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Show that $E^2(XY) \leq E(X^2)E(Y^2)$.

We can assume that $X$ and $Y$ are jointly distributed random variables with finite variances.

I know cov$(X,Y)=E[(X-\mu_x)(Y-\mu_y)]$ and there's a definition that might be of use that states $V(X)=E(X^2)-E(X)^2$.

Any idea where to start to prove this?

complexanalysis
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1 Answers1

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In case @angryavian's insight is unintuitive:

Let $Z\in\{X,\,Y\}$ have mean $\mu_Z$ and standard deviation $\sigma_Z$ and let$$\rho:=\operatorname{Corr}(X,\,Y)\in[-1,\,1].$$Applying sign changes if necessary so means are non-negative,$$\begin{align}E(X^2)E(Y^2)-E(XY)^2&=(\mu_X^2+\sigma_X^2)(\mu_Y^2+\sigma_Y^2)-(\mu_X\mu_Y+\rho\sigma_X\sigma_Y)^2\\&=\sigma_X^2\mu_Y^2+\mu_X^2\sigma_Y^2-2\rho\mu_X\mu_Y\sigma_X\sigma_Y+(1-\rho^2)\sigma_X^2\sigma_Y^2\\&\ge\sigma_X^2\mu_Y^2+\mu_X^2\sigma_Y^2-2\mu_X\mu_Y\sigma_X\sigma_Y\\&=(\mu_X\sigma_Y-\mu_Y\sigma_X)^2.\end{align}$$

J.G.
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  • Okay, that makes sense for the most part, thank you. One quick question though, how do we know $E[X]^2=(\mu^2_x +\sigma^2_x)$? I've only seen it defined as $E[X]^2=(\mu^2_x)$. – complexanalysis Apr 30 '21 at 19:45
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    @complexanalysis Don't confuse $E(X^2)$, which your question contains, with $E(X)^2$, which appears in your formula for $V(X)$. That formula rearranges to $E(X^2)=E(X)^2+V(X)$. On this equation's right-hand side, the first term is a squared mean, while the second term, the variance, is a squared standard deviation. – J.G. Apr 30 '21 at 19:47