Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 44 (Optimal Investment)

An affine function preserves mixtures, soThereforeThe positive mixture weight preserves the strict sign; the common component cancels. This is the independence axiom for lottery preferences for the affine utility on probability measures.

Indifference means equality of the representing utility values. Since $U_0(\lambda )>U_0(\mu )>U_0(\nu )$, setApplying the affine function identity gives $U_0(p\lambda +(1-p)\nu )=U_0(\mu )$, so $\mu \sim p\lambda +(1-p)\nu$. This proves mixture solvability of affine preferences. The displayed weight is unique because the two endpoint utilities differ.

First, equality of two $U_0$ values forces equality of the corresponding $V_0$ values: neither strict comparison holds for $U_0$, and the same is then true for $V_0$ in both directions. Thus $V_0=g\circ U_0$ for a well-defined strictly increasing function $g$ on $J=U_0(\mathcal{P})$. Since $\mathcal{P}$ is a convex set and $U_0,V_0$ are affine functions, $J$ is a real interval andIf $J$ contains $u<v$, let $a=[g(v)-g(u)]/(v-u)>0$ and $b=g(u)-au$. The mixture identity gives $g(w)=aw+b$ for $w\in [u,v]$. For $w>v$, express $v$ as a convex combination of $u$ and $w$ and solve the same identity for $g(w)$; for $w<u$, express $u$ as a mixture of $w$ and $v$. Hence the formula holds on all of $J$, not merely between the chosen anchors. We obtain positive affine uniqueness of affine preference representations:If $U_0$ is constant, the same comparisons force $V_0$ to be constant; choose $a=1$ and the appropriate $b$. If the probability-measure set is empty, the assertion is vacuous. These degenerate cases do not require distinct anchors.

Define $U(x)=U_0({\delta }_x)$, where ${\delta }_x$ is the Dirac measure at $x$. For the full sigma-algebra of subsets of a finite $E$, affineness directly gives $U_0(\lambda )={\sum }_{x\in E}\lambda (\{x\})U(x)$.

The statement also holds for an arbitrary sigma-algebra on a finite set. List each atom of a sigma-algebra as $A_1,\dots ,A_r$ and choose $x_j\in A_j$. Points in one atom have identical Dirac measures on this sigma-algebra, so $U$ is constant on each atom and is measurable. Every probability measure decomposes as $\lambda ={\sum }_j\lambda (A_j){\delta }_{x_j}$, whenceThis is the expected utility representation on a finite measurable space. It does not silently assume that each singleton is measurable.

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.

Let ${\beta }_t$ be the cash position and ${\pi }_t$ the risky shares chosen using information at $t-1$. At the start of the period the portfolio costs ${\beta }_t+{\pi }_t{S}_{t-1}=X_{t-1}$. With cash price fixed at one, its ending value is ${\beta }_t+{\pi }_t{S}_t$. Rebalancing cannot add or remove wealth under a self-financing strategy, soThe holdings are predictable: they are chosen before the current price increment is observed. This is the discounted wealth equation in discrete time with cash as numéraire.

Take the nondegenerate case $\sigma >0$ and an integer trading horizon $T$. For exponential utility, maximize the negative of the exponential loss, equivalently minimize its expectation. Conditional on past information, a predictable choice $\pi$ hasComplete the square:Thus the minimum conditional multiplier is $e^{-\kappa }$ with $\kappa ={\mu }^2/(2{\sigma }^2)$. Backward induction starts with terminal loss $e^{-\gamma x}$ and gives the minimal continuation loss $e^{-\gamma x-\kappa (T-t)}$. Independence of the next increment justifies the same conditional minimization for every history, not just deterministic trading plans. Hence exponential-utility trading with Gaussian increments givesThese are constant numbers of shares, not constant wealth fractions. Strategies with infinite exponential loss have expected utility $-\infty$ and cannot improve this finite value. If $\sigma =0$, the usual formula is inapplicable: with $\mu =0$ trading has no effect, while a nonzero deterministic increment admits unbounded riskless gains and no finite maximizing position.

For a predictable position $\pi$, the current net wealth increment $\pi \Delta S_t+Y_t-y$ is conditionally Gaussian with mean $\pi \mu +a-y$ and variance ${\pi }^2{\sigma }^2+2\pi \rho b\sigma +b^2$. Its exponential-loss multiplier isDifferentiate its quadratic exponent. The unique minimizing position isThe first term is the speculative demand; the second offsets the part of the income correlated with the tradable price increment. Independent period vectors and backward induction justify using this same position at every date. Completing the square gives minimal exponent $-\kappa -\gamma d(y)$, whereTherefore hedging a Gaussian income stream with exponential utility has valueThe residual variance is $b^2(1-{\rho }^2)$; it disappears for perfectly correlated income.

LetPart (c) has $d(y)=\overline{y}-y$. Comparing the optimized values with and without the contract, $V_0^{\mathrm{swap}}=-e^{-\gamma x-T\kappa -\gamma T(\overline{y}-y)}$ and $V_0=-e^{-\gamma x-T\kappa }$, shows that $y>\overline{y}$ makes the swap value strictly more negative. Thus The investor prefers not to enter when $y>\overline{y}$. He is indifferent at equality and prefers entry for a smaller payment.

The periodic utility indifference payment for Gaussian income equals the mean income less the cost of hedging its correlated component and the exponential-utility penalty for the residual unhedgeable variance. Comparing unoptimized strategies would not justify this price threshold.

The marked-to-market value of the current holdings is $X_t=H_t\cdot P_t$. In a self-financing portfolio, purchases are paid for by sales within the portfolio, so rebalancing itself changes no net wealth. Asset price changes contribute $H_t\cdot d{P}_t$, and nonnegative consumption removes $c_{t}dt$ from wealth. HenceThe holdings must be predictable and stochastically integrable. The second identity is the self-financing definition with withdrawals; it is not obtained by incorrectly setting the changes in the holdings themselves to zero.

Apply the Itô product rule and the quadratic covariation rule for a stochastic integral. The finite variation consumption term has zero quadratic covariation with $Z$, soUsing $X=H\cdot P$ and the wealth equation,Thus the deflated wealth equation with consumption isNo finite variation assumption on $H$ is needed: the self-financing equation and the stochastic-integral quadratic covariation rule already incorporate the portfolio's financing constraint.

Use the usual normalization $Z_0=1$, finite deterministic initial wealth $x=X_0\ge 0$, positive state-price density, and nonnegative consumption. The deflated asset prices $ZP$ are local martingales, so predictable stochastic integrability makes $M=\int H\cdot d(ZP)$ a zero-starting local martingale. Integrating part (b) givesA nonnegative local martingale is a supermartingale: localize to martingales, apply Conditional Fatou lemma, and obtain the conditional expectation inequality and integrability. Therefore $N$, and hence $M=N-x$, are supermartingales. In particular,This is supermartingale control of deflated consumption gains. Nonnegative consumption is essential to the lower bound; it is the ordinary meaning of a consumption rate here. Wealth nonnegativity alone would not supply that bound if arbitrary signed cash flows were allowed.

The supermartingale conclusion givesDiscard the nonnegative terminal-wealth term and let $t\to \infty$. The cumulative deflated consumption increases, so monotone convergence theorem yields the state-price budget constraintThe normalization is relevant: with $Z_0\ne 1$ the right side is $Z_0{X}_0$; with random integrable initial deflated wealth its expectation is the unconditional budget bound.

Write $D=\mathbb{E}{\int }_0^{\infty }{e}^{-bt/R}{Z}_t^{1-1/R}dt$ and use the product measure $d\mathbb{P}dt$. For $0<R<1$, the conjugate exponents are $p=1/(1-R)$ and $q=1/R$. Factor the utility integrand asThe Holder inequality on this product measure givesDivide by $1-R$ to obtain the Hölder bound for discounted CRRA consumption:For a finite right side this is ordinary Hölder; on the infinite measure space one can first truncate time and then pass by monotone convergence theorem. If $D=\infty$ and $X_0>0$, the bound is valid but uninformative. If $X_0=0$, the budget forces $c=0$ almost everywhere, so the utility integral is zero; handle this separately instead of writing an undefined $0\cdot \infty$.

When $D$ is finite and the bound is attained, equality requires full budget use and $c_t=(X_0/D)e^{-bt/R}{Z}_t^{-1/R}$. This is the equality pattern of Hölder, not an assertion that every market can finance that process.