TITLE: Inequality $(EX)(EY) \le E(XY)$ - is it true for non-negative random variables?
QUESTION [0 upvotes]: If $X,Y \ge 0$, is it true that $$(EX)(EY) \le E(XY)$$ 
or that 
$$(EX)(EY) \ge E(XY)$$
Believe me, I did my best to solve it. I can only show that none of these inequalities works for any $X,Y$ - however we have assumption on non-negativity...
$X$ and $Y$ are not independent (no information). When it comes to distribution no information also.

REPLY [7 votes]: Neither inequality is true in general, even for non-negative random variables.
For instance, if $X$ is uniformly distributed on $[0,1]$ and $Y=X$, then
$$ \mathbb{E}[XY]=\int_0^1x^2\;dx=\frac{1}{3}>\frac{1}{4}=\mathbb{E}[X]\mathbb{E}[Y] $$
which shows that the second inequality is false, while if we instead let $Y=1-X$ then
$$ \mathbb{E}[XY]=\int_0^1x(1-x)\;dx=\frac{1}{6}<\frac{1}{4}=\mathbb{E}[X]\mathbb{E}[Y] $$
which shows that the first inequality does not hold.
It's worth noting that $\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$ is the covariance of $X$ and $Y$, hence can be either positive or negative in general.
{"set_name": "stack_exchange", "score": 0, "question_id": 1903711}