The question was (now there is an answer):
Suppose that $X \sim N(a_1, \sigma_1^2)$ and $Y \sim N(a_2, \sigma_2^2)$ are normal random variables, possibly dependent. Is $$Z = E( X | Y)$$ a normal random variable?
What do I know? I know, that $Z$ is normal in case, when $(X,Y)$ is a normal vector (e.g., if $X$ and $Y$ are independent). I think than in general case $Z$ is not normal. A counterexample, if it exists, may be found only in case when $X, Y$ are normal, but $(X,Y)$ is not normal. I know two examples, when $X, Y$ are normal, but $(X,Y)$ is not normal. Both of them are useless, but I write it here:
- $Y \sim N(0,1)$, $X = Y$ if $|Y| \le c$ and $X = -Y$ otherwise for some constant $c$. Then $Z = E( X | Y) = X$ because $Y$ is a function of $X$. Hence $Z$ is normal. Unfortunately.
- $Y \sim N(0,1)$, $\xi \sim Bern(\frac12)$, $\xi$ and $Y$ are independent, $X = (2\xi-1) Y$. Then $Z = E( X | Y) = E((2\xi-1) Y|Y) = Y E((2\xi-1)|Y) = Y\cdot 0 = 0 \sim N(0,0).$ Unfortunately. So we need more examples. The answer is below.
Suppose that $p(x)$ - density of $N(0,1)$. Consider the vector $(X,Y)$ with density $2p(x)p(y)I_{x\dot y > 0}$. This example is taken from stats.stackexchange.com/questions/30159/is-it-possible-to-have-a-pair-of-gaussian-random-variables-for-which-the-joint-d
It's easy to see that $X \sim N(0,1)$ and $Y \sim N(0,1)$. As $p(x|y) = 2p(x)I_{x\dot y > 0}$ we have $E(X|Y=y) = c I_{y>0} + (-c) I_{y<0}$ where $c = \int_{0}^{\infty} p(x) dx = E|N(0,1)| > 0$. Hence $Z = E(X|Y) = c \cdot \rm{sgn}(Y)$. $Z$ is a discrete r.v. Q.e.d.