If I know that $E[X^2]$ exist, how can I show that $E[X]$ exist? For me it's obvious, but I don't have idea how to show it, because I don't have any information about distribution. Maybe someone can give a hint?

  • 3
    $\begingroup$ Hint: $1 + X^2 > |X|$ for all $X$. $\endgroup$ – Dilip Sarwate Nov 29 '12 at 15:38

How could it "not exist"?

I assume, $X$ is measurable, then the nonexistence of $E[X]$ would mean that either the positive or the negative part of the integral wants to go to infinity. But then $X^2$ would go even faster as the base is of finite measure.

Let $A:=\{\omega|X(\omega)\ge 0\}=:(X\ge 0)$ and $B:=(X<0)$, then, using that $x\le x^2$ if $x\ge 1$, we have $$\int_A X \le \int_{X\ge 1} X^2+\int_{0\le X<1}1\le E[X^2]+1 $$ And similarly over $B$.

  • $\begingroup$ Yes, exist means that $E[X] < inf$ $\endgroup$ – Johnny Nov 29 '12 at 19:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.