This is a part of an identity that I need to prove in my homework (that is not the identity itself).
$\mathcal{Y}=\{0,1\}$, $(X_{1},Y_{1}),(X_{2},Y_{2})$ are drawn independently from some distribution $\mathcal{D}$ over $\mathcal{X}\times\mathcal{Y}$
So I can assume that $(X_{1},Y_{1}),(X_{2},Y_{2})$ are independent, but I can't assume anything about $P(X_{1}),P(X_{2})$
My question is if the following equation is correct: $P(Y_{1}=y_{1},Y_{2}=y_{2} \mid X_{1}=x_{1},X_{2}=x_{2})=P(Y_{1}=y_{1} \mid X_{1}=x_{1})\cdot P(Y_{2}=y_{2} \mid X_{2}=x_{2})$
Is there proof for that equation? or else, is there a counterexample for this?
If $X_1$ and $X_2$ are also discrete,
\begin{align} &\mathsf{P}(Y_1=y_1,Y_2=y_2\mid X_1=x_1,X_2=x_2) \\[0.5em] &\qquad=\frac{\mathsf{P}(Y_1=y_1,Y_2=y_2,X_1=x_1,X_2=x_2)}{\mathsf{P}(X_1=x_1,X_2=x_2)} \\ &\qquad=\frac{\mathsf{P}(Y_1=y_1,X_1=x_1)\mathsf{P}(Y_2=y_2,X_2=x_2)}{\mathsf{P}(X_1=x_1)\mathsf{P}(X_2=x_2)} \\[0.5em] &\qquad=\mathsf{P}(Y_1=y_1\mid X_1=x_1)\mathsf{P}(Y_2=y_2\mid X_2=x_2). \end{align}