$\mathbb{E}[\mathbb{1}_{\{Y\in B\}}|X]=Q(X,B)$ where $Q(x,B):=\sum\limits_{y\in B}\mathbb{P}(Y=y\mid X=x)$?

Question

Let $X:\Omega\to S_1$ and $Y:\Omega\to S_2$ be two discrete random variables with $P(X=x)>0\;\forall x\in S_1$ and $P(Y=y)>0\;\forall y\in S_2$. I would like to compute the conditional expectation $\mathbb{E}[\mathbb{1}_{\{Y\in B\}}\mid X]$ and in particular show that $\mathbb{E}[\mathbb{1}_{\{Y\in B\}}\mid X]=Q(X,B)$ where $Q(x,B):=\sum\limits_{y\in B}\mathbb{P}(Y=y\mid X=x)$. By the average property of conditional expectation I have to show that $$\mathbb{E}[\mathbb{1}_{\{Y\in B\}}(X)h(X)]=\mathbb{E}[Q(X,B)h(X)]$$ for every $h$ bounded and $\sigma(X)$-measurable. But I have some troubles in expressing $\mathbb{E}[\mathbb{1}_{\{Y\in B\}}(X)h(X)]$. I think $\mathbb{E}[\mathbb{1}_{\{Y\in B\}}(X)h(X)]=\int\mathbb{1}_{y\in B}(x)h(x)d\mu_{X,Y}(x,y)$ but then I am not sure what the joint law $\mu_{X,Y}$ actually is.

Answer 1

Let $h$ be bounded and measurable. Then

$$ \begin{align*} \mathbb E[1_{\{Y\in B\}}h(X)] &=\mathbb E\left[\sum_{x\in S_1}1_{\{Y\in B\}}h(x)1_{\{X=x\}}\right]\\ &=\sum_{x\in S_1}h(x)\mathbb E[1_{\{Y\in B,X=x\}}]\\ &=\sum_{x\in S_1}h(x)\mathbb P(Y\in B,X=x)\\ &=\sum_{x\in S_1}h(x)\mathbb P(X=x)\mathbb P(Y\in B\mid X=x)\\ &=\sum_{x\in S_1}h(x)\mathbb P(X=x)Q(x,B)\\ &=\mathbb E[Q(X,B)h(X)] \end{align*} $$

PS : no need that $Y$ be discrete.