Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 42 (Optimal Investment)

Let $I=(U^{\prime }{)}^{-1}$ be the inverse marginal utility. The derivative $U^{\prime }$ is continuous and strictly decreasing: concavity makes it nonincreasing, and equality at two distinct points would make $U$ affine between them, contradicting strict concavity. The Inada conditions therefore make $U^{\prime }$ a bijection from $(0,\infty )$ onto $(0,\infty )$. For each $y>0$, the derivative of $U(x)-xy$ is positive below $I(y)$ and negative above it. Consequently its unique global maximum is finite andThere is no need to differentiate $I$ to find the first derivative of this utility conjugate. Comparing the maxima at $y$ and $y+h$ gives, for $h>0$,Continuity of $I$, together with the corresponding inequalities for $h<0$, givesThus $V$ is strictly decreasing and strictly convex, since its derivative is strictly increasing. This proves all the asserted properties except the second differentiability, which does not follow from the printed hypotheses.

Here is an explicit Inada utility with vanishing curvature. PutThe positive integral defining $p$ makes $U$ strictly increasing. Also $p^{\prime }=-w$ is negative except at one point, so $p$ is strictly decreasing and $U$ is strictly concave. Near zero $w(s)$ is asymptotic to $s^{-2}$, and at infinity it is asymptotic to $s^{-2}$; thus $p(0+)=\infty$ and $p(\infty )=0$. This smooth utility function satisfies every printed positive-domain assumption, but $U^{\prime \prime }(1)=0$. At $y_0=p(1)=3-4\log 2$, differentiability of $I$ would imply, by differentiating $p(I(y))=y$,which is impossible. Hence $V^{\prime }=-I$ has no finite derivative there. More explicitly, $p(x)-p(1)=-(x-1{)}^3/12+o((x-1{)}^3)$, so its inverse function has a cube-root singularity. The twice-differentiable conclusion is false as printed. With the additional assumption $U^{\prime \prime }(x)<0$ everywhere, the inverse function theorem proves the intended dual differentiability with nonvanishing utility curvature:

For the financial assertions, use the positive-wealth admissible trading strategy class appropriate to the positive-domain utility function, and expectations for which the expressions are defined. A normalized terminal state-price density satisfies $Z_N>0$, $\mathbb{E}{Z}_N=1$, and defines an equivalent martingale measure $Q$ by $dQ=Z_{N}dP$. A self-financing portfolio obeyswhere the holdings are measurable before the next price increment. Under $Q$ the gains are a martingale when integrable; for the usual lower-bounded admissible class, localization gives a supermartingale. In particular the state-price budget constraint is $\mathbb{E}[Z_N{X}_N]\le x$. The positive-domain qualification matters: assumptions only on $U$ at positive arguments do not control its values at nonpositive wealth.

The definition of the utility conjugate gives the pointwise inequalityTaking expectations, using the state-price budget constraint, and then taking the supremum over strategies gives $u(x)\le \mathbb{E}V(y{Z}_N)+yx$ for every $y>0$ and every state-price density. Taking the two infima proves the utility duality with martingale deflators upper bound:If an expectation is infinite, use the same inequality in the extended-real sense whenever the negative parts make it well defined; no subtraction of two infinities is needed.

A complete market is one in which every terminal contingent claim in the specified claim class is attainable by a self-financing portfolio. Bounded terminal claims suffice for the uniqueness argument. In the standard finite-horizon discrete-time market, absence of arbitrage gives existence of an equivalent martingale measure by the fundamental theorem of asset pricing. To prove uniqueness, let $Z_N$ and ${\tilde{Z}}_N$ be two normalized state-price densities. Every bounded claim $H$ has a replicating price $p(H)$, and its replication has the same price under either equivalent martingale measure:Take $H=1_A$ for every terminal event $A$. The integrable difference $Z_N-{\tilde{Z}}_N$ has zero integral over every event, hence vanishes almost surely: its positive and negative sets would otherwise give a nonzero integral. The normalized terminal state-price density is unique up to null sets. Its density process is then uniquely $Z_n=\mathbb{E}[Z_N\mid {\mathcal{F}}_n]$. This uses the usual pricing replication class, where bounded claims have bounded replicating value processes; arbitrary doubling strategies are not an admissible definition of completeness.

For logarithmic utility, direct maximization gives $V(y)=-\log y-1$. In the complete market, write $Z=Z_N$. The unique-density utility conjugate bound is minimized at $y=1/x$, givingThe candidate terminal portfolio wealth is $H^{\ast }=x/Z$, whose cost is exactly $\mathbb{E}[Z{H}^{\ast }]=x$. If completeness includes nonnegative $Q$-integrable claims, it is replicable and attains the bound. There is also an approximation proof using only bounded-claim completeness. DefineThen $0<b_n\le 1$, $b_n\uparrow 1$, $H_n$ is positive and bounded, and $\mathbb{E}[Z{H}_n]=x$. Replicating $H_n$ therefore gives an admissible terminal claim of initial cost $x$. Since $\mathbb{E}(\log Z{)}^{+}\le \mathbb{E}Z=1$, the negative parts of $-\log (Z\vee n^{-1})$ have an integrable common bound. Monotone convergence after adding this lower bound givesConsequently the logarithmic terminal wealth in a complete market isThe value may be $+\infty$ if $\mathbb{E}(\log Z_N{)}^{-}=\infty$; the displayed formula remains meaningful, while attainment by $H^{\ast }$ depends on the precise admissible claim class.

Write $A=\sigma {\sigma }^T$ and let $V(x,t)$ denote the optimal remaining expected reward from positive portfolio wealth $x$ at time $t$. The Hamilton-Jacobi-Bellman equation and terminal condition areThe controls are progressively measurable, the investment integrals exist, consumption is nonnegative and locally integrable, and the resulting portfolio wealth stays in the utility domain. At zero consumption use the right limit of the utility function if needed. This is the usual admissible trading strategy formulation; without restrictions preserving the domain the reward need not be defined.

For the verification bound, assume $V$ is a nonnegative classical solution, continuous through the terminal time, with one time derivative and two wealth derivatives in the interior. Fix any admissible controls and putThe Itô formula givesThe drift is nonpositive because the actual controls cannot exceed the supremum in the Hamilton-Jacobi-Bellman equation. Localize by stopping times on which the stochastic integral is a true martingale, the derivatives and wealth stay in their interior domain, and the accumulated reward is finite. For each such stop ${\tau }_n$,Both utilities and $V$ are nonnegative. Thus the Fatou lemma, continuity, and the terminal condition giveFor admissible paths strictly positive on the closed horizon, the domain-localizing stops eventually reach $T$. If zero wealth is allowed, the same proof uses the continuous boundary extension of $V$ and the utilities there. Nonnegativity is what permits removing the stops without assuming the uncontrolled stochastic term already has zero expectation. This is verification by a nonnegative control supermartingale.

The square-root case needs an additional no-arbitrage qualification because the printed volatility matrix need not be invertible. If there is $v\in \ker {\sigma }^T$ with $v^T\mu >0$, investment $\theta =Kv$ has zero volatility and arbitrarily large positive drift. With any fixed positive consumption, sufficiently large $K$ preserves positive wealth and makes terminal square-root utility tend to infinity. For example, in one dimension $\sigma =0$, $\mu =1$, $\theta =K$, $C=1$ gives $X_s=x+(K-1)(s-t)$, so the value is infinite when $T>t$. A finite separated solution does not exist for all matrices allowed by the printed hypotheses.

The needed condition is $\mu \in \mathrm{Range}A$, equivalent to $\mu$ being orthogonal to $\ker {\sigma }^T$, because $A$ is a symmetric matrix and $\ker A=\ker {\sigma }^T$. Assume this condition and let $B=A^{+}$ be its Moore-Penrose inverse; if $A$ is invertible, simply use $B=A^{-1}$. Put $q={\mu }^{T}B\mu \ge 0$. Seek the square-root investment-consumption value in the formThen $V_x=g/\sqrt{x}$ and $V_{xx}=-g/(2x^{3/2})$. Completing the quadratic investment expression, using $AB\mu =\mu$, shows that its maximum is attained at ${\theta }^{\ast }=2xB\mu$ and has value $qg\sqrt{x}$:Any investment component in $\ker A$ has no effect. For consumption, set $z=\sqrt{c}$. Its contribution is the concave quadratic $2z-(g/\sqrt{x})z^2$, whose maximum is $\sqrt{x}/g$ at $c^{\ast }=x/g^2$. Substituting both maxima into the Hamilton-Jacobi-Bellman equation givesWith $h=g^2$, this becomes $h^{\prime }+qh+1=0$, $h(T)=1$. Solving this linear ordinary differential equation givesIn particular $h\ge 1$, so $g=\sqrt{h}$ stays positive, and the two controls areThese controls are admissible and actually attain the value. Indeed their portfolio wealth satisfiesSince $\Vert 2{\mu }^{T}B\sigma {\Vert }^2=4q$, the explicit positive solution starting at $(x,t)$ isIts finite-horizon moments of every positive order are finite. More directly,Differentiating $g(s)\mathbb{E}\sqrt{X_s}$ and using the equation for $g$ gives $\frac{d}{ds}(g(s)\mathbb{E}\sqrt{X_s})=-\mathbb{E}\sqrt{X_s}/g(s)$. Integrating from $t$ to $T$ therefore yieldsThis proves attainment without relying on an unjustified expectation of a local martingale. When $\mu$ is outside $\mathrm{Range}A$, the singular-volatility drift arbitrage counterexample applies instead.

Fix $t>0$. For $0\le s\le t$, the residual observed path isThis Brownian bridge is independent of $(\lambda ,Y_t)$. Indeed it is independent of $\lambda$, while its covariance with $W_t$ is $sI-(s/t)tI=0$; zero covariance implies independence in the joint Gaussian law. The same argument for every finite collection of bridge times, followed by continuity and generation from rational times, gives independence of the entire bridge. Since the full observed path is determined by its endpoint and this independent bridge, Brownian endpoint sufficiency for a constant drift gives

First suppose the prior covariance matrix $V_0$ is positive definite. The Bayes' theorem multiplies the prior Gaussian density by the conditional observation density, whose dependence on $\lambda$ is $\exp ({\lambda }^T{Y}_t-t|\lambda {|}^2/2)$. Thus the posterior exponent, up to terms not depending on $\lambda$, isCompleting the square gives the Gaussian Brownian drift filter:For invertible $V_0$ these are equivalently $V_t=(V_0^{-1}+tI{)}^{-1}$ and ${\lambda }_t=V_t(V_0^{-1}{\lambda }_0+Y_t)$. At $t=0$ they reduce to the prior parameters.

The boxed formulas remain valid for a singular positive semidefinite matrix $V_0$. To see this directly without inverse prior matrices, the joint Gaussian random vector $(\lambda ,Y_t)$ hasThe latter matrix is invertible for $t>0$. The conditional multivariate normal distribution therefore has meanand covariance matrixAlternatively, subtract this mean from $\lambda$; the resulting Gaussian random vector has zero covariance with $Y_t$ and is therefore independent of it, which proves these conditional parameters.

Write the proposed innovation process in the observable formIt is adapted to the observation filtration and continuous. To prove the martingale property, take $0\le u<v$. Future increments of $W$ are independent of the joint information generated by $\lambda$ and the past of $W$, soAlso the tower property of conditional expectation gives $\mathbb{E}[{\lambda }_s\mid {\mathcal{G}}_u]={\lambda }_u$ for $s\ge u$. Conditional integration is legitimate because $\mathbb{E}|{\lambda }_s{|}^2\le \mathbb{E}|\lambda {|}^2<\infty$. Subtracting the integral proves $\mathbb{E}[{\hat{W}}_v-{\hat{W}}_u\mid {\mathcal{G}}_u]=0$. Removing a continuous finite variation process does not change quadratic covariation, henceThe Lévy characterization of multidimensional Brownian motion now applies: a continuous zero-start martingale with this bracket matrix is a standard $d$-dimensional Brownian motion relative to its given filtration. The observable innovation is a Wiener process in the observation filtration.

Differentiating the posterior formulas yields $V_t^{\prime }=-V_t^2$ andFor the final expression, the printed inverse matrices require the extra assumption that $V_0$ is positive definite. Under that assumption put $A_t=V_t^{-1}=V_0^{-1}+tI$. The Itô formula givessince $A_t{V}_t=I$ and $\mathrm{tr}(A_t{V}_t^2)=\mathrm{tr}{V}_t$. On the other hand,The trace terms cancel in the innovation exponential for Gaussian drift filtering, leavingThus $Z_0=1$ and $Z$ is a positive stochastic exponential, hence a nonnegative local martingale. To obtain the requested supermartingale conclusion explicitly, choose localizing stopping times ${\tau }_n\uparrow \infty$ for which $Z^{{\tau }_n}$ is a true martingale. The Conditional Fatou lemma givesThe same argument at $u=0$ gives $\mathbb{E}{Z}_v\le 1$, supplying integrability. The process is a supermartingale in the observation filtration. No unproved exponential-integrability condition on the random drift is required.

For a singular prior the literal inverse-matrix expression in the PDF is undefined, but there is a natural extension with the same conclusion. Define, for fixed $y$,For positive definite $V_0$, a Gaussian integral shows this reciprocal is exactly the printed expression. For any positive semidefinite $V_0$, the integral is positive and finite, $L_t=-{\Delta }_{y}L/2$, and ${\nabla }_y\log L(t,Y_t)={\lambda }_t$, by the same posterior weighting. The Itô formula consequently gives $d\log L(t,Y_t)={\lambda }_t^{T}d{Y}_t-|{\lambda }_t{|}^{2}dt/2={\lambda }_t^{T}d{\hat{W}}_t+|{\lambda }_t{|}^{2}dt/2$. Its reciprocal satisfies the boxed stochastic differential equation and the same supermartingale proof. This completes the singular-covariance extension while explicitly separating it from the undefined printed formula.

Use the usual stochastic-integrability assumptions implicit in the price stochastic differential equation: the coefficients have progressively measurable versions, the volatility is locally square integrable, and initial prices are finite deterministic positive numbers. A no-consumption admissible trading strategy consists of predictable risky share holdings $\pi$ for which the price stochastic integral exists and the portfolio wealth of the self-financing portfolio isThe remainder $X_t-{\pi }_t^T{S}_t$ is held in the unit cash asset. Admissibility requires $X_t\ge -a$ throughout the horizon for some deterministic finite $a$; the more restrictive nonnegative-wealth convention also works. A finite-horizon arbitrage is such a strategy with $x=0$, $X_T\ge 0$ almost surely and $P(X_T>0)>0$. The lower bound is essential to exclude doubling strategies.

Interpret the stated uniform ellipticity as ${\sigma }_t{\sigma }_t^T\ge \epsilon I$ for a deterministic $\epsilon >0$. If $|{\mu }_t|\le K$, the market price of risk is bounded:The authoritative PDF puts both integrals inside the exponent. With that interpretation,On every finite horizon $T$, boundedness of $\lambda$ gives the Novikov condition:Hence the positive stochastic exponential $Z$ is a true martingale with $\mathbb{E}{Z}_T=1$. Define the equivalent probability measure $Q_T$ by $d{Q}_T=Z_{T}dP$. The Girsanov theorem says that $W_t^Q=W_t+{\int }_0^t{\lambda }_{s}ds$ is a Brownian motion under $Q_T$, up to time $T$. Consequentlyso the prices are local martingales under $Q_T$.

Every admissible zero-initial-capital self-financing portfolio has portfolio wealth which is therefore a $Q_T$-local martingale bounded below by $-a$. The shifted process $X+a$ is a nonnegative local martingale, hence a supermartingale by localization and the Conditional Fatou lemma. In particular ${\mathbb{E}}_{Q_T}{X}_T\le 0$. If the strategy were an arbitrage, nonnegativity of $X_T$ would force $X_T=0$ $Q_T$-almost surely. Equivalence of $P$ and $Q_T$ would then force $P(X_T>0)=0$, a contradiction. The market has no admissible arbitrage on any finite horizon. This proves the exclusion argument itself, rather than just asserting the fundamental theorem of asset pricing.

For the deflated price calculation, let ${\sigma }_{t,i}$ denote the column vector whose transpose is the $i$th row of ${\sigma }_t$. The Itô product rule includes the quadratic covariation termSince ${\sigma }_{t,i}^T{\lambda }_t={\mu }_t^i$, the full product equation isLocalization makes the last stochastic integral a true martingale on each stopped interval, proving each component of the deflated price vector is a local martingale under the original measure. This is the bounded-coefficient asset deflator identity; bounded volatility has not yet been used.

If $\sigma$ is bounded, then ${\sigma }_{t,i}-{\lambda }_t$ is bounded as well. Solving the last linear stochastic differential equation givesThe Novikov condition now holds for this stochastic exponential on every finite horizon, because its bracket is bounded by a deterministic constant times the horizon. ThereforeThe entire deflated price vector is a true martingale when volatility is bounded. This finite-horizon argument does not assert uniform integrability as time tends to infinity, which is not required. Integrable random initial prices can also be handled by the conditional martingale argument; without any integrability assumption on initial prices, a true-martingale assertion would not be meaningful.