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Hello, I am trying to find an upper bound on the expectation value of the product of two random variables.

So suppose x, y are two non-independent random variables, given that I know the distribution of x p(x) and the distribution of y q(y), how can I find an upper bound on E[|x * y |] that is a function of p and q?

I know that Holder's inequality gives an upper bound to my problem in terms of moments of x and y, but this is a poor bound for the problem that I am considering.

Thank you! Best Michele

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closed as not a real question by Yemon Choi, Qiaochu Yuan, Andrés Caicedo, Will Jagy, Bill Johnson Oct 8 '12 at 16:04

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Cauchy-Schwarz? – Yemon Choi Oct 8 '12 at 0:22
Well if it gives you a poor bound for your problem, you need to specify more details. The Cauchy-Schwarz inequality is sharp – Yemon Choi Oct 8 '12 at 1:42
Why the down-votes? I don't think that C-S is sharp for this situation. If you assume that they are non-negative valued, the sharp upper bound is obtained when the variables are monotonically coupled. I'll post a formula for this in a few minutes. – Anthony Quas Oct 8 '12 at 3:04
C-S is sharp if all that you know are the second moments. Here we've got far more information: the entire distribution of the random variables. – Anthony Quas Oct 8 '12 at 3:07
I still think, though, that the question should have included at least some examples of the kinds of distribution that the OP had in mind – Yemon Choi Oct 8 '12 at 3:52

I'll assume that $X$ and $Y$ are non-negative random variables. Let $F_X$ be the cumulative distribution function of $X$ (that is $F_X(t)=\mathbb P(X\le t)$) and $F_Y$ be the cumulative distribution function of $Y$.

In your notation, probably $F_X(t)=\int_0^t p(s)\,ds$ and $F_Y(y)=\int_0^t q(t)\,dt$.

Now define two functions on $[0,1]$: $g_X(x)=\sup\lbrace t\colon \mathbb P(X\le t)\le x\rbrace $ and similarly $g_Y(x)=\sup\lbrace t\colon \mathbb P(Y\le t)\le x\rbrace$. These functions are the increasing rearrangements of $X$ and $Y$. That is these are non-decreasing functions with the property that $m\lbrace x\colon g_X(x)\le t\rbrace =\mathbb P(X\le t)$ and $m\lbrace x\colon g_Y(x)\le t\rbrace = \mathbb P(Y\le t)$.

Now the largest possible value of $\mathbb E XY$ given the distributions is $\int_0^1 g_X(t)g_Y(t)\ dt$. Intuitively the reason for this is that the largest value for the expectation is obtained when the largest values of $X$ are multiplied by the largest values of $Y$. Slightly more precisely imagine you've arranged the $X$ values from largest to smallest. Think of these as "weights" for the $Y$ values. Obviously you get the biggest integral if you weight the big $Y$ values with the biggest weights.

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Dear Anthony, Thank you very much for your answer! A few questions: - Is this bound better than Holder's inequality's bound 𝔼[XY] <= E[X^p]^(1/p)*E[Y^q]^(1/q) with q>1,p>1,1/p+1/q = 1? If it is, is there a way of proving or simply justifying this? - Where can I find a proof of the bound that you suggested? Thanks you Best Michele – Michele Oct 9 '12 at 23:54
The justification is in my answer. For more, you could try Lindvall's book "Lectures on the Coupling Method". This is the best possible bound: If you let $\omega$ be uniformly distributed in the unit interval, then $g_X(\omega)$ has the same distribution as $X$ and $g_Y(\omega)$ has the same distribution as $Y$ and the product of these random variables has the integral in my answer. – Anthony Quas Oct 10 '12 at 5:19
Dear Anthony, Still, it is not clear to me how to prove the inequality that you suggested : E[X*Y] <= \int_{0}^{1} dt g_{X}(t) * g_{Y}(t). Is the proof in the book "Lectures on the Coupling Method"? It it not clear either wether and why this bound is better than the Holder's inequality bound E[X^p]^(1/p)*E[Y^q]^(1/q). Can you prove this? Thanks! Michele – Michele Oct 16 '12 at 1:05
If you know that $X$ is uniformly distributed on the unit interval and $Y$ are is the uniformly distributed random variable on [1,2], then the bound I'm suggesting comes from $g_X(t)=t$, $g_Y(t)=1+t$, so that $\mathbb E XY\le \int (t+t^2)\,dt=5/6$. If you use H\"older's inequality, you get $(1/(p+1))^{1/p}((2^{q+1}-1)/(q+1))^{1/q}$. This is greater than 5/6 for all $1/p+1/q=1$. My bound is attained if $X$ is uniform and $Y=1+X$. In general, my bound is always attained for some joint distribution on $X$ and $Y$. The Holder bound is not always attained. So mine is lower and is best poss. – Anthony Quas Oct 16 '12 at 5:57
Apparently the inequality I'm quoting goes by the name "Hardy-Littlewood inequality". See – Anthony Quas Oct 16 '12 at 6:55

I would try yo apply Hoeffding's Lemma, who used his result to identify the bivariate cdfs with given marginal cdfs that minimize or maximize correlation. Let $(X,Y)$ be a random vector with bivariate cdf $H$, let $F$ and $G$ be their marginal cdfs, respectively. It is well known that a sharp upper bound for $H(x,y)$ is $\min(F(x),G(y))$. By Hoeffding's Lemma we get that $$E(XY)\leq E(X)E(Y)+\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\big[\min(F(x),G(y))-F(x)G(y)\big]dxdy$$

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