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For a continuous random variable $X$, if $E(|X|)$ is finite, is $\lim_{n\to\infty}n P(|X|>n)=0$?

This is a problem I found on the internet, but I'm not sure whether it holds or not.

I know that $n P(|X|>n)<E(|X|)$ holds by Markov inequality, but I can't show that it goes to 0 as $n$ goes to infinity.

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    $\begingroup$ (1) Continuity is not needed. (2) Express the expectation as an integral of the survival function $\Pr(|X|\gt n)$. (3) Consider the contrapositive: what would a nonzero limit imply about the expectation? $\endgroup$
    – whuber ♦
    Commented Oct 5, 2017 at 20:46
  • $\begingroup$ @whuber nice exercise! I think I have a correct answer, but since this looks like self-study, I don't think I should write it here. Can I create a private chat-room and show you my solution, so that you can tell me if it's correct? $\endgroup$
    – DeltaIV
    Commented Oct 6, 2017 at 15:50
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    $\begingroup$ @Delta This is a case where posting your answer would seem fine to me: the OP has a specific sub-question and does not appear just to be trolling for homework answers. $\endgroup$
    – whuber ♦
    Commented Oct 6, 2017 at 15:58
  • $\begingroup$ @whuber this reminds me of the non-existence of a uniform distribution over the natural numbers -- does this mean that while continuity is not needed here, countable additivity is ? $\endgroup$
    – Bill Clark
    Commented Jun 20, 2019 at 15:23
  • $\begingroup$ Does anyone have a reference for the proof of the converse: if $n^{\alpha + \delta}\mathbb{P}(\lvert X \rvert > n) \to 0$ for some $\delta > 0$ then $\mathbb{E}[\lvert X \rvert^\alpha] < \infty $? $\endgroup$
    – Yashaswi Mohanty
    Commented Aug 15, 2023 at 23:08

3 Answers 3

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Look at the sequence of random variables $\{Y_n\}$ defined by retaining only large values of $|X|$: $$Y_n:=|X|I(|X|>n).$$ It's clear that $Y_n\ge nI(|X|>n)$, so $$E(Y_n)\ge nP(|X|>n).\tag1$$ Note that $Y_n\to0$ and $|Y_n|\le |X|$ for each $n$. So the LHS of (1) tends to zero by dominated convergence.

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    $\begingroup$ I think you mean "RHS" in your final sentence, otherwise, good job! $\endgroup$
    – jbowman
    Commented Oct 6, 2017 at 21:45
  • $\begingroup$ @jbowman, s/he means $\mathbb{E}{Y_n} \to 0$ by the dominated convergence theorem (note that $Y_n \to 0$ alone is not sufficient to reach that conclusion). I added the link to DCT on wikipedia $\endgroup$
    – P.Windridge
    Commented Oct 7, 2017 at 11:42
  • $\begingroup$ @P.Windridge - I didn't read carefully enough, and associated the "So the LHS" with equation 1, instead of with the previous sentence. My bad. $\endgroup$
    – jbowman
    Commented Oct 7, 2017 at 16:03
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    $\begingroup$ Note that $Y_n$ is a random variable. $Y_n\rightarrow 0$ in what sense? $\endgroup$
    – YHH
    Commented Feb 3, 2019 at 21:31
  • $\begingroup$ @YHH The convergence is pointwise: For every $\omega$, $Y_n(\omega)\to0$ as $n\to\infty$. $\endgroup$
    – grand_chat
    Commented Feb 4, 2019 at 7:12
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I can provide an answer for a continuous random variable (there is surely a more general answer). Let $Y=|X|$:

$$\mathbb{E}[Y]=\int_0^\infty yf_Y(y)\text{d}y=\int_0^n yf_Y(y)\text{d}y+\int_n^\infty yf_Y(y)\text{d}y\ge\int_0^n yf_Y(y)\text{d}y+n\int_n^\infty f_Y(y)\text{d}y=\dots+n\left(F_Y(\infty)-F_Y(n)\right)=\dots+n(1-F_Y(n))=\int_0^n yf_Y(y)\text{d}y+nP(Y\gt n)$$

Thus

$$0\leq nP(Y\gt n)\le\left(\mathbb{E}[Y]-\int_0^n yf_Y(y)\text{d}y\right)$$

Now,since by hypothesis $\mathbb{E}[Y]$ is finite, we have that

$$\lim_{n\to \infty}\left(\mathbb{E}[Y]-\int_0^n yf_Y(y)\text{d}y\right)=\mathbb{E}[Y]-\lim_{n\to \infty}\int_0^n yf_Y(y)\text{d}y=\mathbb{E}[Y]-\mathbb{E}[Y]=0$$

Then

$$\lim_{n\to \infty}nP(Y\gt n)=0$$

by the sandwich theorem.

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  • $\begingroup$ @P.Windridge could you please check that also my use of the dominated convergence theorem is correct? I have a quantity, $nP(Y>n)$, which is nonnegative, and not greater than a quantity whose limit is 0, thus $\lim_{n\to\infty}nP(Y>n)=0$ in my application of the theorem. Thanks $\endgroup$
    – DeltaIV
    Commented Oct 7, 2017 at 12:01
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    $\begingroup$ @DeltaIV- first, to clarify, "$a_n \le b_n \le c_n$ and $a_n, c_n \to l$ implies $b_n\to l$" is NOT the dominated convergence theorem (usually it's called the sandwich theorem). $\endgroup$
    – P.Windridge
    Commented Oct 7, 2017 at 12:36
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    $\begingroup$ @DeltaIV- no, you don't need DCT, MCT is enough (this includes the possibility that $\mathbb{E} Y = \infty$, but then you can't say $\mathbb{E} Y -\mathbb{E} Y = \infty - \infty = 0$! ) $\endgroup$
    – P.Windridge
    Commented Oct 7, 2017 at 12:42
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    $\begingroup$ No problem. Btw, I know $E[Y]$ is finite by assumption, I was just explaining where you use that assumption (the MCT itself does not require it, unlike the DCT, which @grand_chat used and I hope you looked at :)). $\endgroup$
    – P.Windridge
    Commented Oct 7, 2017 at 17:46
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    $\begingroup$ @P.Windridge ah, ok! I didn't notice that MCT doesn't require the assumption. I did have a look at the DCT, that's why I thought I didn't need it for my proof :) I pay the price of not having being taught about Lebesgue integration at university...for this reason, I'm used to do probability calculus in terms of pdfs, instead than in terms of measures. $\endgroup$
    – DeltaIV
    Commented Oct 7, 2017 at 18:08
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$E\left | X \right |< \infty \Leftrightarrow E\left | X \right |\mathbb{I}_{\left | X \right |>n}\rightarrow 0$ (uniformly integrable)

$E\left | X \right |=E\left | X \right |\mathbb{I}_{\left | X \right |>n}+E\left | X \right |\mathbb{I}_{\left | X \right |\leq n}$

$E\left | X \right |\mathbb{I}_{\left | X \right |>n}\leq E\left | X \right |< \infty $

$E\left | X \right |\mathbb{I}_{\left | X \right |>n}\geq nE\mathbb{I}_{\left | X \right |>n}=nP\left ( \left | X \right |>n\right )$

$E\left | X \right |\mathbb{I}_{\left | X \right |>n} \rightarrow 0 \Rightarrow nP\left ( \left | X \right |>n\right )\rightarrow 0 \Rightarrow P\left ( \left | X \right |>n\right )\rightarrow 0$

i.e. $ \underset{n\rightarrow \infty}{\lim} P\left ( \left | X \right |>n\right )=0$

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