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 $\int_{-\infty}^{\infty}\left|x\right|f\left(x\right)dx\leq\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left|x-y\right|f\left(x\right)f\left(y\right)dxdy$?


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    $\begingroup$ 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... $\endgroup$ – Wadim Zudilin Jun 9 '10 at 6:24
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    $\begingroup$ 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. $\endgroup$ – user4606 Jun 9 '10 at 7:45
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    $\begingroup$ 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. $\endgroup$ – Robin Chapman Jun 9 '10 at 11:12
  • $\begingroup$ I rolled back the original tagging. I am convinced now that the readers of the post are more related to probability. $\endgroup$ – 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)

  • $\begingroup$ 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. $\endgroup$ – Wadim Zudilin Jun 9 '10 at 10:43
  • $\begingroup$ 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. $\endgroup$ – Wadim Zudilin Jun 9 '10 at 23:29
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    $\begingroup$ 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. $\endgroup$ – 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...

  • $\begingroup$ 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. $\endgroup$ – 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.


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