The input $X \in \{0, 1\}$ and label $T \in \{0,1\}$ are binary random variables, and the set of predictors that we consider are the functions $y : \{0, 1\} \rightarrow \{0, 1\}$. Recall the $0$-$1$ loss when predicting $t$ with $y(x)$,
\begin{equation*}
L_{0-1}(y(x), t) = \begin{cases}
0 & \text{if}~~ y(x) = t, \\
1 & \text{if}~~ y(x) \neq t \\
\end{cases}
\end{equation*}
and recall the definition of the expected error,
\begin{equation*}
\mathcal{R}\big[y\big] = E\big[L_{0-1}(y(X), T)\big] = \sum_{t\in\{0,1\}}\sum_{x\in\{0,1\}}L_{0-1}(y(x), t) \cdot P(X=x,T=t)
\end{equation*}
Assume that $P(X = x) > 0$ for all $x\in\{0,1\}$.
Assuming that $P(T=0|X=0) \neq P(T=1|X=0)$ and $P(T=0|X=1) \neq P(T=1|X=1)$, prove that
\begin{equation*}
y^*(x) = \text{arg}~~~\min_{t\in\{0,1\}}~~~P(T=t|X=x)
\end{equation*}
is the unique optimal predictor.
(In other words, prove that $\mathcal{R}\big[y^*\big] \leq \mathcal{R}\big[y\big]$ with equality only if $y^*(x)=y(x)$ for all $x\in\{0,1\}$.
$\textbf{My Attempt:}$
First, I find out the 4 possible predictors for this question are $y_1(x)=x$, $y_2(x)=1-x$, $y_3(x)=0$ and $y_4(x)=1$
Since, by Bayes Law on conditional probability, I have $P(X=x,T=t) = P(T=t|X=x) \cdot P(X=x)$.
Then, I can find out the expected errors for the 4 possible predictors, where
$\mathcal{R}[y_1] = P(T=1|X=0) \cdot P(X=0) + P(T=0|X=1) \cdot P(X=1)$,
$\mathcal{R}[y_2] = P(T=0|X=0) \cdot P(X=0) + P(T=1|X=1) \cdot P(X=1)$,
$\mathcal{R}[y_3] = P(T=1|X=0) \cdot P(X=0) + P(T=1|X=1) \cdot P(X=1)$,
$\mathcal{R}[y_4] = P(T=0|X=0) \cdot P(X=0) + P(T=0|X=1) \cdot P(X=1)$.
But then I am stuck in finding out $\mathcal{R}[y^*]$ and I don't know how to prove next.
$\textbf{So what should I do to complete the prove ?}$