Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / F / F2 / c

Work on the positive orthant, where the real powers defining $g$ are smooth. Its closure under the constraint is the compact weighted simplex $\overline{M}=\{x_i\ge 0,\sum p_i{x}_i=1\}$, since $0\le x_i\le 1/p_i$. The product extends continuously there, is zero on its boundary when $n>1$, and is one at $(1,\dots ,1)$. Thus a maximum is attained in the positive interior.

At an interior maximizer the constraint gradient $(p_1,\dots ,p_n)$ is nonzero. The multiplier equations giveAll coordinates are equal, and the weighted constraint forces each to equal one. Therefore the maximum is one and $\boxed{g(x)\le 1}$ on $M$. For $n=1$ the constraint itself gives $x_1=1$, so the same conclusion holds directly. Equality occurs exactly at the all-ones point.

Set $A={\sum }_i{p}_i{a}_i>0$ and $x_i=a_i/A$. Then ${\sum }_i{p}_i{x}_i=1$, so part 1 gives ${\prod }_i(a_i/A{)}^{p_i}\le 1$. Since ${\sum }_i{p}_i=1$, this becomesEquality holds exactly when $x_i=1$ for all $i$, equivalently all the $a_i$ are equal. This is the weighted arithmetic–geometric mean inequality.