Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 29J (Stochastic Financial Models)

Set $R=1+r$, $a=\mu -R{S}_0$, and $M=(\gamma V{)}^{-1}$. The terminal wealth is $W_1=R{w}_0+{\theta }^T(S_1-R{S}_0)$, a normal random variable with mean $R{w}_0+{\theta }^{T}a$ and variance ${\theta }^{T}V\theta$. Its moment-generating function givesMaximizing this expected exponential utility is equivalent to maximizing the strictly concave quadratic form ${\theta }^{T}a-(\gamma /2){\theta }^{T}V\theta$. Since the nonsingular covariance $V$ is positive definite, the unique stationary point is the unique maximum:

With $x=0$, the budget constraint is $S_0^T\theta =w_0$. On this hyperplane the riskless term is constant, so maximize ${\theta }^T\mu -(\gamma /2){\theta }^{T}V\theta$. The Lagrange multiplier equation gives $\theta =M(\mu -\eta S_0)$. If $S_0\ne 0$, let $d=S_0^{T}M{S}_0>0$ and impose the budget constraint:Positive definiteness makes the objective strictly concave on the hyperplane, proving this stationary point is the global constrained optimum. If $S_0=0$, the cash-free constraint is feasible only when $w_0=0$; in that degenerate case the optimum is $M\mu$. The displayed quotient, including the one in the question, presupposes $S_0\ne 0$.

Assume $S_0\ne 0$ and keep $d=S_0^{T}M{S}_0>0$, $\eta =(S_0^{T}M\mu -w_0)/d$. The constraint $x\ge 0$ is $S_0^T\theta \le w_0$. If the unrestricted optimum in (i) lies in this half-space, it remains optimal. Its cash holding isThus it has strictly positive cash precisely when $\eta <R$. If $\eta >R$, the free optimum violates the constraint, so strict concavity and the Karush-Kuhn-Tucker conditions put the optimum on the boundary, where it is the cash-free solution in (ii). At equality the two solutions coincide. ThereforeIn particular,The multiplier for the inequality is $\max (0,\eta -R)$, explicitly checking the complementary-slackness condition.