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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?

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    @Mario: As long as $X$ and $Y$ are *random variables* (i.e., measurable), the RHS is *always* well-defined. It simply need not be finite (your Cauchy example, for instance).2012-12-18

3 Answers 3

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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 $ f(t) = E((tX+Y)^2) = (E(X^2)) t^2 + 2(E(XY))t + E(Y^2) = at^2 + bt + c. $ 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.

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    A concise proof. make very senseeee^_^2015-09-08
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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 http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality#Probability_theory

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This is known as the Cauchy Schwarz inequality for Random Variables.