Does $|\mathsf{E}(xy)|\leq |\mathsf{E}(x)| \cdot |\mathsf{E}(y)|$ hold?

Question

Does $|\mathsf{E}(xy)|\leq |\mathsf{E}(x)| \cdot |\mathsf{E}(y)|$ hold for any random variable $x$, $y$? Here $\mathsf{E}$ is the expectation.

Answer 1

No, $|\mathsf{E}(xy)|\leq |\mathsf{E}(x)| \cdot |\mathsf{E}(y)|$ does not hold for any random variable $x$, $y$.

Assume that $x$, $y$ are independent r.v., it holds that $|\mathsf{E}(xy)| = |\mathsf{E}(x)\cdot \mathsf{E}(y)| \leq |\mathsf{E}(x)| \cdot |\mathsf{E}(y)|$.

However, one counter example for the dependent random variables $x,y$ can be given as follows

• $x$: $\mathsf P\{x=-1\} = \frac12$, $\mathsf P\{x=+1\} = \frac12$;

• $y$: $y = \begin{cases}+1 &\text{if}\ x>0 \\ -1 &\text{if}\ x<0 \end{cases}$

Then \begin{align*} \mathsf E (xy) &=(+1) \cdot \mathsf P (xy=+1) + (-1)\cdot \mathsf P (xy=-1)\\ & = (+1) \cdot \left(\mathsf P (x=-1, y=-1) + \mathsf P (x=+1, y=+1) \right)\\ & = (+1) \cdot \left(\mathsf P (x=-1) \mathsf P(y=-1| x=-1) + \mathsf P (x=+1)\cdot \mathsf P(y=+1| x=+1) \right)\\ & = (+1) \cdot \left(\frac{1}{2}\cdot 1 + \frac{1}{2}\cdot 1\right)\\ & = +1 \end{align*} where it is impossible for $xy=-1$ (i.e., $\mathsf P(xy=-1)= 0$).

$\mathsf E(x) = 1\cdot 1/2 -1 \cdot 1/2 = 0$. Thus, $1 = |\mathsf{E}(xy)| > |\mathsf{E}(x)| \cdot |\mathsf{E}(y)| = 0$.

Answer 2

simple counterexample here:
if $$E(X) = E(Y) = 0$$ then $$Cov(X, Y) = E[(X - E(X))(Y - E(Y))] = E(XY)$$ if $$|E(XY)| \le |E(X)||E(Y)|$$ holds, which means $Cov(X, Y) = 0$
but $Cov(X, Y) > 0$ is possible if $E(X) = E(Y) = 0$