conditional expectation

[Petrarca]

Assume you have X and Y be independent, both with uniform distribution on [0,1].

What is the conditional expectation of W=(X+Y)^2 , given X

I have the following approach: Let E[W|X=x], but for calculating this expectation I normally need the joint density function??

 

[Jacek Podlewski]

In general case - yes, you do, but here you can omit the use of joint density by clever use of conditional expectation's properties:
E[(X+Y)^2|X]=E[X^2+2XY+Y^2|X]=E[X^2|X]+2E[XY|X]+E[Y^2|X]
taking out the X-measurable terms give
X^2+2XE[Y|X]+E[Y^2|X]
and since Yand X are independent the last expression is equal to
X^2+2XE[Y]+E[Y^2]
and it suffices to compute the moments of Y.

 

[Ito]

Can you explain why you need a joint density function ? Since this is a conditional expectation, the product of XY given X=x would treat X as a constant. You should expand your expression. Use linearity of expectation and compute individual expectations.

 

[Petrarca]

Thank you very much for your help. So for the last step I only need to compute two integrals, right?

E[Y]=the integral from 0 to 1 from x and E[Y^2]=integral from 0 to 1 from x^2

 

[MiloRambaldi]

Can you explain why you need a joint density function ? Since this is a conditional expectation, the product of XY given X=x would treat X as a constant. You should expand your expression. Use linearity of expectation and compute individual expectations.

As Jacek Podlewski showed, under the assumptions on X and Y,
((*)\text{ \quad} E[(X+Y)^2|X=x]=x^2 + 2x E[Y] + E[Y^2]=x^2 + x + \frac13).

Here are some more exercises to further your understanding:

(1) Let X and Y be as before, but without the assumption that they are independent.
Write E[(X+Y)^2|X=x] in terms of the joint density function f(x,y).

(2) Find X, Y ~ U[0,1] such that X and Y are uncorrelated, but (*) fails to hold.

(3) Suppose that (X,Y) is a normal copula with X and Y uncorrelated. Must (*) be true?