Given Random variables $X$ and $Y$ is it true always that;
$$E(XY)^2 \le E(X^2)E(Y^2)$$
Is it easy to prove?
$\endgroup$ 63 Answers
$\begingroup$I'm going to assume you mean$$(E(XY))^2 \le E(X^2)E(Y^2).$$One way to prove this is to realize it's a special case of the Cauchy–Schwarz inequality.
Here's another. Let\begin{align} f(t) = {} & E((tX+Y)^2) \\[8pt] = {} & (E(X^2)) t^2 + 2(E(XY))t + E(Y^2) \\[8pt] = {} & at^2 + bt + c. \end{align}where $t$ is "constant", i.e. not random. Clearly $E((tX+Y)^2)\ge0$ for all real values of $t$. Now recall that for real $a,b,c$, the polynomial $at^2 + bt+c$ remains non-negative as $t$ changes if and only if $a\ge0$ and the discriminant $b^2-4ac\le0$. So$$ b^2-4ac = 4E(XY)^2 - 4E(X^2)E(Y^2). $$So$$ 4(E(XY)^2 - E(X^2)E(Y^2))\le0. $$Divide both sides by $4$ and there you have it.
$\endgroup$ 2 $\begingroup$The expectation of a product of random variables is an inner product, to which you can apply the Cauchy-Schwarz inequality and obtain exactly that inequality. Hence the answer is yes.
See
$\endgroup$ $\begingroup$This is known as the Cauchy Schwarz inequality for Random Variables.
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