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.
2025-01-13 00:02:28.1736726548
Prove: $\operatorname{E}[X^2]<\infty\Longrightarrow \operatorname{E}[X]$ exists
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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) $.
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.