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I don't know how to prove this. Lets assume that $X$ is a discrete random variable. I've just come this far: If we do a direct proof of the implication, then we start with the assumption: $$ \operatorname{E}[X^2]=\sum_{x\in Im(X)} x^2\cdot p_X(x) <\infty. $$ However, I don't know how I can follow from this that $\operatorname{E}[X]$ exists.

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    $\begingroup$ With the Cauchy-Schwarz inequality, you can prove more: if $X, Y \in \mathcal L^2$, then $XY \in \mathcal L^1$. Yours is the special case $Y=1$ $\endgroup$
    – Ant
    Commented Jan 19, 2016 at 19:14
  • $\begingroup$ Sounds interesting. Could you post it as an answer in more detail? $\endgroup$
    – ndrizza
    Commented Jan 19, 2016 at 19:17
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    $\begingroup$ some other answers have been posted in that direction.. Otherwise good old wikipedia is always useful en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality :-) $\endgroup$
    – Ant
    Commented Jan 19, 2016 at 19:23

4 Answers 4

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We have $|X|\leq1+X^2$ and consequently $$\mathbb E|X|\leq1+\mathbb EX^2$$

So $\mathbb EX^2<\infty$ allows the conclusion that $\mathbb EX$ exists.

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  • $\begingroup$ Glad to be of help. $\endgroup$
    – drhab
    Commented Jan 19, 2016 at 19:16
  • $\begingroup$ I'm quite impressed how many topics are touched within this proof: $X\leq Y\Longrightarrow\operatorname{E}[X]\leq\operatorname{E}[Y]$, linearity of expected value, absolute convergence $\Longrightarrow$ convergence. $\endgroup$
    – ndrizza
    Commented Jan 19, 2016 at 19:37
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    $\begingroup$ @ndrizza Is also useful to keep in mind that absolute convergence $\iff$ convergence :) $\endgroup$
    – Ant
    Commented Jan 19, 2016 at 20:26
  • $\begingroup$ A sidenote: I was hesitating for a moment whether the absolute value for X is needed. But it really is - it avoids the case where $\operatorname{E}[X]$ may tend to minus infinity. $\endgroup$
    – ndrizza
    Commented Jan 20, 2016 at 9:47
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I'm a big fan of Jenson's inequality for probability measures, for $f(x)=x^r$ with $r\geq 1$, implies:

$$f(E[X]])\leq E[f(X)],$$

which gives, after rearranging:

$$E[X]\leq E[X^r]^{1/r}$$

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An even more general statement, shown here, is if $$ \mathbb E(|X^r|)< \infty $$ then for any positive integer $s<r$, the $s$-th moment $\mathbb E[|X^s|]$ is also finite (and thus $\mathbb E[X^s]$ exists).

This can be done by noticing that $0<s<r \Longrightarrow |X|^s \le \max(1, |X|^r) $.

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  • $\begingroup$ Thanks. Also very useful to know it in a more general context. $\endgroup$
    – ndrizza
    Commented Jan 19, 2016 at 19:18
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Here's another; Use the Cauchy-Schwarz inequality.

$$E|X| = E|X|\cdot 1 \le \|1\|_2 \|X\|_2 = \left(E|X|^2\right)^{1/2}.$$

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