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http://math.stackexchange.com/questions/104114/is-the-following-function-convex-cap

Let $p=(p_1,\ldots,p_n)$ be a given nondegenerate (i.e., all $p_i> 0$) probability distribution on $n$ points. Define the following function $$\Phi(b_1,\ldots,b_n)=\frac{\left(\sum_{k=1}^n b_k \sqrt{p_k}\right)^2}{\left(\sum_{k=1}^n b_k^2/k\right)}$$ whose domain is the simplex $\{(b_1,\ldots,b_k):\sum_{k=1}^n b_k=1,\quad b_k\geq 0,k=1\ldots n\}.$

Claim 1: This function is convex-$\cap$ and attains a unique maximum on its domain.

I believe that an analytic solution to the maximization problem may be possible, but I am not sure if Lagrange multiplier methods are appropriate, not being an expert on optimization.

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1 Answer 1

1. By Cauchy-Schwarz, $${\left(\sum_{k=1}^n b_k \sqrt{p_k}\right)^2}\le{\left(\sum_{k=1}^n b_ k ^ 2 /k \right)} {\left(\sum_{k=1}^n k\, p_k\right)}\, ,$$ with equality if and only if $b_k=\lambda k \sqrt{p_k}$ for any $k$ and for $\lambda\in\mathbb{R}\\ .$ Here, the constraint implies $\lambda:=1/\sum_{k=1}^n k \sqrt{p_k}\, , $ giving the unique maximum point of $\Phi$ on the simplex; the maximum value is $\sum _ {k=1}^n k \, p_k \, ,$ the first moment of the distribution.

2. The function $\Phi$ is certainly concave on some convex nbd of its maximum point, but it is never concave on the whole simplex if $n\ge2$. Let $u(t):=\Phi(t,1-t,0,\dots,0)$. One finds $u''(0)+4u''(1)=24\sqrt{p_1 p_2 } > 0 $, so $u''$ is strictly positive either at $t=0$ or $t=1$ (which one, depending on the ratio $p_1/p_2$). This implies that $\Phi$ is not concave on the simplex.

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Thanks Pietro for the very nice answer. kodlu –  kodlu Feb 3 '12 at 1:02

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