Does $\mathbb{E}\left[X\right]=\infty$ imply $\mathbb{E}\left[X^{2}\right]=\infty$?

I'm trying to prove that if $\mathbb{E}\left[X\right]=\infty$ then $\mathbb{E}\left[X^{2}\right]=\infty$ for every random variable $X$.

I know that if $X(w)>1$ I'll get that $X^2(w)>X(w)$ so $\mathbb{E}\left[X^{2}\right]\ge\mathbb{E}\left[X\right]$

and if $X(w)\le1$ then $X^2(w)\le X(w)$ so

$\mathbb{E}\left[X^{2}\right]=\sum\nolimits _{w\in\Omega}X^{2}\left(w\right)\cdot p(w)\le\sum\nolimits _{w\in\Omega}1\cdot p(w)=1$

But how do I prove it when some of the $w\in \Omega$ are larger than 1 and some are less ?

Is this even the right way to prove this ?

• Hint: try to condition $\mathbb{E}[X^2]$ on "$X \geq 1$" and "$X < 1$" (or perhaps $|X|$ instead of $X$).
– Gnuk
Commented Dec 26, 2016 at 13:17
• @Arthur Thank you' i've edited the title. Commented Dec 26, 2016 at 13:21
• @HarrySmit could you please tell me how you would handle a radnom variable $Im(X)\in [0,\infty)$ ? This is my problem with this kind of prove. Commented Dec 26, 2016 at 16:19

By Jensen's inequality we know that for convex $f$

$$f\left(\operatorname{E}[X]\right) \leq \operatorname{E}\left[f(X)\right]$$

The fact that $\operatorname{E}[X]=\infty \implies \operatorname{E}\left[ X^2 \right] = \infty$ follows from the observation that $f(x)=x^2$ is convex.

Edit: As referenced in the comments below, for this argument to be rigorous, one needs to multiply inside the expectation by $\mathbf 1(\vert X \vert \leq k)$, apply Jensen's inequality, and then take $k \to \infty$ with an appeal to the monotone convergence theorem.

• I am uncomfortable with the application of Jensen's inequality to infinite quantities. I think a very rigorous proof should use a sequence $X_n$ of random variables with limit $X$, each one having a finite $E(X)$ and a finite $E(X^2).$ Commented Dec 26, 2016 at 14:26
• @JeanMarie Fair -- I should probably multiply inside the expectation by $\mathbf 1( \vert X \vert \leq k)$ and let $k \to \infty$, but that development may demand more than the OP's question suggests they know. I'll edit my answer to include a reference to it for completeness. Commented Dec 26, 2016 at 17:04
• I agree. Sometimes being too rigorous can blur the message... Commented Dec 26, 2016 at 17:15

Proof that if $E[X]=\infty$ then $E[X^2]=\infty$ :
First note that $\int_{-1}^1xf_X(x)dx<=1$ and so if $E[X]=\infty$ then $\int_{\infty}^{-1}xf_X(x)dx+\int_1^{\infty}xf_X(x)dx=\infty$ and so $$E[X^2]>=\int_{\infty}^{-1}x^2f_X(x)dx+\int_1^{\infty}x^2f_X(x)dx>=\int_{\infty}^{-1}xf_X(x)dx+\int_1^{\infty}xf_X(x)dx=\infty$$

Note that $$0\leq\text{Var} X=E(X-EX)^{2}=EX^2-(EX)^2$$ so that $$(EX)^2\leq EX^2.$$ Hence $$EX=\infty \implies EX^2=\infty.$$

• But if $E[X] = \infty$, $Var[X]$ doesn't exist. The ordinary rules of algebra don't apply to $\infty$. Commented Jun 21, 2018 at 11:25

By the Cauchy-Schwarz inequality the contrapositive is available: $$E|X|\leq\sqrt{EX^2}$$ implies that if $EX^2<\infty$ then even $E|X|<\infty$.

Edit: If it was not already clear, $E|X|<\infty$ leads to $EX<\infty$.

• It's hard to tell from the little you've written whether this is even a contrapositive (i.e. an equivalent) for the claim asked about in the Question. If possible, please clarify and justify with additional detail. Commented Jan 15, 2017 at 21:06
• Sure thing, now it's done.
– mbe
Commented Jan 15, 2017 at 21:11
• I'm doubtful of your use of absolute values, which do not appear in the Question (those are square brackets, indicating the scope of the expectation operator). Commented Jan 15, 2017 at 21:15

It is a fact that for finite measure spaces, every $L^2$ function is $L^1$. (In fact, every $L^p$ function is $L^q$ when $p>q\geq1$.) Your claim follows from the observation that $X^2$ is nonnegative, so $E[X^2]$ is either finite or infinite.