Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 39 (Advanced Financial Models)

A one-period arbitrage is a portfolio of initial-information-measurable holdings $(h_B,h_S)$ with value $V_t=h_B{B}_t+h_S{S}_t$ such thatThe holdings are fixed during the period, so this is a self-financing strategy. A portfolio costing a negative amount with nonnegative terminal value also gives an arbitrage: invest its initial surplus in the strictly positive bank account to obtain a zero-cost portfolio with a strict gain. The same definition applies when further traded assets are added.

An equivalent martingale measure is a probability measure $Q$ equivalent to the objective measure $P$, meaning the two have exactly the same null events, under which the discounted stock ${\tilde{S}}_t=S_t/B_t$ is a martingale. In this one-period model the condition isThe discounted bank account is identically one. Integrability of the discounted stock is part of the martingale condition. In an augmented market the discounted price of every additional asset must satisfy the same condition.

Suppose $Q$ is an equivalent martingale measure and a zero-cost self-financing strategy has holdings $(h_B,h_S)$. Its discounted terminal value has conditional expectationIf it were an arbitrage, $V_1/B_1$ would be nonnegative because $B_1>0$. Equivalence implies that it is strictly positive on an event of positive $Q$-probability. A nonnegative random variable with zero conditional expectation must be zero almost surely, a contradiction. Thus $\boxed{\text{an equivalent martingale measure excludes arbitrage}}$. The calculation uses the usual integrable one-period strategies, and also works conditionally for finite initial-information-measurable holdings.

The European call option pays one in the state $S_1=5$ and zero in the other two states. Short one call, buy one third of a stock, and borrow one half unit of cash. Writing the option quote as $c=1/2$, the holdings and initial value areThe terminal value is $-1/2+S_1/3-(S_1-4{)}^{+}$. At stock prices $5,3,2$ it is respectivelyIt is strictly positive in every state, so this portfolio is an arbitrage.

Let $c$ be the initial call price. Write the risk-neutral probabilities of stock prices $5,3,2$ as $q_5,q_3,q_2$. The stock's martingale equation and the normalization give $q_2=2q_5$ and $q_3=1-3q_5$. The call pays only in the first state, so its martingale equation is $q_5=c$. ThusAll three objective probabilities are positive. This defines an equivalent martingale measure precisely when $0<c<1/3$, and part (iii) proves that every such price excludes arbitrage.

To prove necessity without appealing to an existence theorem, consider the other prices explicitly. If $c\le 0$, buy one call and hold $-c$ units of cash. Its initial value is zero and terminal value is $(S_1-4{)}^{+}-c$, nonnegative and positive with positive probability. If $c\ge 1/3$, take holdings $(h_B,h_S,h_C)=(c-1,1/3,-1)$. The initial value is $c-1+1-c=0$ and terminal values at $5,3,2$ are $(c-1/3,c,c-1/3)$. They are nonnegative and the middle-state value is strictly positive. This includes the endpoint $c=1/3$. Consequently

A complete market is one in which every contingent claim in the pricing class can be replicated: for each finite maturity $T$ and terminal ${\mathcal{F}}_T$-measurable payoff $H$, there are an initial value and an admissible self-financing strategy whose terminal value is $H$ almost surely. It is enough for the complete market characterization to include all claims with bounded discounted payoffs. Thus every such payoff is attainable by trading the existing assets, rather than needing an additional independent security.

Suppose $Q_1,Q_2$ are equivalent martingale measures. Fix a finite maturity $T$ and an event $E\in {\mathcal{F}}_T$. Because the market is a complete market, replicate the claim $H=B_T{1}_E$, whose discounted payoff is bounded. For a self-financing strategy, the discounted value satisfiesUnder either pricing measure its conditional mean increment is zero, since $h_t^S$ is known at time $t$ and the discounted stock is a martingale. In the standard admissible discrete-time pricing class these finite-horizon values are integrable; a constant lower bound on discounted wealth also gives integrability successively from the conditional increment formula. Therefore replication and taking expectations giveThe replicating portfolio and its initial cost are the same for both measures, so $Q_1(E)=Q_2(E)$. Since $E$ was arbitrary, the measures agree on ${\mathcal{F}}_T$, for every finite $T$, and hence on the sigma-algebra generated by the market filtration. Thus $\boxed{\text{the equivalent martingale measure is unique}}$. This is the measure version of complete markets have unique state-price densities.

Let $Q$ be the unique equivalent martingale measure. The risk-neutral pricing formula isFor the usual nonnegative strike $K$ and maturities $u\ge t$, the monotonic bank account gives $K/B_u\le K/B_t$. ThereforeThe second inequality follows directly since the conditional expectation of a positive part is at least both zero and the conditional expectation of its argument; it is also conditional Jensen's inequality. Taking expectations proves $C(u,K)\ge C(t,K)$, the maturity monotonicity of calls with nonnegative strikes.

If $K_1\le K_2$, then $(S_T-K_1{)}^{+}\ge (S_T-K_2{)}^{+}$ pointwise; positive discounting and expectation prove monotonicity of a European call price in strike. For $0\le a\le 1$, the pointwise inequalitygives convexity of a European call price in strike after the same operations. HenceThe maturity assertion uses $K\ge 0$; it need not hold for a negative strike. For example, in the deterministic complete market $S_t=B_t=e^{rt}$ with $r>0$, a negative-strike call has price $1-K{e}^{-rT}$, which decreases with $T$. The monotonicity assertions are non-strict, as equality can occur.

Relative to a filtration $({\mathcal{F}}_t)$, a supermartingale is an adapted process $U_t$ with $\mathbb{E}|U_t|<\infty$ for every $t$ andBy the tower property this also gives $\mathbb{E}[U_u\mid {\mathcal{F}}_t]\le U_t$ for every $u\ge t$.

For a stopping time $\tau$, $U_{t\wedge \tau }$ is adapted: split according to $\{\tau =j\}$ for $j<t$ and $\{\tau \ge t\}$, all events belonging to ${\mathcal{F}}_t$. It is integrable because $|U_{t\wedge \tau }|\le {\sum }_{j=0}^t|U_j|$. Moreover,The indicator is ${\mathcal{F}}_t$-measurable, so conditioning and the supermartingale property giveThus $\boxed{(U_{t\wedge \tau }{)}_t\text{ is a supermartingale}}$. This proves stopping preserves supermartingales in discrete time without assuming $\tau$ itself bounded or integrable.

Use the tree's natural filtration. The stopping times take values in the process horizon $\{0,1,2\}$. Construct a dominating supermartingale $Y$ by backward comparison, the finite-horizon Snell envelope. At time two put $Y_2={\xi }_2$. At time one the continuation values areCompare these with immediate rewards $11$ and $7$ to obtain $Y_1=12$ on the upper branch and $Y_1=7$ on the lower branch. Finally setThese definitions explicitly give $Y_t\ge {\xi }_t$, $Y_0=\mathbb{E}{Y}_1$, and $Y_1\ge \mathbb{E}[Y_2\mid {\mathcal{F}}_1]$. Thus $Y$ is a supermartingale. By part (ii), its stopped process is a supermartingale, so for every allowed $\tau$,The reward process itself need not be a supermartingale; the constructed majorant is what proves the bound.

Continue at time zero. At time one stop on the lower branch and continue on the upper branch:This is a stopping time since the branch decision is measurable at time one. Its expected reward isIt therefore attains the upper bound in part (iii).

Use the usual Black-Scholes model with $\sigma >0$ and the filtration generated by $W$. SetFor $u>t$, independence and the Gaussian distribution exponential moment formula give ${\mathbb{E}}_P[Z_u/Z_t\mid {\mathcal{F}}_t]=1$. Thus $Z$ is a strictly positive martingale of mean one, defining an equivalent measure $Q$ on every finite horizon by $dQ/dP{|}_{{\mathcal{F}}_t}=Z_t$.

To verify the measure change directly, put ${\hat{W}}_t=W_t+\lambda t$. For any real $a$ and $h=u-t$,The conditional characteristic function is that of $N(0,h)$ and does not depend on past information. Therefore $\hat{W}$ has independent increments with Gaussian distribution and continuous paths, and is a Brownian motion under $Q$. Substituting $dW=d\hat{W}-\lambda dt$ gives $d{S}_t=S_t(rdt+\sigma d{\hat{W}}_t)$. Hence $S_t/B_t$ is a geometric Brownian motion that is a martingale and $Q$ is an equivalent martingale measure.

The form is also forced. By the Brownian martingale representation theorem, any density process that is a martingale has $d{Z}_t=a_{t}d{W}_t$. The change of measure identity makes $Z_t{S}_t/B_t$ a local martingale under $P$. Its drift, calculated by Itô formula, is $(S_t/B_t)[Z_t(\mu -r)+\sigma a_t]dt$, which must vanish. Thus $a_t=-\lambda Z_t$. Solving $d{Z}_t=-\lambda Z_{t}d{W}_t$ with $Z_0=1$ gives exactly the displayed stochastic exponential. Equivalence here is on each finite horizon, as usual for Black-Scholes pricing.

Let $\Phi$ denote the standard normal distribution cumulative distribution function, and $\varphi$ its density. Put $\tau =T-t$. For $K>0$ and $t<T$, defineUnder the risk-neutral measure, conditional on ${\mathcal{F}}_t$,Thus $Q(S_T<K\mid {\mathcal{F}}_t)=\Phi (-d_2)$. The identity $e^{az}\varphi (z)=e^{a^2/2}\varphi (z-a)$ for the standard Gaussian distribution probability density function givesSplit the payout according to whether the stock exceeds the floor and discount. The risk-neutral pricing process isIts discounted value is ${\mathbb{E}}_Q[{\xi }_T/B_T\mid {\mathcal{F}}_t]$, a true martingale. Thus adding this asset preserves the equivalent martingale measure and gives no arbitrage. This is the guaranteed terminal stock floor in the Black-Scholes model. For $K\le 0$, positivity of the stock makes the claim simply $S_T$, and the price is ${\xi }_t=S_t$.

Let $\varphi$ denote the standard Gaussian distribution probability density function. Directly from the definitions, $S\varphi (d_1)=K{e}^{-r\tau }\varphi (d_2)$. Differentiating the price in part (ii), the terms involving derivatives of $d_1,d_2$ cancel, yielding $V_S=\Phi (d_1)$. Therefore, in units of the two traded assets, the replicating strategy isIts value is $h_t^B{B}_t+h_t^S{S}_t=V(t,S_t)$. To check self-financing explicitly, differentiation also givesHence $V_t+rS{V}_S+\frac{1}{2}{\sigma }^2{S}^2{V}_{SS}=rV$. Applying Itô formula under the physical measure givesThus the holdings form a self-financing strategy and their terminal value is the required payout. Holdings at the single maturity instant can be assigned by their limiting values; the continuous-time trading gains do not depend on that assignment. If ${\pi }_t$ denotes wealth fractions instead, the stock fraction is $S_t\Phi (d_1)/V(t,S_t)$ and the bank fraction is $K{e}^{-r\tau }\Phi (-d_2)/V(t,S_t)$. For $K\le 0$, hold one stock and no bank-account units.

Let $Q$ be the given risk-neutral measure, and write $B_t=B_0{e}^{rt}$. Take $V$ to be classical on $t<T$ and continuous at its prescribed terminal value. By Itô formula and the pricing equation for a local volatility model,soThe discounted added asset is a local martingale, as is $S_t/B_t$, while $B_t/B_t=1$. Consequently $Q$ is an equivalent local martingale measure for all three assets.

Here is the arbitrage argument, including the usual admissible trading strategy condition. If $X$ is the discounted wealth of a self-financing strategy, it is a stochastic integral against the discounted asset prices, plus its initial value, and hence a local martingale. Assume $X_t\ge -c$ for a fixed $c$, as required for an admissible trading strategy. For a localizing sequence ${\tau }_n$, the stopped process $X^{{\tau }_n}$ is a martingale. Apply the conditional Fatou lemma to $X_{t\wedge {\tau }_n}+c\ge 0$: for $s\le t$, continuity and the stopped martingale identity give ${\mathbb{E}}_Q[X_t+c\mid {\mathcal{F}}_s]\le X_s+c$. Thus $X$ is a supermartingale. An arbitrage would have $X_0=0$, $X_T\ge 0$ and positive probability of $X_T>0$. Equivalence of $P,Q$ makes that probability positive under $Q$, contradicting ${\mathbb{E}}_Q[X_T]\le 0$. Therefore the augmented market has no admissible arbitrage.

Nonnegativity of $V$ gives the needed price process, but does not by itself make its discounted local martingale a true martingale. If $V$ is bounded, the latter follows by bounded localization, and its price is also $B_t{\mathbb{E}}_Q[g(S_T)/B_T\mid {\mathcal{F}}_t]$. The arbitrage proof above applies to the stated nonnegative classical solution without assuming that additional bound.

The delta replication from a local-volatility pricing equation uses the asset-unit holdingsTheir wealth is exactly $V(t,S_t)$. Moreover,where the final equality is the calculation in part (i). Thus the holdings are a self-financing strategy with terminal wealth $g(S_T)$, proving replication. Since $V\ge 0$, this replicating strategy has nonnegative wealth and is an admissible trading strategy. The equality of trading gains is unchanged by replacing the drift with its physical-measure value: the stock drift in Itô formula and in the stock holding changes by the same amount. If ${\pi }_t$ denotes wealth fractions, then, wherever $V>0$, its stock fraction is $S_t{V}_S/V$ and its bank fraction is $1-S_t{V}_S/V$.

The diffusion coefficient becomes $S\sigma (S)=\sqrt{S}$, so the stock is a driftless square-root diffusion, with zero taken as an absorbing boundary. Substitution of $V(t,S)=e^{A(t)S+B(t)}$ into the pricing equation for a local volatility model givesEquating the constant and linear terms and imposing the terminal payoff givesThus $1/A(t)=(1+t)/2$ and $B(t)=0$, and the price isThis is the exponential claim in a driftless square-root model. At the absorbing boundary its value is $1$, matching the payoff at zero.

The payoff in this part is unbounded, unlike the payoff assumed in part (i), so it is useful to verify that the proposed price is a finite conditional expectation. For $h>0$, starting from stock value $s$, defineThen $F_v+\frac{1}{2}x{F}_{xx}=0$, $F(h,x)=e^{-ux}$, and $0\le F\le 1$. By Itô formula, $F(v,S_v)$ is a bounded local martingale, hence a true martingale. It follows thatThis is also the Laplace transform of a sum of $N$ independent variables with exponential distribution of rate $2/h$, where $N$ has Poisson distribution of mean $2s/h$ and is independent of the summands: its transform is $\exp [(2s/h)((2/h)/(2/h+u)-1)]$. Uniqueness of the Laplace transform therefore gives the compound Poisson transition law of a driftless square-root diffusion. The exponential distribution jump has exponential moment $\mathbb{E}[e^Y]=(2/h)/(2/h-1)$ whenever $h<2$. Summing over the Poisson distribution yieldsFor the remaining horizon $h=1-t\le 1$, this is finite and, by the Markov property, is precisely ${\mathbb{E}}_Q[e^{S_1}\mid {\mathcal{F}}_t]=e^{2S_t/(1+t)}$. Thus the displayed process is indeed the risk-neutral pricing value and a true martingale. The replicating strategy from part (ii) has $h_t^S=[2/(1+t)]V(t,S_t)$ and $h_t^B=[1-2S_t/(1+t)]V(t,S_t)/B_t$.

Let $Q$ denote the given unique equivalent martingale measure. A zero-coupon bond paying one at $T$ has risk-neutral pricing valueWrite $h=T-t$. Split the integrated short rate into its known and future parts:By integrating the future Brownian motion increments in the opposite order, the last integral equals ${\int }_t^T(T-u)d{\hat{W}}_u$. It is independent of ${\mathcal{F}}_t$ and is a Gaussian random variable of mean zero and varianceThe Gaussian exponential moment formula $\mathbb{E}[e^{aZ}]=e^{a^2\mathrm{Var}(Z)/2}$ for centered $Z$ therefore gives the bond price in this Gaussian short-rate model with a deterministic shift:The instantaneous forward rate is the negative maturity derivative of the logarithm of the bond price. ConsequentlyIn particular, $P_t(t)=1$ and $f_t(t)=g(t)+\sigma {\hat{W}}_t=r_t$, as required. The derivative formula is pointwise for continuous $g$ and almost everywhere for locally integrable $g$.

Let $F_0(T)$ be the observed initial instantaneous forward rate curve. Since ${\hat{W}}_0=0$, part (i) gives $f_0(T)=g(T)-{\sigma }^2{T}^2/2$. Thus the forward-curve calibration of a shifted Brownian short rate is achieved by the deterministic choiceSubstitution gives $f_0(T)=F_0(T)$ at every maturity where the curve is defined classically. It also reproduces the corresponding initial zero-coupon bond curve:There is no shape restriction on the initial curve beyond the regularity needed to define these integrals and forward derivatives; for continuous curves the match is pointwise, and for locally integrable curves the forward identity has its almost-everywhere interpretation.