,

$\displaystyle\dfrac{\partial
e_{b}(\theta,\sigma_{q}^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}$
$\displaystyle=\dfrac{\partial[w^{\star
2}(\theta)(\mu_{\theta}-\theta)^{2}]}{\partial(\mu_{\theta}-\theta)^{2}}$
$\displaystyle=2(\mu_{\theta}-\theta)^{2}w^{\star}(\theta)\dfrac{\partial
w^{\star}(\theta)}{\partial(\mu_{\theta}-\theta)^{2}}+w^{\star 2}(\theta)$
$\displaystyle=w^{\star}(\theta)\left[2(\mu_{\theta}-\theta)^{2}\dfrac{\partial
w^{\star}(\theta)}{\partial(\mu_{\theta}-\theta)^{2}}+w^{\star}(\theta)\right]$
Substitute Equation 9 into the above equation
$\displaystyle=w^{\star}(\theta)\left[2(\mu_{\theta}-\theta)^{2}\cdot\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}+w^{\star}(\theta)\right]$
$\displaystyle\text{where}\
\Delta=\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$
With some simplifications
$\displaystyle=w^{\star}(\theta)\cdot\dfrac{\sigma_{\theta}^{2}[((\mu_{\theta}-\theta)^{2}+\sigma_{\theta}^{2})(\sqrt{\Delta}-\sigma_{\theta}^{2})-4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})]}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$

Let $d(\theta)=|\mu_{\theta}-\theta|$. Then,

$\dfrac{\partial e_{b}(\theta,\sigma_{q}^{\star}(\theta))}{\partial
d(\theta)^{2}}=w^{\star}(\theta)\cdot\dfrac{\sigma_{\theta}^{2}[(d(\theta)^{2}+\sigma_{\theta}^{2})(\sqrt{\Delta}-\sigma_{\theta}^{2})-4\gamma(d(\theta)^{2}-\sigma_{\theta}^{2})]}{4\sqrt{\Delta}(d(\theta)^{2}-\sigma_{\theta}^{2})^{2}}$

Now, we want to show that when $d(\theta)\geq\tau_{d}(\gamma,\sigma_{\theta})$
and $d(\theta)$ is finite, $\dfrac{\partial
e_{b}(\theta,\sigma_{q}^{\star}(\theta))}{\partial d(\theta)^{2}}$ has at most
one zero point with respect to $d(\theta)$. This is equivalent to showing

$\dfrac{(d(\theta)^{2}+\sigma_{\theta}^{2})(\sqrt{\Delta}-\sigma_{\theta}^{2})-4\gamma(d(\theta)^{2}-\sigma_{\theta}^{2})}{\sqrt{\Delta}(d(\theta)^{2}-\sigma_{\theta}^{2})^{2}}$

has at most one zero point with respect to $d(\theta)$. Let
$\widehat{d(\theta)}$ denote a solution of $d(\theta)$ such that

$\dfrac{(d(\theta)^{2}+\sigma_{\theta}^{2})(\sqrt{\Delta}-\sigma_{\theta}^{2})-4\gamma(d(\theta)^{2}-\sigma_{\theta}^{2})}{\sqrt{\Delta}(d(\theta)^{2}-\sigma_{\theta}^{2})^{2}}=0$

First, let the nominator
$(d(\theta)^{2}+\sigma_{\theta}^{2})(\sqrt{\Delta}-\sigma_{\theta}^{2})-4\gamma(d(\theta)^{2}-\sigma_{\theta}^{2})=0$,
we get:

$\displaystyle(d(\theta)^{2}+\sigma_{\theta}^{2})(\sqrt{\Delta}-\sigma_{\theta}^{2})=4\gamma(d(\theta)^{2}-\sigma_{\theta}^{2})$
$\displaystyle\implies$
$\displaystyle(d(\theta)^{2}+\sigma_{\theta}^{2})\sqrt{\Delta}=4\gamma(d(\theta)^{2}-\sigma_{\theta}^{2})+\sigma_{\theta}^{2}(d(\theta)^{2}+\sigma_{\theta}^{2})$
$\displaystyle\implies$
$\displaystyle\sqrt{\Delta}=\dfrac{4\gamma(d(\theta)^{2}-\sigma_{\theta}^{2})+\sigma_{\theta}^{2}(d(\theta)^{2}+\sigma_{\theta}^{2})}{d(\theta)^{2}+\sigma_{\theta}^{2}}$
$\displaystyle\implies$ $\displaystyle
1=\dfrac{4\gamma(d(\theta)^{2}-\sigma_{\theta}^{2})}{(d(\theta)^{2}+\sigma_{\theta}^{2})^{2}}+\dfrac{2\sigma_{\theta}^{2}}{d(\theta)^{2}+\sigma_{\theta}^{2}}$
$\displaystyle\implies$
$\displaystyle(d(\theta)^{2}+\sigma_{\theta}^{2})^{2}=4\gamma(d(\theta)^{2}-\sigma_{\theta}^{2})+2\sigma_{\theta}^{2}(d(\theta)^{2}+\sigma_{\theta}^{2})$
$\displaystyle\implies$
$\displaystyle(d(\theta)^{2}-\sigma_{\theta}^{2})(d(\theta)^{2}+\sigma_{\theta}^{2}-4\gamma)=0$

So the candidates of $\widehat{d(\theta)}$ are
$\widehat{d(\theta)}=\sigma_{\theta}$, and
$\widehat{d(\theta)}=\sqrt{4\gamma-\sigma_{\theta}^{2}}$ if
$4\gamma\geq\sigma_{\theta}^{2}$

Furthermore, by using L’Hôpital’s rule, one can get

$\lim_{\widehat{d(\theta)}\to\sigma_{\theta}}\dfrac{(d(\theta)^{2}+\sigma_{\theta}^{2})(\sqrt{\Delta}-\sigma_{\theta}^{2})-4\gamma(d(\theta)^{2}-\sigma_{\theta}^{2})}{\sqrt{\Delta}(d(\theta)^{2}-\sigma_{\theta}^{2})^{2}}=\dfrac{2\sigma_{\theta}^{2}\gamma-4\gamma^{2}}{\sigma_{\theta}^{6}}$

which is zero if and only if $\sigma_{\theta}^{2}=2\gamma$. This means when
$\sigma_{\theta}^{2}\neq 2\gamma$, there is no real $\widehat{d(\theta)}$ if
$4\gamma<\sigma_{\theta}^{2}$ or
$\widehat{d(\theta)}=\sqrt{4\gamma-\sigma_{\theta}^{2}}$ if
$4\gamma\geq\sigma_{\theta}^{2}$.

And when $\sigma_{\theta}^{2}=2\gamma$,
$\widehat{d(\theta)}=\sqrt{4\gamma-\sigma_{\theta}^{2}}=\sigma_{\theta}$, so
we also only have one solution for $\widehat{d(\theta)}$.

Thus, $\dfrac{\partial e_{b}(\theta,\sigma_{q}^{\star}(\theta))}{\partial
d(\theta)^{2}}$ has at most one zero point with respect to $d(\theta)$.

In addition, if $d(\theta)>\sigma_{\theta}$,
$(d(\theta)^{2}+\sigma_{\theta}^{2})(\sqrt{\Delta}-\sigma_{\theta}^{2})-4\gamma(d(\theta)^{2}-\sigma_{\theta}^{2})>(d(\theta)^{2}-\sigma_{\theta}^{2})(\sqrt{\Delta}-\sigma_{\theta}^{2})-4\gamma(d(\theta)^{2}-\sigma_{\theta}^{2})=(d(\theta)^{2}-\sigma_{\theta}^{2})(\sqrt{\Delta}-\sigma_{\theta}^{2}-4\gamma)$,
which is positive if
$d(\theta)>\max\\{\sigma_{\theta},\sqrt{|4\gamma-\sigma_{\theta}^{2}|}\\}$.
This means that for any
$d(\theta)>\max\\{\sigma_{\theta},\sqrt{|4\gamma-\sigma_{\theta}^{2}|}\\}$,
$\dfrac{\partial e_{b}(\theta,\sigma_{q}^{\star}(\theta))}{\partial
d(\theta)^{2}}$ is positive. Because we have shown $\dfrac{\partial
e_{b}(\theta,\sigma_{q}^{\star}(\theta))}{\partial d(\theta)^{2}}$ has at most
one zero point with respect to $d(\theta)$, the intermediate value theorem
implies that when $|\mu_{\theta}-\theta|>\tau_{d}(\gamma,\sigma_{\theta})$,
$\dfrac{\partial e_{b}(\theta,\sigma_{q}^{\star}(\theta))}{\partial
d(\theta)^{2}}$ is either always positive, or negative for small
$|\mu_{\theta}-\theta|$ and then positive for large $|\mu_{\theta}-\theta|$.

Hence, we either have $e_{b}(\theta,\sigma_{q}^{\star}(\theta))$ first
increases in $|\mu_{\theta}-\theta|$ for
$|\mu_{\theta}-\theta|<\tau_{d}(\gamma,\sigma_{\theta})$, then decreases and
finally increases in $|\mu_{\theta}-\theta|$ for
$|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma,\sigma_{\theta})$; or
$e_{b}(\theta,\sigma_{q}^{\star}(\theta))$ first increases in
$|\mu_{\theta}-\theta|$ for
$|\mu_{\theta}-\theta|<\tau_{d}(\gamma,\sigma_{\theta})$, then there is a jump
at $|\mu_{\theta}-\theta|=\tau_{d}(\gamma,\sigma_{\theta})$
($e_{b}(\theta,\sigma_{q}^{\star}(\theta))$ jumps to a smaller value), and
then $e_{b}(\theta,\sigma_{q}^{\star}(\theta))$ monotonically increases in
$|\mu_{\theta}-\theta|$.

##  11 More Detailed Version of Theorem 4.3.

In this section, we present a more detailed description of Theorem 4.3 :
[Theorem 4.3](Full version) When $\Gamma\rightarrow+\infty$, the variance of
the population output is lower than the variance of the population
preferences, $Var(\theta^{\star})<Var(\theta)$, and strictly decreases in the
cost of human-AI interactions $\gamma$. When $\Gamma<+\infty$,
$\lim_{\gamma\rightarrow 0}Var(\theta^{\star})=Var(\theta)$ and
$\lim_{\gamma\rightarrow+\infty}Var(\theta^{\star})<Var(\theta)$. In addition,
$Var(\theta^{\star})<Var(\theta)$ if $\gamma\geq\sigma_{q}^{2}/2$ or
$\Gamma\leq\hat{\Gamma}$ or $\Gamma\geq\tilde{\Gamma}$ for some
$\hat{\Gamma}>0,\tilde{\Gamma}>0$. The full proof is provided in Section 8.2.
The last sentence in this detailed version is the additional part compared
with the version that we presented in the main text. In particular, we show
that the population variance of the output is strictly less than the
population variance of the preferences if $\gamma$ is sufficiently large or
$\Gamma$ is outside an interval $(\hat{\Gamma},\tilde{\Gamma})$.

Figure 10: We use $\mu_{\theta}=0,\sigma_{\theta}=1,\gamma=0.1$. The black
dashed vertical lines are at $d(\theta)=\tau_{d}$. The orange curve is above
the blue curve for some $\theta$ with $d(\theta)>\tau_{d}$ but close to
$\tau_{d}$, showing that
$E[(\theta^{\star}-\mu_{\theta})^{2}|\theta]\geq(\theta-\mu_{\theta})^{2}$ for
these user $\theta$.

In Figure 10, we show why it is possible that the population variance of the
output can be larger than the population variance of the preferences when
$\gamma<\sigma_{q}^{2}/2$ and $\Gamma\in(\hat{\Gamma},\tilde{\Gamma})$. By the
tower property of conditional expectation, we know
$Var(\theta^{\star})=E[E[(\theta^{\star}-\mu_{\theta})^{2}|\theta]]$ (Notice
that $E[\theta^{\star}]=\mu_{\theta}$ is shown in the proof of Theorem 4.3).
So if $E[(\theta^{\star}-\mu_{\theta})^{2}|\theta]<(\theta-\mu_{\theta})^{2}$,
we must have $Var(\theta^{\star})<Var(\theta)$. However, it is possible that
$E[(\theta^{\star}-\mu_{\theta})^{2}|\theta]\geq(\theta-\mu_{\theta})^{2}$ for
some $\theta$ whose $d(\theta)$ are close to $\tau_{d}$. Since $\tau_{d}$ is
the root of a transcendental equation, it is complicated to find the closed
form of this region. Despite this possibility, it is actually hard to find a
scenario such that $Var(\theta^{\star})>Var(\theta)$ in our numerical tests.

Intuitively, the users with $d(\theta)>\tau_{d}$ but close to $\tau_{d}$ will
send a small amount of information. Since the information is always noisy, it
will also add more randomness and uncertainty to the outputs. So these users
have a higher $E[(\theta^{\star}-\mu_{\theta})^{2}|\theta]$ and can
“contribute” more to the population variance of outputs. However, there are
also many users who simply accept the default outputs (i.e.,
$d(\theta)<\tau_{d}$) and users with preferences that are far from the mean.
They have a lower $E[(\theta^{\star}-\mu_{\theta})^{2}|\theta]$, thereby
reducing the population variance of outputs. These two counter-forces interact
with each other, leading to a change in the variance.