[Math] If $E(X^2)$ exists, $E(X)$ exists


How do I prove this? Is it even true?

I proved the obvious cases, but don't know how to prove/disprove the $0\lt x\lt 1$ case.

IF $E(X^2)$ exists, $\sum_{x\in\omega}x^2P(x)\lt \infty$. Therefore:

$$\forall x \le 0:\space \sum_{x\in\omega}xP(x) \le \sum_{x\in\omega}x^2P(x) = E(X^2)\lt \infty \\
\forall x\ge 1:\space \sum_{x\in\omega}xP(x) \le \sum_{x\in\omega}x^2P(x) = E(X^2)\lt \infty$$

Thank you.

Best Answer

I will assume that "exists" means that the expectation in question is finite.

Then, since $|x|< 1+x^2$ for all $x \in \mathbb R$, $E[|X|] < 1+E[X^2] < \infty$, and so $E[X]$ also exists.

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