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  1. Is the following claim correct:

    If $\mathrm{E} |X|^2$ exists, then $\mathrm{E} X$ also exists, because $$ \mathrm{E} X \leq \mathrm{E} |X| \leq \sqrt{\mathrm{E} |X|^2} $$ by Jensen's inequality.

  2. Is the following claim correct:

    If $\mathrm{E} |X|^2$ exists and is finite, then $\mathrm{E} X$ also exists and is finite, because $$ \mathrm{E} X \leq \mathrm{E} |X| \leq \sqrt{\mathrm{E} |X|^2} $$ by Jensen's inequality.

I guess the arguments are not right, because Jensen's inequality assumes $X$ to be integratable, i.e. $\mathrm E |X| < \infty$.

Thanks.

The first statement itself is wrong, and the Second is correct.

Tim
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3 Answers3

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Inequality $\left|X\right|\leq1+X^{2}$ gives you $\mathbb{E}\left|X\right|\leq1+\mathbb{E}X^{2}$

drhab
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$\newcommand{\E}{\operatorname{E}}$

Suppose $\E(X^2)<\infty$. Then $$ \E |X| \le \E\left.\begin{cases} 1 & \text{if }|X|\le 1 \\ X^2 & \text{if }|X|>1 \end{cases}\right\} \le 1 + \E(X^2)<\infty. $$ Hence $\E(X)$ exists (and is finite).

Michael Hardy
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  • (1) when you write $E|X|$, do you assume it already exists? (2) Is Jensen's inequality not useful to the problem? – Tim Nov 03 '14 at 18:38
  • @Tim : $\operatorname{E}|X|$ exists within the set $[0,\infty]$ (closed at both ends of the interval) as long as $X$ is in fact measurable. That is true of all non-negative random variables. The only time $\operatorname{E}X$ fails to exist within $[-\infty,+\infty]$ is when the integrals of the positive and negative parts are both infinite (as with the Cauchy distribution). ${}\qquad{}$ – Michael Hardy Nov 03 '14 at 18:43
  • Yes, I remember now. What about Jensen's ineq? Is it not useful here? – Tim Nov 03 '14 at 18:45
  • You can use Jensen's inequality, but the way above seems a bit simpler. – Michael Hardy Nov 03 '14 at 18:49
  • By the way, can Cauchy Schwarz be used here? To prove $\mathrm{E} |X| \leq \sqrt{\mathrm{E} |X|^2}$? I seemed to see that C-S could be proved by Jensen, but forgot it now. – Tim Nov 03 '14 at 19:14
  • Hi Michael, I think we might have learned from similar references before, about existence of lebesgue integral and finitness of lebesgue integral. I am confused about other peoples posts here http://stats.stackexchange.com/questions/91512/how-can-a-distribution-have-infinite-mean-and-variance/91515?noredirect=1#comment233491_91515 and here http://math.stackexchange.com/questions/1004772/for-e-x-ex2-to-exist-do-we-need-ex-to-exist-and-be-finite. I would like to know your thoughts. Thanks. – Tim Nov 03 '14 at 22:19
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Your second statement is a standard result in measure theory, proven with Hölder’s inequality: If $\mu(\Omega) < \infty, 0 < p < q \leq \infty$, then $L^q(\Omega,A,\mu) \subset L^p(\Omega,A,\mu)$.

Concentrate on the $<\infty$ part. Let $f\in L^q$. Set $r=q/p$, take $s$ as the appropriate Hölder conjugate.

$\int |f|^p = \int |f|^p \cdot 1 \leq \left(\int |f|^{pr} \right)^{1/r} \cdot \left(\int 1^s\right)^{1/s} = \lVert f \rVert_q ^{q/r} \cdot \mu(\Omega)^{1/s}$.

Fye
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  • Thanks. Is Jensen's or Cauchy-Schwarz useful to prove the special case in my problem when p=1 and q=2? – Tim Nov 03 '14 at 18:47
  • The inequalities that I know would already assume the finite mean (or in other words integrability) of $X$. But maybe there are other versions that only assume quasi-integrability. The explanations of @Michael Hardy and drhab seem to be the simplest for this ($p=1,q=2$) case. – Fye Nov 03 '14 at 18:57