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- Thread starter Usagi
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- Thread starter
- #1

- Jan 26, 2012

- 890

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

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- #3

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!

- Jan 26, 2012

- 890

It is the assumed homoskedasticity (that is what it means).

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!

CB