# Conditional Expectation

• Dec 7th 2007, 07:27 PM
matheye
Conditional Expectation
Hi there, could you help me out with the following question,

X~N(0,s1), Y~N(0,s2), cov(X,Y)=r, What is the law of E[X-Y|2X+Y]?

Many thanks
• Dec 8th 2007, 10:35 AM
Constatine11
Quote:

Originally Posted by matheye
Hi there, could you help me out with the following question,

X~N(0,s1), Y~N(0,s2), cov(X,Y)=r, What is the law of E[X-Y|2X+Y]?

Many thanks

I may be misinterpreting your notation, but:

Put $Z=2X+Y$ then $Y=2X-Z$, and $X-Y=-X+Z$, so

$
E(X-Y|[2X+Y=Z]) = \int (-x+Z) \frac{1}{\sqrt{2\pi}\sigma_1}e^{\frac{-x^2}{2\sigma_1}} dx=-\bar{X}+Z
$

ZB
• Dec 8th 2007, 04:26 PM
matheye
Thank you very much, ZB.

However, in your notation, Z itself is still a normal RV. We do not know the law (distribution) of Z-X|Z, although both are normal, so we are not able to evaluate the integral.

I was thinking about working out cov(X-Y,2X+Y) first. Under some value of r (correlation), we can say X-Y and 2X+Y are not correlated and so indepedent. Then X-Y and 2X+Y are joint normal. So we can work out E[E(X-Y|2X+Y)] and Var[E(X-Y|2X+Y)]. But the problem is this only works out a solution for certain value of corrletation r, not for a universal case.

Could anybody offer some insight?

Thanks,

MC
• Dec 9th 2007, 08:22 PM
matheye
Was struggling to show X-Y and 2X+Y are jointly Gaussian. Now notice the statement that if an arbitary linear combination of two Gaussian RVs is Gaussian then the joint of two is Gaussian. As we are able to show that any linear combination of X-Y and 2X+Y is Gaussian, the problem can be solved now.

Any critics on this?

Thanks,

MC