True or false, prove or give a counterexample
If $Y = \alpha +\beta X+\epsilon$:
and $E[\epsilon\mid X] = 0$ then $E[X^2\epsilon] = 0 $
and $E[X\epsilon] = 0$ then $E[X^2\epsilon] = 0$
and $E[\epsilon \mid X] = 0$ then $\epsilon \perp\!\!\!\perp X$
I have some problems regarding problem 3. Could someone verify?
1: True
Since $E[X^2\epsilon] = E(E[X^2\epsilon\mid X]) = E(X^2 E[\epsilon\mid X]) = 0$
2: False
Counterexample:
Let $X\stackrel{d}{=} N(0,\sigma^2)$ and $\epsilon$ be a constant then $E[X\epsilon] = 0$ but $E[X^2] = 1$ and $E[X^2\epsilon] = \epsilon$.
3: False
Would the following meet the problem statement?
Let $\epsilon\mid X \stackrel{d}{=} N(0, X^2)$, then $E[\epsilon\mid X] = 0$ but they are not independent.