We will use this later when discussing security with preexisting partial knowledge and additional gained knowledge in the message exchange.

\[lemma:sub-subset\] If the trace distance between $P_X$ and the uniform distribution is ${\varepsilon}$, then for any subsets ${\mathcal{X}}''\subseteq{\mathcal{X}}'\subseteq{\mathcal{X}}$, $$\label{eq:sub-subset}
  \Big|P_X({\mathcal{X}}''\,|\,{\mathcal{X}}') - \frac{|{\mathcal{X}}''|}{|{\mathcal{X}}'|}\Big| 
  \le \frac{|{\mathcal{X}}|}{|{\mathcal{X}}'|}{\varepsilon}.$$ Also, there are subsets which reach the bound.

The conditional probability can be written $$\label{eq:sss2}
\begin{aligned}
  P_X({\mathcal{X}}''\,|\,{\mathcal{X}}') &= \frac{P_X({\mathcal{X}}'')}{P_X({\mathcal{X}}')} 
  = \frac{P_X({\mathcal{X}}'')}{P_X({\mathcal{X}}'') + P_X({\mathcal{X}}'\setminus{\mathcal{X}}'')}
  = \left(1+\frac{P_X({\mathcal{X}}'\setminus{\mathcal{X}}'')}{P_X({\mathcal{X}}'')}\right)^{-1}.
\end{aligned}$$ To bound this from above, we need an upper bound for $P_X({\mathcal{X}}'')$ and a lower bound for $P_X({\mathcal{X}}'\setminus{\mathcal{X}}'')$, both of which can be obtained using Lemma \[lemma:subset\], $$\label{eq:sss3}
    P_X({\mathcal{X}}'')\le \frac{|{\mathcal{X}}''|}{|{\mathcal{X}}|} + {\varepsilon};\quad
  P_X({\mathcal{X}}'\setminus{\mathcal{X}}'') \ge \frac{|{\mathcal{X}}'\setminus{\mathcal{X}}''|}{|{\mathcal{X}}|}
  - {\varepsilon}.$$ These give us the upper bound $$\label{eq:sss5}
\begin{aligned}
  P_X({\mathcal{X}}''\,&|\,{\mathcal{X}}')
  = \left(1+\frac{P_X({\mathcal{X}}'\setminus{\mathcal{X}}'')}{P_X({\mathcal{X}}'')}\right)^{-1} 
  \le\left(1+\frac{\frac{|{\mathcal{X}}'\setminus{\mathcal{X}}''|}{|{\mathcal{X}}|} -
      {\varepsilon}}{\frac{|{\mathcal{X}}''|}{|{\mathcal{X}}|} + {\varepsilon}} \right)^{-1} =
  \frac{|{\mathcal{X}}''|}{|{\mathcal{X}}'|} + \frac{|{\mathcal{X}}|}{|{\mathcal{X}}'|}{\varepsilon}.
\end{aligned}$$ Similarly, from Lemma \[lemma:subset\] we also know that $$\label{eq:sss6} 
  P_X({\mathcal{X}}'')\ge\frac{|{\mathcal{X}}''|}{|{\mathcal{X}}|} - {\varepsilon};\quad
  P_X({\mathcal{X}}'\setminus{\mathcal{X}}'') \le \frac{|{\mathcal{X}}'\setminus{\mathcal{X}}''|}{|{\mathcal{X}}|} 
  + {\varepsilon}.$$ These give us the lower bound $$\label{eq:sss8}
\begin{aligned}
  P_X({\mathcal{X}}''\,&|\,{\mathcal{X}}') = \left(1+\frac{P_X({\mathcal{X}}'\setminus{\mathcal{X}}'')}
    {P_X({\mathcal{X}}'')}\right)^{-1} 
  \ge\left(1+\frac{\frac{|{\mathcal{X}}'\setminus{\mathcal{X}}''|}{|{\mathcal{X}}|} + {\varepsilon}}
    {\frac{|{\mathcal{X}}''|}{|{\mathcal{X}}|} - {\varepsilon}} \right)^{-1} =
  \frac{|{\mathcal{X}}''|}{|{\mathcal{X}}'|} - \frac{|{\mathcal{X}}|}{|{\mathcal{X}}'|}{\varepsilon}.
\end{aligned}$$ This proves the inequality. The bound can be reached in several ways, for example when $({\mathcal{X}}_+\cup{\mathcal{X}}_-)\subseteq{\mathcal{X}}'$ and ${\mathcal{X}}''={\mathcal{X}}_+$.$\square$

Using the above theorem, we can derive a bound for the trace 