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Let $X_1,X_2,...,X_n$ and $Y$ be random variables. I know that:

$$\label{aaa}\tag{I} E\left[\sum_{j=1}^n X_j \Bigg | Y \right]=\sum_{j=1}^n E\left[ X_j \Big | Y \right] $$

Now, suppose that $Y$ takes values $1,2, 3,...$.

How to prove the following? $$E\left[\sum_{j=1}^Y X_j \Bigg | Y \right]=\sum_{j=1}^Y E\left[ X_j \Big | Y \right] $$

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Because $\sum_{j = 1}^Y E[X_j|Y]$ is $\sigma(Y)$-measurable, to show it is the conditional expectation as desired, it is sufficient to show for any $n \in \{1, 2, \ldots\}$, it holds that (this is because any $\sigma(Y)$-set can be written as the union of sets of the form $\{Y = n\}$). \begin{align} \int_{\{Y = n\}}\sum_{j = 1}^Y X_j dP = \int_{\{Y = n\}}\sum_{j = 1}^Y E[X_j|Y]dP. \end{align}

Indeed, \begin{align} LHS &= \int_\Omega \sum_{j = 1}^Y X_j I_{\{Y = n\}}dP = \int_\Omega \sum_{j = 1}^n X_j I_{\{Y = n\}}dP \\ &= \sum_{j = 1}^n\int_{\{Y = n\}}X_jdP \\ &= \sum_{j = 1}^n\int_{\{Y = n\}}E[X_j | Y]dP \\ &= \int_{\{Y = n\}}\sum_{j = 1}^n E[X_j|Y]dP = \int_{\{Y = n\}}\sum_{j = 1}^Y E[X_j|Y]dP = RHS. \end{align}

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  • $\begingroup$ Thanks. I only need understand the following $$\sum_{j = 1}^n\int_{\{Y = n\}}X_jdP = \sum_{j = 1}^n\int_{\{Y = n\}}E[X_j | Y]dP$$. $\endgroup$ Nov 21, 2022 at 5:39
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    $\begingroup$ I got it, thank you! $\endgroup$ Nov 21, 2022 at 5:43

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