Proving conditional expectation

[Usagi]

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

 

[CaptainBlack]

Hi guys, assume we have an equality involving 2 random variables U and X such that $$E(U|X) = E(U)=0$$, now I was told that this assumption implies that $$E(U^2|X) = E(U^2)$$. 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]

Hi CB,

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

http://img444.imageshack.us/img444/6892/asdfsdfc.jpg
I do not understand how they came to the conclusion that $$\sigma^2 = E(u^2|x) \implies \sigma^2 = E(u^2)$$

Thanks for your help!

 

[CaptainBlack]

Hi CB,

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

http://img444.imageshack.us/img444/6892/asdfsdfc.jpg
I do not understand how they came to the conclusion that $$\sigma^2 = E(u^2|x) \implies \sigma^2 = E(u^2)$$

Thanks for your help!

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

CB

 

[CellCoree]

I can provide a proof for the equality E(U^2|X) = E(U^2) given the assumption E(U|X) = E(U)=0.

First, we know that the conditional expectation of a random variable U given another random variable X is defined as E(U|X) = ∫ U(x) f(x) dx, where f(x) is the conditional probability density function of X.

Using this definition, we can rewrite E(U^2|X) as ∫ U^2(x) f(x) dx. Similarly, E(U^2) can be written as ∫ U^2(x) g(x) dx, where g(x) is the probability density function of U.

Now, since we are given the assumption E(U|X) = E(U)=0, this implies that the expected value of U is equal to 0 for all values of X. In other words, the probability density function of U, g(x), is a constant function with a value of 0.

Substituting this into our equation for E(U^2), we get ∫ U^2(x) (0) dx = 0, which means that E(U^2) = 0.

Using the same logic, we can see that for E(U^2|X), the integral will also evaluate to 0 since U(x) is multiplied by the same constant function of 0. Therefore, E(U^2|X) = 0.

Therefore, we have proven that E(U^2|X) = E(U^2) given the assumption E(U|X) = E(U)=0.