3
$\begingroup$

I am struggling to prove a very intuitive property about iterated expectations.

We know that $$\mathbb E[X] = \mathbb E[\mathbb E[X\vert Y]].$$

I want to argue the following:

\begin{align*} \mathbb E[XY \vert Z ] = \mathbb E[X \mathbb E[Y \vert X,Z] \vert Z] \end{align*}

But I am having a hard time establishing this using the definition of conditional expectations.

To prove this would entail proving,

\begin{align*} \int_S XY d\mathbb P = \int_S X \mathbb E[Y \vert X,Z] d\mathbb P \end{align*} for all $S \in \sigma(Z)$.

This would be true if, for example, $Y = \mathbb E[Y \vert X,Z]$ which seems way demanding and is, most likely, not true.

Could anyone help please?

$\endgroup$

2 Answers 2

1
$\begingroup$

In its more general version, the tower property of conditional expectation tells you that if $\mathcal G,\mathcal H$ are two sub-sigma algebras of $\mathcal F$ with $\mathcal H \subseteq \mathcal G$, then for any random variable $\tilde X$ on $(\Omega,\mathcal F,\mathbb P)$, we have $$\mathbb E[\tilde X\mid\mathcal H] =\mathbb E \big[\mathbb E[\tilde X\mid\mathcal G]\mid\mathcal H\big]$$

Applying the above with $\tilde X := XY$, $\mathcal G :=\sigma(X,Z)$ and $\mathcal H := \sigma(Z)$ we immediately get $$\mathbb E[XY \mid Z ] = \mathbb E[ \mathbb E\big[XY \mid X,Z] \mid Z\big] $$ Which is almost the desired result. All that is left is to notice that since $X$ is $\sigma(X,Z)$-measurable, we have $\mathbb E\big[XY \mid X,Z]=X\mathbb E\big[Y \mid X,Z]$ and the desired result follows.

$\endgroup$
0
$\begingroup$

I will explain the general idea and let you fill in the small details.

For the conditional expectation, you have the property $\mathbb{E}[XY] =\mathbb{E}[X\cdot\mathbb{E}[Y|X]]$, also called the tower property.

The discrete case gives you a really good intuition about why it is true: if $X,Y$ are discrete random variables, then $\mathbb{E}[XY]= \sum_{k,n\in \mathbb{N}} x_ky_n\Pr[X=x_k,Y=y_n]$.By using conditional probability, we get $\mathbb{E}[XY]=\sum_{k\in \mathbb{N}} x_k\Pr[X=x_k]\sum_{n\in \mathbb{N}}y_n\Pr[Y=y_n|X=x_k]$. We write $\mathbb{E}[Y|X=x_k] = \sum_{n\in \mathbb{N}}y_n\Pr[Y=y_n|X=x_k]$. We can define $\mathbb{E}[Y|X]$ as the random variable that gets the value $\mathbb{E}[Y|X=x_k]$ whenever $X=x_k$. Hence, since $\mathbb{E}[XY] = \sum_{k\in \mathbb{N}} x_k \mathbb{E}[Y|X=x_k] \Pr[X=x_k]$, we can deduce $\mathbb{E}[XY] = \mathbb{E}[X \cdot \mathbb{E}[Y|X]]$.

The identity $\mathbb{E}[XY|Z] = \mathbb{E}[X\cdot \mathbb{E}[Y|X,Z] |Z]$ is really just the tower property in disguise - only now we work with the conditional expectation of $Z$ instead of an actual expectation.

We can apply $\mathbb{E}[XY] =\mathbb{E}[X\cdot\mathbb{E}[Y|X]]$ on the conditional expectation of $Z$, by writing $\mathbb{E}[XY|Z] = \mathbb{E}[X\cdot \mathbb{E}[\mathbb{E}[Y|Z]|X]|Z]$. Since $\mathbb{E}[\mathbb{E}[Y|Z]|X]] = \mathbb{E}[Y|X,Z]$, we are done.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .