Suppose the $Y = a + bX +U$, where $X$ and $U$ are random variables and $a$ and $b$ are constants. Assume that $E[U|X] = 0$ and that $Var[U|X] = X^2$
I'm having trouble proving this:
c. Is $U$ independent of $X$? Why?
d. Show the $E[U] = 0$ and that $Var[U] = E[X^2] $
For part c, I got: $A$ is independent to $B$ iff $P(AB)=P(A)*P(B)$. $E(U|X)=0$, so $E(U and X)=0$, but I'm a little bit confused where to go from here.
For part d, I got proving Expected Value, but I'm totally stumped on the second part. Any help would be appreciated!
I'm not sure why did you bring up the first line $Y=aX+bY+U$ as it doesn't add anything for the subject, maybe you forgot to add something related to $Y$?
c. Your definition of independency is the definition for events, not for random variables. As for question, it may be if $X^2$ is constant almost surely, otherwise no, as the conditional variance of $U$ under the condition $X^2$ is constants if the $U$ and $X^2$ are independent (follows immediately from a shown below). For example for constant $X^2%$ consider $P(X=1)=P(X=-1)=1/2$ and let $U$ be of standard normal distribution independent of X. It satisfies both the conditions and $U$ is independent of $X$.
d. We have $E[U|X]=0$ then by applying Expected Value for the both sides of equation, we have $E(E[U|X])=E[U]=0$.
For the second part of d, we can try to show that:
a) $Var [U|X]=E[ U^2|X]- E^2[U|X]$
$$Var [U|X]=E[(U-E[U|X])^2|X]=E[U^2-2UE[U|X]+E[U|X]^2|X]=E[U^2|X]-2E[U|X]E[U|X]+E[U|X]^2=E[ U^2|X]-E^2[U|X]$$ b) $Var\ U = E[Var(U|X)]+Var(E(U|X)$ $$E[Var(U|X)]+Var(E(U|X)]=E[E[U^2|X]-E^2[E[U|X]]+E[E^2[U|X]-E^2[E[U|X]]=EU^2-E^2U$$ The second part of d follows from b) immediately as the right-hand side is known.