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.
 A: 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.
A: 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. 
