Showing $E|X|$ is finite!

Does knowing that for a iid $X_j$'s, $EXj = 0$ and $EX_j^2 < \infty$ indicate $E|X_j|$ is finite too? How to show that.

Can we say: $EX_j^2 = E|X_j|^2 < \infty \rightarrow \text{then } E|X_j| < \infty$ I appreciate your help.

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You can use Jensen's inequality for $|X|$, for example: $E|X|^2 - (E|X|)^2 \geq 0$ to create an upper bound for $E|X|$ – yoki Nov 29 '12 at 5:30
Cool...thanks... – Sam Nov 29 '12 at 5:37
No need for jensen's inequality. Use $Var(|X|) \geq 0$. Its the same thing but you get it from here also. – Gautam Shenoy Nov 29 '12 at 6:02

This is a matter of definition: E(X) is undefined when X is not integrable. Hence, to assume that E(X) = 0 is, in particular, to assume that X is integrable, which is needed for E(X) to exist, that is, to assume that (X is measurable and that) E(|X|) is finite.

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It's not necessary for $E[X_j^2]$ to be finite either. Just knowing $E[X_j]$ is finite is sufficient. For instance, consider $$X = n \quad \mbox{w.prob} \quad c\frac{1}{n^3} \quad \forall n \geq 1$$ where $c=\sum \frac{1}{n^3}$ which converges by the way.

Here $E[X^2]$ is infinite but the mean is finite if you compute it.

To show that $E[X] < \infty$ implies $E[|X|] < \infty$, observe $X=X^+ - X^-$, where $X^+ = \max(X,0)$ and $X^+ = \max(-X,0)$. Then $$E[X] = E[X^+ - X^-] = EX^+ - EX^-$$ But the way Lebesgue integrals are defined, E[X] is finite if and only if EX^+ and EX^- are finite. This is due to construction. So it is disallowed for either term to be infinite. Thus $$E[|X|] = E[X^+] + E[X^-] < \infty$$

So all in all, you're condition for $E[X^2] < \infty$ is superfluous.

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