Conditional expectation $E(X\mid XY)$ when $(X,Y)$ is standard normal

Question

I have to calculate $\mathbb{E}(X|X*Y)$ with X,Y being independent and standard normal distributed. I got at tip in this post (Conditional expectation on components of gaussian vector), that I should use the definition and Bayes Thm to solve the problem. I played around a bit, but I just don't get it :(

May anyone give me another hint?

Have you made use of the information that $X$ and $Y$ are independent? — Mico, Oct 06 '13 at 12:57

Did · Accepted Answer · 2015-11-09T14:41:59.610

11

As every conditional expectation, $E[X\mid XY]=w(XY)$ for some measurable function $w$. Recall that:

Conditional expectations depend only on the joint distribution of the random variables considered, in the sense that, if $E[X\mid XY]=w(XY)$, then $E[X'\mid X'Y']=w(X'Y')$ for every $(X',Y')$ distributed like $(X,Y)$.

Choosing $(X',Y')=(-X,-Y)$ above, one gets $X'Y'=XY$ hence $$w(XY)=E[-X\mid XY]=-E[X\mid XY]=-w(XY).$$ Thus, $$E[X\mid XY]=0. $$ One sees that $E[X\mid XY]=0$ for every centered gaussian vector $(X,Y)$, neither necessarily independent nor standard.

Still more generally:

Let $\Xi$ denote any centered gaussian vector, $u$ an odd measurable function such that $u(\Xi)$ is integrable and $v$ an even measurable function. Then, $$E[u(\Xi)\mid v(\Xi)]=0.$$

edited Nov 09 '15 at 14:41

answered Oct 06 '13 at 12:38

Did

284,245

Thanks a lot Did, that clearly answeres my question! Although this technique may not apply to a problem like e.g. $\mathbb{E}(Z|XY)$ if $(X,Y,Z)$ is gaussian. Btw, what is an "even" measurable function? – Mr. Barrrington Oct 08 '13 at 14:00
If (X,Y,Z) is centered, the argument works. "Even": u(-x)=u(x). "Odd": u(-x)=-u(x). – Did Oct 08 '13 at 17:27

score 0 · Answer 2 · edited Apr 13 '17 at 12:20

0

You know that $$ \mathbb{E}\left(X\mid XY=\alpha\right) = \int_{\mathbb{R}}\beta f_{X\mid XY}\left(\beta\mid\alpha\right)d\alpha $$ Now, using Bayes theorem $$ f_{X\mid XY}\left(\beta\mid\alpha\right) = \frac{f_{XY\mid X}(\alpha\mid\beta)f_X(\beta)}{f_{XY}(\alpha)}. $$

THIS FOLLOWING IS NOT TRUE

But, $$ f_{XY\mid X}(\alpha\mid\beta) = f_Y\left(\frac{\alpha}{\beta}\right). $$ and you can find the distribution of the product $XY$ here.

All you need to do now is to substitute the last results in $f_{X\mid XY}$, and calculate the integral. I'm not sure whether there is a simple closed form result to this integral.

edited Apr 13 '17 at 12:20

Community

1

answered Oct 06 '13 at 11:22

user91011

1,138

1

"But, fXY∣X(α∣β)=fY(αβ)". Sorry but this is not so. – Did Oct 06 '13 at 12:27
and indeed I think that the distribution of XY is not given in the linked topic. I don't even think it is easy to calculate, right? – Mr. Barrrington Oct 08 '13 at 14:02

score -1 · Answer 3 · answered Oct 06 '13 at 11:18

you need to compute the integral : $$ \int_{-\infty}^{\infty} (x f_{x|xy}) dx $$ where $f_{x|xy}$ is the conditional pdf. To find this pdf, $$f_{x|xy} = \frac{f_{x,xy}}{f_{xy}}$$ Try to do a transformation with $U=XY$ and $V=.....$ that can help you to compute the expectation

Conditional expectation $E(X\mid XY)$ when $(X,Y)$ is standard normal

3 Answers3

Linked

Related