Checking Conditional Probability on Discrete Random Variable

Question

Let $Y: \left( \Omega, \Sigma \right) \rightarrow ( \mathbb{R}, \mathrm{Borel})$ be a discrete random variable. I would like to check whether \begin{align} P(A | Y) = \sum_i 1_{Y = y_i} P(A | Y = y_i) \end{align} holds. I.e. I want to show that for all $B$ that lie in the sigma algebra generated by $Y$, we have \begin{align} \mathbb{E} \left( P(A | Y ) 1_B \right) = P(A \cap B). \end{align} How would I go about that?

Answer 1

Taking the right side of your first equation as an alternative definition of the left, you have \begin{align} \mathbb{E}(P(A|Y)1_B)&=\mathbb{E}\left(\sum_i1_B1_{Y=y_i}P\big(A|Y=y_i\big)\right)\\ &=\sum_i\mathbb{E}\big(1_{B\cap\{Y=y_i\}}\big)P\big(A|Y=y_i\big)\\ &=\sum_iP\big(B\,\cap\{Y=y_i\}\big)\frac{P\big(A\cap\{Y=y_i\}\big)}{P\big(Y=y_i\big)}\\ &=\sum_{i:y_i\in Y(B)}P\big(A\,\cap\{Y=y_i\}\big)\\ &=P(A\cap B)\ . \end{align}