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How to prove that:

$$ \left(\sum_{i=1}^n w_i n_i \sqrt{\dfrac{y_i(1-y_i)}{n_i+1}}\right)^2 \leq \dfrac{\left(\sum_{i=1}^n w_i n_i y_i\right)\left(\sum_{i=1}^n w_i n_i (1-y_i)\right)}{(\sum_{i=1}^n w_i n_i+1)} $$ where $w_i\geq0$, $\sum_{i=1}^n w_i=1$, $n_i>0$ and $y_i \in (0,1)$ for $i=1,\dots,n$, with $n>1$?

I have verified numerically that it should hold, but I cannot still find an elegant way to show it.

The formula comes from an inequality for the variance of a convex combination of beta-distributed variables.

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1 Answer 1

up vote 14 down vote accepted

Consider some random variables $X$ and $Y$ such that, for every $i$, $(X,Y)=(n_iy_i,n_i(1-y_i))$ with probability $w_i$. The OP asks a proof of an inequality equivalent to $$ E(g(X,Y))\le g(E(X),E(Y)), $$ where, for every nonnegative $x$ and $y$, $$ g(x,y)=\sqrt{\frac{xy}{x+y+1}}. $$ The second partial derivatives $\partial^2_{xx}g$ and $\partial^2_{yy}g$ are negative and the determinant of the Hessian matrix of $g$ is $(xy+x+y)/(4xy(x+y+1)^3)$, which is positive. Hence both eigenvalues of the Hessian matrix are negative, the function $g$ is concave on its domain, and Jensen's inequality yields the result.

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Very nice solution! However, concavity is not determined by the signs of the individual second derivatives; $\partial^2_{xx}g$ and $\partial^2_{yy}g$ could both be negative and yet the Hessian could have a positive eigenvalue. (The sign of $\partial^2_{xy}g$ doesn't affect the eigenvalues of the Hessian.) However, according to Wolfram, the determinant of the Hessian is $\frac{xy+x+y}{4xy(x+y+1)^3}>0$, so the function is indeed concave on its domain. –  joriki Mar 1 '11 at 19:53
    
@joriki: You are right, I modified my post, thanks. –  Did Mar 1 '11 at 20:33
    
Thank you very much for the help, it is a very elegant solution –  vatna Mar 2 '11 at 9:24

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