Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 15 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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.