8

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 ?

Sri-Amirthan Theivendran
  • 31,116
limitless
  • 283

5 Answers5

13

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.

Theoretical Economist
  • 6,472
  • 3
    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 $E(X^2).$ – Jean Marie Dec 26 '16 at 14:26
  • 3
    @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. – Theoretical Economist Dec 26 '16 at 17:04
  • 2
    I agree. Sometimes being too rigorous can blur the message... – Jean Marie Dec 26 '16 at 17:15
3

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 $$E[X^2]>=\int_{\infty}^{-1}x^2f_X(x)dx+\int_1^{\infty}x^2f_X(x)dx>=\int_{\infty}^{-1}xf_X(x)dx+\int_1^{\infty}xf_X(x)dx=\infty$$

user1131274
  • 1,678
2

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

Sri-Amirthan Theivendran
  • 31,116
1

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 $EX<\infty$.

mbe
  • 947
  • 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. – hardmath Jan 15 '17 at 21:06
  • Sure thing, now it's done. – mbe Jan 15 '17 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). – hardmath Jan 15 '17 at 21:15
0

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.

See also this discussion on Cross-Validated.

Community
  • 1
symplectomorphic
  • 18,604