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How can I prove that $$ \mathbb{E} [X^2] \geq \mathbb{E} [|X|]^2 $$ This resembles the Cauchy-Schwarz inequality a lot but I'm unable to prove it with the usual method (i.e. when there are two random variables and you choose to set $f(\alpha)=(X-\alpha Y)^2$. Any help is greatly appreciated!

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    $\begingroup$ This is a special case of Jensen's inequality. $\endgroup$ – carmichael561 Apr 9 '16 at 21:08
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From Jensen's inequality we know that, whenever f is a convex function and X is a random variable we have: $$ E(f(X)) \geq f(E(X))$$.

In this context $f(X) = X^2$, which is a convex function and hence we are done.

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  • $\begingroup$ Yes, but I was wondering if considering the absolute value of the R.V. doesn't change things a little bit... $\endgroup$ – james42 Apr 9 '16 at 21:21
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    $\begingroup$ @james42 $\lvert X\rvert^2 = X^2$ $\endgroup$ – Calculon Apr 9 '16 at 21:33
  • $\begingroup$ Abby and @Calculon $E[|X|]^2 = E[X]^2$? $\endgroup$ – BCLC Apr 9 '16 at 21:48
  • $\begingroup$ @BCLC Jensen: $E[\lvert X\rvert^2] \geq E[\lvert X\rvert]^2$. The convex function underneath is $x \mapsto \lvert x \rvert^2$. $\endgroup$ – Calculon Apr 9 '16 at 21:50
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    $\begingroup$ @BCLC It is Jensen's equality. Instead of $X$ consider $\lvert X \rvert $ as the random variable. $\endgroup$ – Calculon Apr 9 '16 at 21:55
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For $a,b\in\mathbb R$ and $t\in(0,1)$, we have \begin{align} (ta^2+(1-t)b^2) - (ta+(1-t)b)^2 &= ta^2 + (1-t)b^2 -t^2a^2 -2t(1-t)ab -(1-t)^2b^2\\ &= t(1-t)a^2 -2t(1-t)ab + t(1-t)b^2\\ &= t(1-t)(a-b)^2\\ &>0, \end{align} so $x\mapsto x^2$ is a convex function. It follows from Jensen's inequality that $$\mathbb E[X^2]\geqslant \mathbb E[X]^2 $$ and $$\mathbb E[X^2]=\mathbb E[(-X)^2]\geqslant \mathbb E[-X]^2, $$ and so $$\mathbb E[X^2]\geqslant \mathbb E[|X|]^2. $$

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$Var(|X|) = E[|X|^2] - E[|X|]^2 \geq 0 $ and this implies the result

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