# Conditional expectation of product of two random variables

I'm quite new to statistics and I'm struggling with this problem.

Two random variables $$X$$ ~ Normal$$(1, 2)$$ and conditioning on $$X = x$$, $$Y$$ ~ Normal$$(x + 2, 3)$$. I want to find $$\mathbb{E}[XY]$$. This is what I've tried:

\begin{align}\mathbb{E}[XY] &= \mathbb{E}[\mathbb{E}[XY|X]]\\&=\mathbb{E}[X\mathbb{E}[Y|X]]\end{align}

I have two questions:

1. Is moving $$X$$ from inside the inner expectation out to the outer expectation considered valid? And why is that?
2. Suppose the answer to 1 is yes, how would I proceed from here? I'm thinking of replacing $$\mathbb{E}[Y|X]$$ with $$X + 2$$. Am I on the right track?

Yes, moving $$X$$ out of the conditional expectation is right. This is a general property of conditional expectations: $$E(XY|\mathbb G)=XE(Y|\mathbb G)$$ if $$X$$ is measurable w.r.t. $$\mathbb G$$.

By hypothesis the mean of $$Y$$ given $$X$$ is $$X+2$$. Hence $$EXY=E(X(X+2))=EX^{2}+2EX =5$$.

• Thank you very much. Does that property have a name? Where can I read more about it? Sep 23, 2019 at 9:14
• It doesn't have a name but many books on Probability Theory prove it. You can look at Billingsley, Breiman, Chung etc. Sep 23, 2019 at 9:16
• @KaboMurphy I guess the answer will be $E[XY]=E[X^2 + 2X]=1^2+2+2\times1=5$.
– MUB
Nov 12, 2019 at 17:22
• @MUB Thank you for the correction. Nov 12, 2019 at 23:12

1) Yes, it is valid.

Let us work on probability space $$(\Omega,\mathcal A,P)$$ and let $$\mathcal V$$ denote a sub $$\sigma$$-algebra of $$\mathcal A$$.

By definition we have for suitable random variable $$Y$$ defined on that space: $$\mathbb E[\mathbf1_AY]=\mathbb E[\mathbf1_A\mathbb E[Y\mid \mathcal V]]\text{ for every }A\in\mathcal V\tag1$$

Now in the special case $$\mathcal V=\sigma(X)$$ where $$X$$ denotes a random variable on the space we have: $$\mathcal V=\{\{X\in B\}\mid B\in\mathcal B\}$$ where $$\mathcal B$$ denotes the Borel $$\sigma$$-algebra on $$\mathbb R$$.

That means that $$(1)$$ can be rewritten as:$$\mathbb E[\mathbf1_B(X)Y]=\mathbb E[\mathbf1_B(X)\mathbb E[Y\mid X]]\text{ for every }B\in\mathcal B\tag2$$

We recognize this as a statement about indicator functions $$1_B:\mathbb R\to\mathbb R$$ but fortunately it can be proved that this goes further. It can be proved (give it a try) that it works for any suitable Borel-measurable function $$f:\mathbb R\to\mathbb R$$ leading to:$$\mathbb E[f(X)Y]=\mathbb E[f(X)\mathbb E[Y\mid X]]\tag3$$

So put $$(3)$$ in your "probability-backpack".

A special case is then the application of this on the identity function, leading to:$$\mathbb E[XY]=\mathbb E[X\mathbb E[Y\mid X]]$$and this is applied in your question.

2) Yes, you are on the right track.