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Proving conditional expectation

Usagi

Member
Jun 26, 2012
46
Hi guys, assume we have an equality involving 2 random variables U and X such that [tex]E(U|X) = E(U)=0[/tex], now I was told that this assumption implies that [tex]E(U^2|X) = E(U^2)[/tex]. However I'm not sure on how to prove this, if anyone could show me that'd be great!
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Hi guys, assume we have an equality involving 2 random variables U and X such that [tex]E(U|X) = E(U)=0[/tex], now I was told that this assumption implies that [tex]E(U^2|X) = E(U^2)[/tex]. However I'm not sure on how to prove this, if anyone could show me that'd be great!
Not sure this is true. Suppose \(U|(X=x) \sim N(0,x^2)\), and \(X\) has whatever distribution we like.

Then \(E(U|X=x)=0\) and \( \displaystyle E(U)=\int \int u f_{U|X=x}(u) f_X(x)\;dudx =\int E(U|X=x) f_X(x) \; dx=0\).

Now \(E(U^2|X=x)={\text{Var}}(U|X=x)=x^2\). While \( \displaystyle E(U^2)=\int E(U^2|X=x) f_X(x) \; dx= \int x^2 f_X(x) \; dx\).

Or have I misunderstood something?

CB
 

Usagi

Member
Jun 26, 2012
46
Hi CB,

Actually the problem arose from the following passage regarding the homoskedasticity assumption for simple linear regression:


I do not understand how they came to the conclusion that [tex]\sigma^2 = E(u^2|x) \implies \sigma^2 = E(u^2)[/tex]

Thanks for your help!
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Hi CB,

Actually the problem arose from the following passage regarding the homoskedasticity assumption for simple linear regression:


I do not understand how they came to the conclusion that [tex]\sigma^2 = E(u^2|x) \implies \sigma^2 = E(u^2)[/tex]

Thanks for your help!
It is the assumed homoskedasticity (that is what it means).

CB