Timeline for Upper bound on expectation value of the product of two random variables
Current License: CC BY-SA 3.0
8 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Oct 17, 2012 at 22:48 | comment | added | Anthony Quas | Sorry Michele. I don't have time to do tutorials in all of this stuff. I'd suggest you seek out someone nearby who can help you. | |
Oct 17, 2012 at 13:25 | comment | added | James | Dear Anthony, Thank you for your reply. It is now clear to me why that bound is the best possible. I looked the Hardy-Littlewood inequality, that reads "If f and g are nonnegative and measurable functions that vanish at infinity, then \int f(x) g(x) dx <= \int f*(x) g*(x) dx, where f*, g* are the symmetric decreasing rearrangements of f,g respectively". I don't understand how this inequality can be applied to E[XY], where there is an additional weight factor p(x,y) in the integral that I want to estimate: E[XY] = \int dX dY p(X,Y) X*Y. May you please clarify this? Thanks! | |
Oct 16, 2012 at 6:55 | comment | added | Anthony Quas | Apparently the inequality I'm quoting goes by the name "Hardy-Littlewood inequality". See math.toronto.edu/almut/rearrange.pdf | |
Oct 16, 2012 at 5:57 | comment | added | Anthony Quas | 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. | |
Oct 16, 2012 at 1:05 | comment | added | James | 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 | |
Oct 10, 2012 at 5:19 | comment | added | Anthony Quas | 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. | |
Oct 9, 2012 at 23:54 | comment | added | James | 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 | |
Oct 8, 2012 at 3:22 | history | answered | Anthony Quas | CC BY-SA 3.0 |