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If I know that $E[X^2]$ exist, how can I show that $E[X]$ exist? For me it's obvious, but I don't have idea how to show it, because I don't have any information about distribution. Maybe someone can give a hint?

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    $\begingroup$ Hint: $1 + X^2 > |X|$ for all $X$. $\endgroup$ – Dilip Sarwate Nov 29 '12 at 15:38
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How could it "not exist"?

I assume, $X$ is measurable, then the nonexistence of $E[X]$ would mean that either the positive or the negative part of the integral wants to go to infinity. But then $X^2$ would go even faster as the base is of finite measure.

Let $A:=\{\omega|X(\omega)\ge 0\}=:(X\ge 0)$ and $B:=(X<0)$, then, using that $x\le x^2$ if $x\ge 1$, we have $$\int_A X \le \int_{X\ge 1} X^2+\int_{0\le X<1}1\le E[X^2]+1 $$ And similarly over $B$.

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  • $\begingroup$ Yes, exist means that $E[X] < inf$ $\endgroup$ – Johnny Nov 29 '12 at 19:59

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