# Thread: E(X^2) exists =>E(E(X|Y) ^2) exists?

1. ## E(X^2) exists =>E(E(X|Y) ^2) exists?

True/False?

Let X,Y by random variables. If X2 has an expectation (not necessarily finite), then (E(X|Y))2 has an expectation.
In other words, if E(X2 ) exists, does E(E(X|Y) E(X|Y)) exist?

2. ## Re: E(X^2) exists =>E(E(X|Y) ^2) exists?

If X itself hasn't an expectation, the condition expectation is not well-defined.

3. ## Re: E(X^2) exists =>E(E(X|Y) ^2) exists?

But X does have an expectation in this case, since X^2 does.

4. ## Re: E(X^2) exists =>E(E(X|Y) ^2) exists?

I meant a finite expectation (of the absolute value for $\displaystyle$X). If the expectation of $\displaystyle X^2$ is finite the result holds using Jensen's inequality. So we have to deal with the case $\displaystyle E[X^2]=+\infty$.

5. ## Re: E(X^2) exists =>E(E(X|Y) ^2) exists?

Could you please indicate how Jensen's inequality can be used to prove the claim under the assumption that E(X^2) is finite?

6. ## Re: E(X^2) exists =>E(E(X|Y) ^2) exists?

By Jensen's inequality $\displaystyle (E(X\mid Y))^2\leq E(X^2\mid Y)$ so $\displaystyle E[(E(X\mid Y))^2]\leq E[ E(X^2\mid Y)]=E[X^2]$.