Independent bidimensional random variables.

Question

Let $(X_i,Y_i)$ be independent bidimensional random variables, is the following equality true?: $$E\Big[X_1X_2\cdots X_n\mid Y_1,Y_2,\cdots,Y_n \Big]=\prod_1^n E\left[X_i\mid Y_i\right]$$

Answer 1

Yes. For each $k$, let $A_k \in \sigma(Y_k)$ and set $A := \bigcap_{k=1}^n A_k$. For convenience, define $X := X_1X_2\dots X_n$ and $Y := (Y_1,Y_2,\dots, Y_n)$. Then applying the definition of conditional independence and the mutual independence of $\{(X_i,Y_i)\}_{i=1}^n$, $$E\left[\mathbb{I}_{A}E[X|Y]\right] = E\left[\mathbb{I}_{A}X\right] = E\left[\prod_{k=1}^n \mathbb{I}_{A_k}X_k\right] = \prod_{k=1}^nE\left[ \mathbb{I}_{A_k}X_k\right] = \prod_{k=1}^nE\left[ \mathbb{I}_{A_k}E[X_k|Y_k]\right] = E\left[\prod_{k=1}^n\mathbb{I}_{A_k}E[X_k|Y_k]\right] = E\left[\mathbb{I}_{A}\prod_{k=1}^nE[X_k|Y_k]\right].$$

Because $\sigma(Y_1,\dots,Y_k)$ is generated by sets of the form $\bigcap_{k=1}^n A_k$ where each $A_k \in \sigma(Y_k)$, it follows that for all $B \in \sigma(Y)$,

$$E\left[\mathbb{I}_{B}E[X|Y]\right] = E\left[\mathbb{I}_{B}\prod_{k=1}^nE[X_k|Y_k]\right].$$

Thus,

$$E[X|Y] = \prod_{k=1}^nE[X_k|Y_k]\text{ almost surely.}$$