Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question about a proof of the tower property of conditional expectation that I found in Rosenthal's "A First Look at Rigorous Probability".

Proposition: Let $Y$ be a random variable with finite mean, and $\mathcal{G}_1 \subseteq \mathcal{G}_2$ be two sub-$\sigma$-algebras. The with probability $1$, \begin{equation} E(E(Y|\mathcal{G}_2)|\mathcal{G}_1) = E(Y|\mathcal{G}_1). \end{equation}

Are both of the following approaches sufficient to prove the statement?

(1) Suppose I define $X = E(Y|\mathcal{G}_2)$ and $Z = E(Y|\mathcal{G}_1)$. Then I must show under the given assumptions that \begin{equation} E(X|\mathcal{G}_1) = Z. \end{equation} So I must show that (i) $Z$ is a $\mathcal{G}_1$ measurable function and (ii) that \begin{equation} E(Z\mathbb{1}_G) = E(X\mathbb{1}_G) \qquad \forall G \in \mathcal{G}_1. \end{equation}

(2) But in the textbook Rosenthal shows that (i) $E(X|\mathcal{G}_1)$ is a $\mathcal{G}_1$ measurable function and (ii) that \begin{equation} E(E(X|\mathcal{G}_1)\mathbb{1}_G) = E(Y\mathbb{1}_G) \qquad \forall G \in \mathcal{G}_1. \end{equation}

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

Yes, both are sufficient. In (1) you think of computing $E[X \mid \mathcal{G}_1]$, and you show it is given by $E[Y \mid \mathcal{G}_1]$. In (2) you go the other way: you think of computing $E[Y \mid \mathcal{G}_1]$, and you show it is given by $E[X \mid \mathcal{G}_1]$. Either way you get the desired equality.

share|improve this answer
add comment

$E(Y\mathbb{1}_G)=E(X\mathbb{1}_G)$ by the definition of conditional expectation.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.