Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 39 (Advanced Financial Models)

Use a strictly positive bank account as numéraire, and write all prices and payoffs in discounted units. There are $d$ risky assets with deterministic initial price vector $S_0$, terminal price vector $S_1\in L^2(P)$, and gains $Y=S_1-S_0$. A self-financing portfolio with initial capital $x$ and constant risky holdings $\theta$ has discounted terminal wealth $x+{\theta }^{T}Y$. The initial amount in the bank account is $x-{\theta }^T{S}_0$, so $x$ denotes total initial capital, not just the cash holding. Assume the usual absence of arbitrage, and hence existence of an equivalent martingale measure; wherever a general probability space is used below, explicitly assume that such a measure exists and makes the asset payoffs integrable.

The attainable claims are precisely the linear subspaceThe finite-dimensional subspace is closed, so every square-integrable contingent claim $H$ has a unique orthogonal projection onto $\mathcal{A}$. This is its optimal one-period least-squares hedging payoff. Holdings are unique after redundant assets have been removed. Putand suppose the remaining covariance matrix $C$ is invertible. Expanding the squared error givesFirst minimize over $x$, then complete the square in $\theta$. ThusThe residual $L=H-x_{\ast }-{\theta }_{\ast }^{T}Y$ satisfiesIndeed, every other squared error equals this minimum plusThis proves optimality, and also shows that zero error is equivalent to attainability of a European contingent claim. If initial capital $x$ is prescribed instead, the normal equations for linear least squares give $\mathbb{E}[Y{Y}^T]\theta =\mathbb{E}[Y(H-x)]$; one must not use the free-capital formula without this adjustment.

A dominated martingale measure is a probability measure $Q\ll P$ such that ${\mathbb{E}}_{Q}Y=0$. Its Radon-Nikodym derivative $Z$ satisfies $Z\ge 0$, ${\mathbb{E}}_{P}Z=1$, and ${\mathbb{E}}_P[ZY]=0$. An equivalent martingale measure additionally requires $Z>0$ almost surely. Every integrable attainable claim $H=x+{\theta }^{T}Y$ then has price $x={\mathbb{E}}_{Q}H$, independent of which such measure is chosen. For an unattainable contingent claim, different dominated martingale measures can give different prices; least-squares approximation chooses a projection criterion rather than exact replication.

Write $Y=m+M$, where $M=Y-m$ is the centered martingale part of the one-period gains. The minimal martingale measure in a one-period market changes the means of the traded gains to zero while preserving all square-integrable mean-zero directions $L$ orthogonal to $M$. Its candidate density isDirect computation gives ${\mathbb{E}}_P{Z}_{\ast }=1$ and ${\mathbb{E}}_P[Z_{\ast }Y]=m-C{C}^{-1}m=0$. If ${\mathbb{E}}_{P}L=0$ and ${\mathbb{E}}_P[ML]=0$, then ${\mathbb{E}}_P[Z_{\ast }L]=0$, as required. Conversely, an $L^2$ density with this preservation property is orthogonal to the orthogonal complement of $\mathrm{span}\{1,M_1,\dots ,M_d\}$, so it lies in that span. Its normalization and martingale constraints force exactly $Z_{\ast }$. Moreover,Every other square-integrable signed pricing density has the form $Z_{\ast }+D$, with $D$ orthogonal to $1$ and $Y$, hence to $Z_{\ast }$. Thereforeso the minimal density is also the minimum-norm one-period pricing weight. This coincidence concerns one period; it is not an unrestricted assertion about dynamic hedging models.

Positivity needs checking. If $Z_{\ast }>0$ it defines an equivalent martingale measure; if merely $Z_{\ast }\ge 0$, it defines a dominated martingale measure. Otherwise it defines only a signed martingale measure. For example, take $S_0=2$ and $Y=-1,1,3$ with physical probabilities $1/100,89/100,10/100$. Then $m=59/50$, $C=1019/2500$, and $Z_{\ast }=-4350/1019$ in the third state. There is nonetheless an equivalent martingale measure assigning probabilities $3/5,3/10,1/10$, whose mean gain is zero. Thus absence of arbitrage does not guarantee positivity of the least-squares density, and a least-squares initial cost need not be an arbitrage-free price for an unattainable payoff.

Finally prove completeness and uniqueness of dominated martingale measures. Suppose first that every bounded contingent claim is attainable. For any event $A$, replicate $1_A=x_A+{\theta }_A^{T}Y$. Under any dominated martingale measure, $Q(A)=x_A$. Thus all such measures agree on every event. Existence follows from our no-arbitrage assumption, giving uniqueness.

Conversely, fix an equivalent martingale measure $Q_0$, and let $k=\dim \mathcal{A}$. If the probability space admits $k+1$ disjoint events $A_1,\dots ,A_{k+1}$ of positive probability, choose a basis $a_1,\dots ,a_k$ of $\mathcal{A}$, including $1$. The $k\times (k+1)$ matrix with entries ${\mathbb{E}}_{Q_0}[a_i{1}_{A_j}]$ has a nonzero null vector $(c_j)$. Consequently the bounded, nonzero random variablesatisfies ${\mathbb{E}}_{Q_0}h=0$ and ${\mathbb{E}}_{Q_0}[hY]=0$. For $0<|\epsilon |<1/\Vert h{\Vert }_{\infty }$, defineThese are distinct equivalent martingale measures, because their densities are positive and their gain moments remain zero. Thus uniqueness rules out such a partition. A probability space with no $k+1$ disjoint positive-probability events consists, modulo null sets, of at most $k$ atoms: repeatedly split any non-atomic positive event to obtain a larger finite partition. On $N$ atoms the payoff space has dimension $N$, while $\mathcal{A}$ has dimension $k\le N$. Hence $N=k$ and $\mathcal{A}$ is the whole payoff space. Every contingent claim is attainable. This provesHere completeness means all bounded claims, or all square-integrable claims in the $L^2$ formulation. The existence/no-arbitrage hypothesis matters: on a one-state space, cash and a stock with a sure positive discounted gain span every payoff, but no dominated martingale measure can make that gain have zero mean.

Let the one-period riskless gross return be $R>0$, with $0<d<R<u$, and let the physical up probability be $\pi \in (0,1)$. The binomial market has risk-neutral probabilityUnder its equivalent martingale measure $Q$, successive multipliers $J_k$ are independent, taking $u,d$ with probabilities $q,1-q$. The risk-neutral pricing function has the extensionFor every fixed positive multiplier $A$, the function $s\mapsto f(As)$ is convex when $f$ is convex. Taking a positive weighted sum preserves convexity. More explicitly, for $0\le \lambda \le 1$,Thus the pricing extension is convex, and so are its values on the attainable stock-price grid.

The replicating portfolio in a binomial market holdsshares over the next interval. If $r+1<n$, put $A={\Delta }_{r+1}(us)$ and $D={\Delta }_{r+1}(ds)$. These are the secant slopes of $g_{r+2}$ on $[uds,u^{2}s]$ and $[d^{2}s,uds]$, respectively. For a convex function, the slope of a secant on a later adjacent interval is no smaller than that on the earlier interval: apply the defining convexity inequality to the middle point $uds$, then rearrange. Hence $A\ge D$. Using backward option pricing twice givesThe two coefficients are positive and sum to one. Therefore convex-payoff delta monotonicity in a binomial market yieldsIn particular an up move cannot reduce the stock holding. “Increases” here means nondecreases: an affine payoff gives equality, so strict increase cannot be claimed for every convex payoff.

For the expected utility maximization problem, let $P$ be the physical measure and $Z=dQ/dP$ on terminal paths. If a path has $j$ up moves, its binomial-market probability density isEvery terminal payoff is replicable in this complete market. Buying a positive payoff $H$ with initial wealth $w_0$ therefore imposes the state-price budget constraint ${\mathbb{E}}_{Q}H=R^n{w}_0$. Since the utility function is increasing, the full budget is used. Its derivative and second derivative arePut $a=\gamma /(\gamma -1)$. The first-order condition $v^{\prime }(H_{\ast })=\lambda Z$ for a positive multiplier $\lambda$ gives $H_{\ast }=C{Z}^{-a}$. The budget determines $C$ uniquely. Independence of the successive moves givesThus the optimal claim, as a function of the number of up moves, isBecause $S_n=S_0{u}^j{d}^{n-j}$ uniquely determines $j$, this is also a terminal-stock-price payoff, attainable by backward option pricing and claim replication. It rewards states relatively more probable under the investor's physical measure than under the pricing measure. If $\pi =q$, the optimum is the riskless payoff $R^n{w}_0$.

To prove global optimality rather than merely assert a stationary point, use the tangent inequality for the strictly concave function $v$:Taking physical expectations makes the last term zero for any fully financed competing claim, and nonpositive for any claim costing less. Therefore ${\mathbb{E}}_{P}v(H)\le {\mathbb{E}}_{P}v(H_{\ast })$, with equality only when $H=H_{\ast }$ almost surely. Finiteness causes no difficulty on the finite binomial tree, and $H_{\ast }>0$ in every terminal state.

Subtract the boundary slope from the Brownian motion: $X_t=B_t-bt$. Then $T_{a,b}$ is the first-passage time of $X$ to the fixed level $a$. The zero-drift Brownian reflection principle givesHere $\Phi$ is the standard normal distribution function. To change the drift to $-b$, use the Girsanov theorem on each finite horizon, with density $L_t=\exp (-b{B}_t-b^{2}t/2)$. On $\{T_a\le t\}$, conditional expectation at the stopped time gives $\mathbb{E}[L_t\mid {\mathcal{F}}_{T_a}]=L_{T_a}$, because the density is a true martingale. Since $B_{T_a}=a$, the drifted Brownian first-passage density is consequentlyThere may be additional mass at infinite hitting time; the displayed density only describes finite passage. With $e^{-\theta \infty }=0$, apply the given zero-drift Brownian first-passage Laplace transform:The zero-discount limit is $1$ for $b\le 0$ and $e^{-2ab}$ for $b>0$, so the hitting probability is less than one precisely when the line moves away with positive slope.

For a direct verification of the linear-boundary Brownian first-passage distribution, defineBoth terms tend to zero as $t\downarrow 0$. Let $\varphi$ be the standard normal density. The identityfollows by expanding the two squares. Differentiating the two normal probabilities then givesSince this is the previously derived hitting density and both distribution functions start at zero,This proof applies to every real $b$, including the defective distribution when $b>0$.

For the barrier digital put, use the dividend-free Black-Scholes model with constant interest rate $\rho$ and spot volatility $\sigma >0$. Under the risk-neutral measure,Set $T=t_0$ andReaching the barrier $c$ is equivalent to $W_t^Q$ reaching $a+bt$. The contingent claim pays one only on the survival event $\{T_{a,b}>T\}$. Risk-neutral pricing therefore givesThis is a path-survival price, not the price of a terminal digital put option. Continuity of the stock paths and the absence of an atom in the finite-horizon maximum imply that alternative touching conventions change no price. As $c\to \infty$, the value tends to $e^{-\rho T}$; as $c\downarrow S_0$, it tends to zero. Both checks agree with the survival interpretation.

Take a non-dividend-paying stock with physical dynamics $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$ and continuous-time bank account $B_t=e^{\rho t}$. Write $T=t_0$ and assume the candidate claim price $p(x,t)$ is $C^{2,1}$ before maturity, with enough growth control for the hedge to be admissible. The terminal condition is $p(x,T)=f(x)$.

The Itô formula gives the claim's price changeA self-financing portfolio worth $p$ with $\Delta$ shares has bank-account value $p-\Delta S_t$. Its gain isTo replicate the claim, the Brownian coefficients must agree, forcing the delta hedge $\Delta =p_x$. Equating the remaining drift coefficients cancels the physical drift $\mu$ and yieldsThe cancellation explains why the Black-Scholes equation contains the interest rate, not the physical stock drift. Conversely, if an admissible solution of this equation is given, holdings $p_x$ in the stock and $(p-x{p}_x)/B_t$ in the bank account have value $p$ and gains exactly $dp$. They are therefore a self-financing portfolio replicating the terminal payoff. Absence of arbitrage forces the claim to have this price. Merely decomposing the current value into arbitrary holdings would not justify self-financing; the gain identity and delta hedge are essential.

Now suppose the claim holder receives the cash rate $k(S_t,t)$ before $T$. The price $p$ is the ex-dividend value, so its total gain is $dp+kdt$. The replicating portfolio pays the same amounts out of its wealth, with no external injections. Thus its gain identity isAgain the Brownian coefficients force $\Delta =p_x$. Comparing drifts now gives the Black-Scholes equation with claim dividends:In particular the cash-flow source has a positive sign in the left-hand side. It is not an underlying-stock dividend yield and does not replace the drift $\rho x$ by a dividend-adjusted drift.

For another justification, under the risk-neutral measure $Q$, discount the claim's cum-dividend gain. The Itô formula shows thatexactly when the displayed equation holds. Subject to the usual integrability making this a true martingale, conditional expectation at maturity givesPositive interim payments therefore add positive value, which also checks the source sign. For instance, a constant payment rate $k_0$ and zero terminal payoff have price $k_0(1-e^{-\rho (T-t)})/\rho$, interpreted as $k_0(T-t)$ at $\rho =0$; substitution satisfies $p_t-\rho p+k_0=0$.

Let $T=t_0$, $\tau =T-t$, $x=S_t>0$, and $\sigma >0$. Use a dividend-free Black-Scholes model with constant interest rate $\rho$. The terminal payoff is assumed independent of the parameters being varied. In addition to twice differentiability, impose enough integrability to differentiate the pricing formula; for example, $f,f^{\prime },f^{\prime \prime }$ of polynomial growth suffice. Twice differentiability alone is not enough: the smooth payoff $f(x)=e^{x^2}$ has infinite expectation under every nondegenerate lognormal distribution.

Under the risk-neutral measure $Q$, future stock values conditional on $S_t=x$ have the formClaim replication, or equivalently the martingale property of discounted value in the complete market, givesThe replicating strategy holds the option delta $p_x$ shares and bank-account value $\beta =p-x{p}_x$. No physical drift parameter appears: the Girsanov theorem replaces the physical drift by $\rho$ under the pricing measure, and the replication argument cancels it directly. The risk-neutral pricing operator is linear in the payoff and preserves pointwise payoff inequalities.

Here is a verification of the Black-Scholes equation using the integral itself. Differentiation in spot givesDifferentiating with respect to $\tau$ givesFor $G(Z)=S_T{f}^{\prime }(S_T)$, the Gaussian integration by parts identity $\mathbb{E}[ZG]=\mathbb{E}[G^{\prime }]$ yieldsSubstitution cancels the $-{\sigma }^2/2$ term and provesSince $p_t=-p_{\tau }$, this is precisely the Black-Scholes equation. As $\tau \downarrow 0$, $M\to 1$ and growth control permits passage to the limit, giving $p(x,T)=f(x)$. The delta hedge then supplies the admissible replicating gains under the same regularity assumptions.

The Black-Scholes parameter sensitivities describe the dependence on the model parameters without assuming a particular payoff. If $f$ is increasing, then $p_x\ge 0$. If $f$ is convex, then $p_{xx}\ge 0$, so its price is convex in spot. The option gamma $p_{xx}$ measures the response of the option delta to spot changes.

For the interest rate, the distribution of the future stock also changes with $\rho$, so looking only at discounting gives the wrong general answer. Differentiation gives the option rhoThus a short bank-account position gives positive rate sensitivity, while a positive bank-account position gives negative rate sensitivity. For a cash payoff the result is negative; for the payoff $f(x)=x$ it is zero.

For spot volatility, differentiate the lognormal multiplier and then use Gaussian integration by parts:Consequently a convex payoff has nonnegative option vega. This monotonicity can also be understood by a mean-preserving increase of dispersion in the discounted stock. Strict positivity requires nonzero curvature seen with positive probability; an affine payoff has zero option vega.

Time enters through the remaining maturity $\tau$. Holding spot fixed, the Black-Scholes equation givesUnder the conventional assumption $\rho \ge 0$, a convex payoff and a short bond position $\beta <0$ imply $p_t\le 0$. The inequality is strict if $\rho >0$, or if $p_{xx}>0$ with positive volatility. Thus the price at fixed stock value is nonincreasing in calendar time, and nondecreasing in maturity, under these hypotheses. It is not a claim that the realized stochastic price path decreases: that path still has Brownian exposure $\sigma S_t{p}_{x}d{W}_t^Q$.

The printed conclusion needs the nonnegative-rate qualification; it is false for arbitrary negative rates. For $K>0$, the twice-differentiable convex payoff $f(x)=x-K$ hasThis is an exact counterexample, not a failure of the pricing formula. Also, with $\rho =0$ and an affine payoff, “decreasing” can only mean nonincreasing. Without payoff monotonicity or the specified bond-position sign, neither the spot nor rate nor time sensitivity has a universal sign. Any additional payoff parameter, such as a strike, is differentiated through $f$; pointwise payoff ordering, rather than a generic parameter rule, determines its price effect.

An interest rate model must describe a whole term structure of interest rates, not just one future scalar rate. Let $P(t,T)$ be the time-$t$ price of a unit zero-coupon bond maturing at $T$. For a sufficiently regular maturity curve, define the instantaneous forward rate and short rate bySince $P(t,t)=1$, integration givesHere $B$ is the continuous-time bank account. Thus modelling the two-parameter field $f(t,T)$ determines both the bonds and the discounting account. A Gaussian random field means that every finite collection of its values is jointly Gaussian; merely making each individual forward rate normally distributed is insufficient.

A useful arbitrage-free construction works under a money-market risk-neutral measure $Q$. Take independent Brownian motions $W^j$, deterministic coefficient sequences ${\sigma }_j(t,T)$, and a deterministic initial forward curve $f(0,T)$. With the coefficients viewed as vectors in a separable Hilbert space $H$, writeFor infinitely many factors, the last term means the $L^2$ limit of ${\sum }_j{\sigma }_j(t,T)d{W}_t^j$, not an assumption that the infinite vector of Brownian motions is itself $H$-valued. Require ${\int }_0^t\Vert \sigma (s,T){\Vert }_H^{2}ds<\infty$ and the maturity/time regularity needed for stochastic integration and exchange of integrals. Deterministic drift and volatility make this a Gaussian forward-rate field.

The field covariance is determined byFor every finite set of maturities, $(c_t(T_i,T_j))$ must be positive semidefinite. A one-factor model has rank-one instantaneous covariance, while additional or infinitely many factors permit richer correlations along the yield curve. Equivalently one can drive the curve with a maturity-correlated Brownian field $Z_t(T)$ satisfying $d⟨Z(T),Z(U){⟩}_t=c(T,U)dt$. A Brownian sheet gives the exampleMultiplication by deterministic maturity-dependent amplitudes, or deterministic increasing changes of the two sheet coordinates, gives further Gaussian covariance structures. A covariance specification alone is not sufficient: the forward drift must also obey absence of arbitrage.

To derive that drift restriction, putDifferentiate the moving lower endpoint in $\log P(t,T)=-{\int }_t^{T}f(t,u)du$ to obtainThe Itô formula therefore givesDiscounted bond prices must be martingales under $Q$. Their drift vanishes precisely when ${\int }_t^T\alpha (t,u)du=\frac{1}{2}\Vert \Sigma (t,T){\Vert }_H^2$. Differentiating in maturity gives the Heath-Jarrow-Morton model restrictionFor scalar amplitudes $v(t,T)$ multiplying a field with covariance kernel $c(T,U)$, this becomes $\alpha (t,T)=v(t,T){\int }_t^{T}v(t,u)c(T,u)du$. The extra maturity correlation factor cannot in general be dropped.

The complete Gaussian specification is consequentlyIts mean and covariance follow immediately from deterministic integration and the Itô isometry. The initial yield curve is an input, so fitting initial bond prices requires choosing $f(0,T)$ rather than changing a few scalar parameters. With the displayed drift, the discounted bond obeysOn finite horizons, deterministic square-integrable bond volatility gives an expectation-one stochastic exponential, hence a true discounted bond price martingale. This supplies an arbitrage-free pricing model for admissible portfolios in any finite collection of these bonds. Under a physical measure, a market price of risk vector changes the drift: if $d{W}_t^Q=d{W}_t^P+{\lambda }_{t}dt$, then the physical forward drift is $\alpha (t,T)+⟨\sigma (t,T),{\lambda }_t{⟩}_H$. Historical mean changes must therefore not be inserted in place of the risk-neutral drift restriction. Gaussianity under both measures additionally requires suitable deterministic drift changes; state-dependent risk premiums need not preserve Gaussianity under the physical measure.

There is also a useful general consistency test for a proposed Gaussian field beyond this independent-increment construction. DefineIn the filtration generated by a Gaussian field, conditional distributions of such linear functionals are Gaussian. For $s\le t\le T$, the martingale requirement is therefore exactlyThis follows by taking the conditional exponential moment and equating it to $e^{-L_s^T}$. It constrains both the deterministic mean and the covariance-induced conditional mean. Matching only unconditional discounted bond expectations does not suffice. For the Brownian-field construction, the Heath-Jarrow-Morton model restriction realizes this consistency condition through local drift cancellation.

Gaussian structure makes risk-neutral pricing particularly tractable. As a finite-factor example, take the Vasicek model under $Q$:Solving this Ornstein-Uhlenbeck process gives, for $u\ge t$,Hence, with $\tau =T-t$ and $B(\tau )=(1-e^{-a\tau })/a$, the conditional Gaussian integral $I={\int }_t^T{r}_{u}du$ hasThe exponential moment identity for a normal variable now gives the zero-coupon bond price $P(t,T)={\mathbb{E}}_Q[e^{-I}\mid {\mathcal{F}}_t]=A(\tau )e^{-B(\tau )r_t}$, whereThis outlines the derivation of the familiar exponential-affine form without assuming a deterministic future discount rate. More general forward-field models need not reduce to a finite-dimensional Markov short-rate process.

For another application, a call at time $T$ on a unit maturity-$U$ bond, with $t<T<U$ and strike $K>0$, is priced under the T-forward measure, using $P(t,T)$ as numéraire. The bond-price ratio $F_s=P(s,U)/P(s,T)$ is a martingale under that measure and has deterministic volatility $-(\Sigma (s,U)-\Sigma (s,T))$. It is consequently lognormal, with integrated log-varianceIntegrating the lognormal payoff gives the Gaussian bond-option formulaAt $V=0$ use $(P(t,U)-KP(t,T){)}^{+}$. The same idea prices other suitable Gaussian-rate derivatives through forward measures and normal exponential moments.

The attractions are exact initial-curve fitting, explicit covariance control across maturities, and tractable bond and option prices. The limitations are also structural: Gaussian short rates and instantaneous forward rates generally can be negative; finite-factor specifications restrict maturity correlations; and deterministic volatilities cannot reproduce arbitrary stochastic changes in rate dispersion. An infinite-factor model allows richer covariance but typically leaves risks that cannot be spanned by finitely many traded bonds. Thus Gaussianity is a useful modelling choice, not by itself a guarantee of positivity, market completeness, or absence of arbitrage.