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\begin{theorem}
\label{th:eoo}
If $\hat{Y}$ satisfies \eo for $Y$ and $S$ then the balanced accuracy of \aia in \ref{tm:hard} is $\frac{1}{2}$ iff $Y$ is independent of $S$ or $\hat{Y}$ is independent of $Y$.
\end{theorem}
\begin{proof}
Let $\attackmodel$ be the attack model trained for AS: $\hat{S}=a\circ \hat{Y}$.
By the total probability formula
{\footnotesize
\begin{align*}
P(\hat{S}=0|S=0)=&P(\hat{S}=0|S=0Y=0)P(Y=0|S=0)\\
+&P(\hat{S}=0|S=0Y=1)P(Y=1|S=0)
\end{align*}
}
and as well
{\footnotesize
\begin{align*}
P(\hat{S}=1|S=1)=&P(\hat{S}=1|S=1Y=0)P(Y=0|S=1)\\
+&P(\hat{S}=1|S=1Y=1)P(Y=1|S=1)
\end{align*}
}
Then we substitute those terms in the definition of the balanced accuracy of $\targetmodel$.
{\footnotesize
\begin{align*}
&\frac{P(\hat{S}=0|S=0)+P(\hat{S}=1|S=1)}{2}\\
=&\frac{1}{2}+\frac{1}{2}\left(P(Y=0|S=0)-P(Y=0|S=1)\right)\\
&\left(P(\hat{Y}\in \attackmodel^{-1}(\{1\})|S=1Y=0) -
P(\hat{Y}\in \attackmodel^{-1}(\{1\})|S=1Y=1)\right)
\end{align*}
}
The balanced accuracy is equal to 0.5 if and only if $P(Y=0|S=0)=P(Y=0|S=1)$
or $\forall \attackmodel~P(\hat{Y}\in \attackmodel^{-1}(\{1\})|S=1Y=0)=P(\hat{Y}\in \attackmodel^{-1}(\{1\})|S=1Y=1)$.
The first term indicates that $Y$ is independent of $S$ and the second term indicates that $S=1$ the $\targetmodel$ random guess utility.
We can do the same computing for $S=0$ and obtain a similar conclusion.
\end{proof}
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