Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 19 / c

Introduce a second multiplier $b$ and useThe stationary equations become $-\log x_i-1-\lambda -b{y}_i=0$. Writing $a=e^{-1-\lambda }>0$ givesThe two constraints determine the constants throughEquivalently, $a=1/Z(b)$ with $Z(b)={\sum }_i{e}^{-b{y}_i}$, and $-d\log Z/db=Y$. This is the maximum entropy distribution with a prescribed finite mean.

For nonconstant $y_i$, a strictly positive solution exists exactly when ${\min }_i{y}_i<Y<{\max }_i{y}_i$. The mean's derivative with respect to $b$ is $-{\mathrm{Var}}_b(y)<0$, so $b$ is unique in that case. At either extreme mean no strictly positive solution exists. If all $y_i=y_0$, feasibility requires $Y=y_0$ and the extra constraint is redundant: $x_i=1/n$, while $a,b$ are not separately identifiable. The negative entropy Hessian ensures that any feasible interior stationary point is the unique constrained maximum.