Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 44 / 2

The Inada conditions give $U^{\prime }(0+)=\infty$ and $U^{\prime }(\infty )=0$. Differentiability and strict concavity make $U^{\prime }$ continuous and strictly decreasing, with range $(0,\infty )$. Hence the inverse marginal utility $I=(U^{\prime }{)}^{-1}$ exists. The unique maximizer in the utility conjugate is $x=I(y)$, andThe derivative formula does not require differentiability of $I$: compare the optimizing values at $y$ and $y+h$ to squeeze the difference quotient between $-I(y)$ and $-I(y+h)$, and use continuity of $I$. Thus the dual is continuously differentiable, strictly decreasing, and strictly convex, since $-I$ is strictly increasing.

For the requested second-derivative assertion, a curvature hypothesis is missing. Under the intended nondegeneracy $U^{\prime \prime }(x)<0$ for every $x>0$, inverse differentiation givesThis proves the intended dual differentiability with nonvanishing utility curvature. If $U$ is only twice differentiable, it gives pointwise twice differentiability of the dual; continuous second derivatives additionally follow when $U\in C^2$.

Literal strict concavity does not imply nonvanishing curvature. An explicit Inada utility with vanishing curvature isIts derivative is positive, strictly decreasing, tends to infinity at zero and to zero at infinity. It is smooth and strictly concave, but $U^{\prime \prime }(1)=0$. At $y_0=U^{\prime }(1)=1/3$, a finite derivative of $I$ would contradict differentiation of $U^{\prime }(I(y))=y$, giving $0\cdot I^{\prime }(y_0)=1$. Hence ${\hat{U}}^{\prime }$ is not differentiable there. As printed, the twice-differentiable-dual claim is false; it is valid with $U^{\prime \prime }<0$. The remaining differentiability, monotonicity and strict-convexity conclusions above hold under the printed hypotheses.

Admissible terminal wealth must lie in $(0,\infty )$, the domain of the utility; alternatively extend $U$ by $-\infty$ outside that domain. A one-period state-price density has $Z>0$ and $\mathbb{E}[Z{P}_1]=P_0$, componentwise. Therefore for $W=H\cdot P_1$, $\mathbb{E}[ZW]=H\cdot P_0=x$.

The dual definition gives the pointwise inequality $U(w)\le \hat{U}(z)+zw$ for $w,z>0$. Apply it with $z=yZ$ and take expectations on the finite state space:Equality holds precisely when $U^{\prime }(w)=z$, because that is the unique maximizing payoff in the dual definition. If the feasible portfolio $H^{\ast }$ has $U^{\prime }(H^{\ast }\cdot P_1)=y^{\ast }{Z}^{\ast }$, thenThe same right side bounds every competing feasible portfolio. Thus $H^{\ast }$ is optimal. This is the one-period marginal-utility certificate of optimality; it does not require the erroneous second-differentiability assertion in part (a).

Let $z_u,z_d$ be the state-price-density values at the higher and lower risky payoff. Pricing the two assets givesSolving yieldsBoth values are positive, and the two independent equations make the solution unique. The payoff matrix has nonzero determinant, so this is a complete two-state market. The actual Arrow state prices, including physical probabilities, are $q_u=5/9$ and $q_d=5/18$. Their sum is $5/6$, the discount factor. Normalizing gives risk-neutral probabilities $2/3,1/3$. Thus $Z$ itself is not a probability density of mean one: its mean is $5/6$ because the riskless asset earns interest.

The stated portfolio has initial wealth $5$ and terminal payoffs $W_u=4\cdot 6-3\cdot 7=3$ and $W_d=4\cdot 6-3\cdot 4=12$. For CRRA utility, $U^{\prime }(w)=w^{-R}$. The market is complete, and an interior optimum in its positive payoff domain has marginal utility proportional to the unique state-price density. ThereforeThe Lagrange multiplier is $y^{\ast }=9/(10\sqrt{3})$, and both state equations $U^{\prime }(W)=y^{\ast }Z$ are satisfied. Part (b) verifies sufficiency and strict concavity gives uniqueness of the optimal payoff. This illustrates risk aversion recovered from an optimal two-state payoff.