Compare $\mathbb{E}[XY]\mathbb{E}[XY]$ with $\mathbb{E}[X]\mathbb{E}[XY^2]$ $\newcommand{\E}{\mathbb{E}}$So this was a question asked to me in an interview where $X$ and $Y$ are two random variables and I was asked to compare the $\E[XY]\E[XY]$ with $\E[X]\E[XY^2]$ . The interviewer didn't give any more details. I am not exactly sure what he wanted.
 A: Something can be said if $X\ge 0$: By Cauchy-Schwarz,
$$
\left(\Bbb E[XY]\right)^2=\left(\Bbb E[\sqrt{X}\cdot\sqrt{X}Y]\right)^2\le\Bbb E[X]\cdot\Bbb E[XY^2].
$$
A: I don't think that much can be said about the relationship between the two quantities in general, other than what Michael Hardy has already said. As the following examples show, it's not true that one of the quantities is always larger than the other:
For instance, if $X$ is a random variable taking the values $\pm 1$, each with probability $\frac{1}{2}$, and $Y=X$, then $\mathbb{E}[XY]\mathbb{E}[XY]=1$ while $\mathbb{E}[X]\mathbb{E}[XY^2]=0$.
On the other hand, if $X$ and $Y$ are independent and $\mathbb{E}[Y]=0,\mathbb{E}[X]\neq0$, $Y\neq 0$, then $\mathbb{E}[XY]\mathbb{E}[XY]=0$ while $\mathbb{E}[X]\mathbb{E}[XY^2]=\mathbb{E}[X]^2\mathbb{E}[Y^2]>0$.
A: Here is another example. Notice first that
$$ (\Bbb{E}(XY))^2 - \Bbb{E}X \Bbb{E}(XY^2) = \operatorname{Cov}(X, Y)^2 - (\Bbb{E}X)^2 \operatorname{Var}(Y) - \Bbb{E}[(X - \Bbb{E}X)(Y - \Bbb{E}Y)^2].$$
Now let $(X, Y)$ be a multivariate Gaussian variable. Then the last term vanishes and we have
$$ (\Bbb{E}(XY))^2 - \Bbb{E}X \Bbb{E}(XY^2) = \operatorname{Cov}(X, Y)^2 - (\Bbb{E}X)^2 \operatorname{Var}(Y). \tag{*}$$
Notice that neither $\operatorname{Cov}(X, Y)$ nor $\operatorname{Var}(Y)$ depends on the $\Bbb{E}X$. Thus we are free to choose $\Bbb{E}X$ without changing other terms, and $\text{(*)}$ can attain any values $ \leq \operatorname{Cov}(X,Y)^2$. In particular, it can be both positive and negative.
A: Maybe he was looking for you to give an approximation and solve the problem. If $X$ and $Y$ are independent then the expectations become $E[X]^2 E[Y]^2$ and $E[X]^2 E[Y^2]$ comparing the two we see the latter is larger by Jensen's inequality.
A: Note that
$$\mathbb E(X)\mathbb E(XY^2)-\mathbb E(XY)^2=V(XY)-coV(X,XY^2) $$
If $X$ and $Y$ are independent 
$$\mathbb E(X)\mathbb E(XY^2)-\mathbb E(XY)^2= \mathbb E(X)^2V(Y) $$
