Prove $E[Y|X] = 3 \to Var(XY) \geq Var(X)$. How to do this? My main concern is the conditional expectation. I am not quite sure what to do with it. Taking the expectation of both sides of the hypothesis suggests $E[Y] = 3$, but I don't think this is right since $E[Y|X]$ is a random variable...What do I do with it?
Anyway, I tried rewriting the conclusion as:
$E[(XY)^2-(3X)^2] = E[X^2(Y-3)(Y+3)] \geq [E(XY)-3E(X)][E(XY)+3E(X)]$, but I don't know if that helps.
I thought of some other stuff but don't really know if they are helpful.
$E[XY] = E[ \ E[XY|Y] \ ] = E[ \ Y \ E[X|Y] \ ] = E[3Y] = 3E[Y]$
$E[Y \ E[Y|X] \ |X] = 9$
Help please?
Assume that $E(Y\mid X)=c$.
Then $E(Y^2\mid X)\geqslant E(Y\mid X)^2=c^2$ hence $E((XY)^2)=E(X^2E(Y^2\mid X))\geqslant c^2E(X^2)$. Since $E(XY)=E(XE(Y\mid X))=cE(X)$, one gets $\mathrm{var}(XY)\geqslant c^2E(X^2)-(cE(X))^2=c^2\mathrm{var}(X)$.
Thus, $E(Y\mid X)=3$ implies $\mathrm{var}(XY)\geqslant9\mathrm{var}(X)$.