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Claim: Let $X$ be a random variable whose support is $[0,r]$. Then $\text{Var }X<\frac{r^2}{4}.$

I'm not sure how to prove this claim, but the intuitive idea is that you saturate the inequality by calculating the variance when $P(X=0)=P(X=r)=\frac{1}{2}$.

I want to find a bivariate generalization of the above theorem. Specifically, something like the following:

Claim: Let $X$ and $Y$ be random variables with a joint compact support of area $A$. Then $\text{Var }X \cdot \text{Var }Y - \text{Cov}^2(X,Y) < ???$

I would like to figure the value of '???' as a function of $A$ if it even exists. And also perhaps prove the claim if it is true. Can you help?

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Clearly by definition of variance, if you move some probability mass farther away from the mean then this will increase variance. So all the probability mass is split between $0$ and $r$. Let $p = p(r)$. Then the mean is $pr$ and the expectation of the square is $pr^2$. Thus you want to maximize $pr^2 - (pr)^2 = r^2p(1-p)$. You can show the maximum occurs at $p = 1/2$ in which case the variance is $r^2/4$. I'm not sure about the two variable generalization but it seems like the maximum should occur when the two variables are independent, i.e. your support is a rectangle. If that's true, then you get the same answer for your variance quantity if $A$ is fixed regardless of the side lengths of your rectangle, and I believe the answer for the maximum will be $A^2/16$.

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