MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that $f\left(x\right)\geq0$ is continuous on $\left[-\infty,\infty\right]$ and $\int_{-\infty}^{\infty}f\left(x\right)dx=1$. Is it true that $\int_{-\infty}^{\infty}\left|x\right|f\left(x\right)dx\leq\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left|x-y\right|f\left(x\right)f\left(y\right)dxdy$?


share|cite|improve this question
Since there is no background for the problem, why do you tag "probability"? It's (classical analysis). Why don't you do the change of variable $\hat y=y-x$ yourself? Of course, you have to do it carefully by replacing the improper integral by a proper one... – Wadim Zudilin Jun 9 '10 at 6:24
As Robin suggested, this problem could be solved by using probability and expection. I tried to integrate them by removing the absolute value and changing variables, but it seems hard. – user4606 Jun 9 '10 at 7:45
Wadim, it's clearly a question (but not a very interesting one) about continuous probability distributions, even if not phrased in that language. It's also going to be a race between the proposer providing more context/conditions to make the question more relevant/interesting and closure. – Robin Chapman Jun 9 '10 at 11:12
I rolled back the original tagging. I am convinced now that the readers of the post are more related to probability. – Wadim Zudilin Jun 10 '10 at 2:49

As suggested by Robin, we would need to prove that if $X$ and $Y$ are independent random variables with the same distribution, the inequality $E|X|\leq E|X-Y|$ holds.

This is certainly NOT true in general; for example suppose $X$ and $Y$ have some constant non-zero value; then the RHS is 0 and the LHS is positive.

However, if you add the condition that $E(X)=0$ then the result is true. (In your notation, $\int_{-\infty}^{\infty}xf(x)=0$ ).

In that case write $p=P(X\geq 0)$, and write $X_+$ for the positive part of $X$ and $X_-\leq 0$ for the negative part of $X$.

We have $0=EX=E|X_+|-E|X_-|$, so $E|X_+|=E|X_-|$.

Hence also $E|X|=E|X_+|+E|X_-|=2E|X_+|$.

Now $E|X-Y| \geq E(|X-Y|I(Y<0)I(X\geq 0)) + E(|X-Y|I(X<0)I(Y \geq 0))$

$=2 E(|X-Y|I(Y<0)I(X\geq0))$

$=2 E(|X_+|I(Y<0)I(X\geq0)) + 2E(|Y_-|I(Y<0) I(X\geq 0))$

$=2(1-p)E(|X_+|I(X\geq 0)) + 2p E(|Y_-|I(Y<0))$

$=2(1-p)E|X_+| + 2pE|Y_-|$



(using at various times the facts that $X$ and $Y$ are independent and that $X$ and $Y$ have the same distribution)

share|cite|improve this answer
James, I vote for your scepticism! The condition $\int_{\mathbb R}xf(x)\ dx=0$ is to replace the parity of $f(x)$, so your proof is essentially the same. And I now show why this is not true in general. – Wadim Zudilin Jun 9 '10 at 10:43
James, I can't get the voting logic: if I give my students your solution, they would clearly complain. Anyway, seeing a very strong preference of yours by the MO community, I have to delete mine. – Wadim Zudilin Jun 9 '10 at 23:29
Hi Wadim, what would they complain about? (Maybe the fact that it's written out in terms of random variables rather than integrals; just because I'd seen Robin's answer and I was feeling lazy and it seemed easier to write it out that way... no doubt my answer could be shortened.) I think your response was fine, no need to delete it. The answer I gave was somewhat more general since it covers any case where the distribution has mean 0, not just those which are symmetric around 0. – James Martin Jun 10 '10 at 16:25

This can be rephrased as follows.

Let $X$ and $Y$ be independent random variables with the same, continuous, distribution. Is it true that $E(X)\le E(|X-Y|)$.

  1. Is this likely?

  2. Is it true for a discrete distribution?

  3. If one has a discrete counterexample, can one convert it to a continuous counterexample?

Added I now realise that there was some absolute value signs in the original integral (these come out badly on my screen) and I should have written $E(|X|)\le E(|X-Y|)$. Plus ca change...

share|cite|improve this answer
Sorry, I keep imagining the question written with the other inequality! the function I wrote below (that is 1/x supported on [1,n] and normalized) gives indeed n-1 and 2n+o(n), that is, it satisfies the better inequality that you are suggesting. – Pietro Majer Jun 9 '10 at 8:57

As it is mentioned in James Martin's answer, if one does not make additional assumptions, the statement is wrong.

One correct statement is: if $X$ and $Y$ are independent r.v.'s with finite expectations, and $\mathbf{E}Y=0$, then $\mathbf{E}f(X)\le \mathbf{E}f(X+Y)$ for any convex function $f$ such that these expectations are finite.

Proof: The sequence of two r.v.'s $X$ and $X+Y$ is a martingale. Therefore, the sequence $f(X), f(X+Y)$ is a submartingale, and the claim follows.

Of course, $f$ should be taken to be the absolute value function, and $Y$ should be replaced by $-Y$, if we want this statement to look more like the statement in question.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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