# Bound on $E[X Y]$, where $Y$ is bounded in terms of $X$

I have two random variables $X \in [0,1], Y \in [a,1]$, and I would like to lower-bound the expectation $E[X Y]$, preferably in terms of $E[X]$ and $E[Y]$. I would accept anything better than the trivial $E[X Y] \geq a E[X]$.

In general, the worst possible distribution would be if $X, Y$ were anti-correlated, that is, when $Y$ is large then $X$ is small. However, I also have bounds on these namely that with certainty $$X \geq Y^k$$ for some constant $k \geq 1$. Does this give any way to lower-bound $E[X Y]$?

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You need to say something more about the distributions. If the distributions can be discrete X can be identically 0 and there is no lower bound above 0. If you require X and Y to have absolutely continuous distributions you may need to require that the density has support over the entire interval [0, 1]. Otherwise X can be concentrated on [0, a] for some arbitrary small a meaning that E[XY] will be less than a. –  Michael Chernick Jul 4 '12 at 18:38
@Michael: But in this case $E[X], E[Y]$ will also be very small. I only want to bound $E[X Y]$ in terms of $E[X], E[Y]$. –  David Harris Jul 4 '12 at 18:47
Are you allowing degenerate discrete distributions? in that case or if you allow X concentrated on [0,a] then E[XY] cannot be greater than E[X] which can be zero in the first case and less than a in the second. –  Michael Chernick Jul 4 '12 at 18:57
If $X \ge Y^k$, then $XY \ge Y^{k+1}$ and so $E[XY] \ge E[Y^{k+1}]$. I don't know what to do with $E[Y^{k+1}]$ though. –  echoone Jul 4 '12 at 19:06
echoone: $E[Y^{k+1}]\geq E[Y]^{k+1}$ by Hölder's inequality, and this is sharp if we ignore $E[X]$. –  Colin McQuillan Jul 4 '12 at 20:26