TITLE: Questions regarding non-degenerate random vector
QUESTION [0 upvotes]: For each of the statements below, it is possible to find a non-degenerate random vector (i.e with $Var (X)> 0$ and $Var (Y)> 0)$ satisfying the described conditions? If so, give an example. If not, explain why?

Y perfectly predictable given X, but a probability associated with X is not influenced by $Y$. In other words: $E[Y|X]= Y$ , but $P(X \in A | Y) = P (X \in A)$, for every event A. 
$Var(X) = Var (Y) = Var (Y | X = x)$, for all x, and $E (Y | X) = X$

What Ive been thinking so far: For the first one the answer is no, because The variance of $Var(Y)=0$. Y is a number.
The second one I still thinking.

REPLY [1 votes]: Since $P(X\in A|Y)=P(X\in A)$ for all $A$ it tells you that $X$ is independent of $Y$ (and therefore, $Y$ is also independent of $X$). Thus $E(Y|X)=E(Y)$. But you are given $E(Y|X)=Y$ which would imply $Y=E(Y)$ and as $E(Y)$ is a constant, this shows $Y$ is degenerate. Note that $X$ need not be degenerate. Once $Y$ is degenerate, $Y$ becomes independent of any random variable and the above conditions hold.
$Var(Y)=E(Var(Y|X))+Var(E(Y|X))\geq E(Var(Y|X))$. You are given that $Var(Y|X)=Var(Y)$ which is a constant so $E(Var(Y|X))=Var(Y)$. So equality holds in the above inequality. However equality can hold if and only if $Var(E(Y|X))=0$ which would imply $E(Y|X)$ is degenerate at its expected value i.e. $E(Y|X)=E(E(Y|X))=E(Y)$. But you also wrote $E(Y|X)=X$ which would imply $X=E(Y)$ which again shows $X$ is degenerate. Thus $Var(X)=0$ and thus $Var(Y)=0$ showing $Y$ is also degenerate.
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