Skip to main content
Cross Validated

Return to Answer

added 304 characters in body
Source Link
Zhanxiong
Zhanxiong
  • 21.2k
  • 2
  • 46
  • 88

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}^n X_j I_{\{Y = n\}}dP \\ &= \sum_{j = 1}^n\int_{\{Y = n\}}X_jdP \end{align}\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}

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 \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}^n X_j I_{\{Y = n\}}dP \\ &= \sum_{j = 1}^n\int_{\{Y = n\}}X_jdP \end{align}

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}

deleted 5 characters in body
Source Link
User1865345
User1865345
  • 10.3k
  • 12
  • 23
  • 40

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 \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}^n X_j I_{{Y = n}}dP \ &= \sum_{j = 1}^n\int_{{Y = n}}X_jdP \

\end{align}\begin{align} LHS &= \int_\Omega \sum_{j = 1}^n X_j I_{\{Y = n\}}dP \\ &= \sum_{j = 1}^n\int_{\{Y = n\}}X_jdP \end{align}

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 \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}^n X_j I_{{Y = n}}dP \ &= \sum_{j = 1}^n\int_{{Y = n}}X_jdP \

\end{align}

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 \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}^n X_j I_{\{Y = n\}}dP \\ &= \sum_{j = 1}^n\int_{\{Y = n\}}X_jdP \end{align}

Source Link
Zhanxiong
Zhanxiong
  • 21.2k
  • 2
  • 46
  • 88

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 \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}^n X_j I_{{Y = n}}dP \ &= \sum_{j = 1}^n\int_{{Y = n}}X_jdP \

\end{align}