Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 211 (Advanced Financial Models)

The natural filtration of $S$ is the natural Brownian filtration: since $\sigma >0$, the equation $W_t=(\log (S_t/S_0)-\mu t)/\sigma$ recovers the entire Brownian motion history from the stock history. For $s<t$, independent increments and the moment-generating function of a normal distribution giveAll these expectations are finite. Since $S_s>0$, the martingale identity holds for every $s<t$ exactly when the last exponential equals one. The required logarithmic drift isThis is the distinction between the drift of $\log S$ and the drift of $S$: here the latter is zero.

Write $h=T-t$. Conditional on ${\mathcal{F}}_t^S$, independent increments of the Brownian motion givewhere $Y$ has the standard normal distribution and is independent of ${\mathcal{F}}_t^S$. Factoring $S_t$ out of the positive part identifies the normalized Black-Scholes call function:The European call option payoff is integrable, because $0\le (S_T-K{)}^{+}\le S_T$ and $\mathbb{E}{S}_T=S_0$. The tower property of conditional expectation therefore proves that $C$ is a true martingale. At $t=T$, the boundary value $F(0,m)=(1-m{)}^{+}$ gives $C_T=(S_T-K{)}^{+}$. The answer is the conditional payoff martingale

First work with the filtration ${\mathcal{G}}_t={\mathcal{F}}_t^W\vee \sigma (1_{\{u\le \tau \}}:0\le u\le t)$, which records the Brownian motion and whether default has occurred. On $\{s>\tau \}$ all subsequent defaultable stock prices vanish. On $\{s\le \tau \}$, the memoryless property of the exponential distribution gives survival from $s$ to $t$ with conditional probability $e^{-\lambda (t-s)}$. The residual survival event is independent of the future independent increments of the Brownian motion. Thus, for deterministic $s<t$,Also $\mathbb{E}{\hat{S}}_t=e^{\lambda t}{e}^{-\lambda t}{S}_0=S_0$, so this is a true martingale, with no localization needed. Its natural filtration is contained in $\mathcal{G}$; applying the tower property of conditional expectation to the displayed identity yields the requested natural-filtration martingale property:The printed convention keeps the stock alive at $t=\tau$. It is a left-continuous convention at that one random time. Replacing it by $1_{\{t<\tau \}}$ gives the usual càdlàg zero-recovery default model; since $\mathbb{P}(\tau =t)=0$ at each fixed $t$, the fixed-time martingale calculations above are unchanged.

A martingale with the specified terminal value must equal its conditional expectation. If ${\hat{S}}_t=0$, default has already occurred and the terminal European call option payoff is zero. If ${\hat{S}}_t>0$, put $h=T-t$. Conditional on survival to $T$,where $Y$ has the standard normal distribution; the conditional probability of this survival is $e^{-\lambda h}$. The conditional expectation of the payoff in the enlarged filtration used in part (c) is consequentlyThis expression is already measurable with respect to the natural filtration of $\hat{S}$, so the tower property of conditional expectation gives the same value there. The survival factor cancels the compensating growth factor:The separate zero branch avoids division by zero. At $t=T$ it gives the required payoff, using $F(0,m)=(1-m{)}^{+}$; integrability follows from $({\hat{S}}_T-K{)}^{+}\le {\hat{S}}_T$.

The short rate is the limiting instantaneous forward rate at the present maturity. The continuously compounded zero-coupon bond price is obtained by integrating the instantaneous forward rate curve in its maturity variable. Both requested relations areIn particular $P(T,T)=1$, and $f(t,T)=-{\partial }_T\log P(t,T)$. The short rate here is instantaneous, rather than the one-period rate used in a discrete-time bank account.

For fixed maturity $T$, define $b_t^T={\int }_t^T\sigma (t,u)du$ and $I_t^T={\int }_t^{T}f(t,u)du$. The stochastic Fubini theorem and the moving lower endpoint giveThe factor $1/2$ is the integral over one of the two triangles in the square $[t,T{]}^2$. Applying the Itô formula to $P(t,T)=e^{-I_t^T}$, its quadratic-variation correction cancels that factor:Set $D_t=e^{-{\int }_0^t{r}_{s}ds}$, the reciprocal of the continuous-time bank account. The Itô product rule then givesIf $|\sigma |\le M$, then $|b_t^T|\le MT$ on this finite horizon. The Novikov condition holds, so this stochastic exponential is a true martingale, not merely a local martingale. The discounted price is thereforeThe authoritative PDF discounts to $t$ in this part. The TeX transcription's upper endpoint $T$ would include future short rates and is incorrect here.

Since $P(T,T)=1$, part (b) shows that the density process of the T-forward measure isIts terminal expectation is one and it is strictly positive, so the stated Radon-Nikodym derivative defines an equivalent probability measure. By the Girsanov theorem,is a Brownian motion under ${\mathbb{Q}}_T$. Substitution of $d{W}_t=d{W}_t^T-b_t^{T}dt$ cancels the entire instantaneous forward rate drift:Because the integrand is deterministic and bounded, its Itô integral is square-integrable on $[0,T]$. With the usual fixed initial instantaneous forward rate curve, the requested true martingale is

Use the Radon-Nikodym derivative at $T_1$ and the discounted bond price martingale from part (b):For the variance, put $b_t^i={\int }_t^{T_i}\sigma (t,u)du$ and $d_t=b_t^2-b_t^1={\int }_{T_1}^{T_2}\sigma (t,u)du$, for $t\le T_1$. The ratio $R_t=P(t,T_2)/P(t,T_1)$ has, by the Itô formula,Under the T-forward measure ${\mathbb{Q}}_{T_1}$, $d{W}_t=d{W}_t^{T_1}-b_t^{1}dt$, so its drift becomes $-d_t^2/2$. Since $R_{T_1}=P(T_1,T_2)$,The first two terms are deterministic; the Itô isometry supplies the variance of the last. The two requested answers areIn fact the terminal zero-coupon bond price has a log-normal distribution under this forward measure, with the deterministic negative half-variance correction ensuring the displayed mean.

Choose any strictly positive pricing kernel $Z\in \mathcal{Z}$. The assumed componentwise expectations and the linearity of expectation giveThe integrand is nonnegative, so its expectation is also nonnegative and hence zero. A nonnegative random variable with zero expectation vanishes almost surely. Since $Z>0$ almost surely, this forces $H\cdot P=0$ almost surely. Both conclusions follow:The strict positivity of the pricing kernel is essential: a kernel allowed to vanish could miss a positive payoff on its zero set. No normalization $\mathbb{E}Z=1$ is assumed in this abstract market; that normalization follows only if a unit-priced unit-payoff cash asset is included.

For fixed $\gamma$, differentiation under the integral sign is justified by bounded $P$ and $X$. At the unconstrained minimizer $H_{\gamma }$, the first-order condition isDefine the strictly positive, bounded pricing kernelThe first-order condition says $\mathbb{E}[Z_{\gamma }P]=p$, so $Z_{\gamma }\in \mathcal{Z}$. When taking the requested partial derivative, hold $H$ fixed and only afterwards set $H=H_{\gamma }$. Writing $A_{\gamma }=H_{\gamma }\cdot p-x$, we obtainThe assumed dual bound applies to this particular pricing kernel. ThusNo derivative of the map $\gamma \mapsto H_{\gamma }$ is needed; confusing a partial derivative with a derivative along the minimizing path would add an unnecessary hypothesis.

Let $M={\sup }_{\gamma >1}{F}_{\gamma }(H_{\gamma })<\infty$. Choose ${\gamma }_j\to \infty$. Boundedness of the minimizers and the Bolzano-Weierstrass theorem yield a subsequence, still indexed by $j$, with $H_{{\gamma }_j}\to H^{\ast }\in {\mathbb{R}}^n$. Each $H_{{\gamma }_j}$ is deterministic, hence so is $H^{\ast }$. The first exponential term givesand passing to the limit gives $H^{\ast }\cdot p\le x$.

To obtain the superhedging inequality, suppose instead that $\mathbb{P}(X>H^{\ast }\cdot P)>0$. There is then an $\epsilon >0$ such that the event $E=\{X-H^{\ast }\cdot P\ge \epsilon \}$ has positive probability. Let $\Vert P\Vert \le B$ almost surely. For all sufficiently large $j$,Consequently $X-H_{{\gamma }_j}\cdot P\ge \epsilon /2$ on $E$, givinga contradiction. The limiting portfolio meets both constraints:Bounded $P$ is used precisely to turn convergence of deterministic holdings into a uniform bound on the payoff error.

Let $h$ be the stock holding and let $c$ be the initial cost, so the cash holding is $b=c-10h$. Terminal portfolio wealth at stock price $s$ is $c+h(s-10)$. Dominating the European call option payoff at the three possible prices givesAdding the two endpoint inequalities yields $2c\ge 1$. Equality is attained by $h=1/2$ and $c=1/2$, which imply $b=-9/2$. The corresponding terminal wealth is $0,1/2,1$ at $s=9,10,11$, respectively, compared with the required payoff $0,0,1$. The cheapest super-replication strategy isThe middle-state excess shows why this is superhedging rather than exact claim replication. The endpoint bound proves global minimality, without relying on the physical state probabilities.

A complete market can replicate every contingent claim in its specified payoff class: for every terminal ${\mathcal{F}}_T$-measurable payoff $X$, there is an admissible self-financing portfolio whose terminal wealth equals $X$ almost surely. Equivalently, in an arbitrage-free finite-state discrete-time model the equivalent martingale measure is unique. In notation,Market completeness is a replication property. The following price comparisons also use the usual absence of arbitrage, which the source leaves implicit: without it the initial cost of claim replication need not be unique. For example, in a deterministic one-period model with cash worth one at both dates and a stock worth one initially but zero finally, all terminal claims are replicable using cash. Adding any stock holding changes the initial cost without changing the terminal payoff. This is a complete model with arbitrage, so it has no well-defined unique replication price.

Under the usual no-arbitrage interpretation from part (a), let $Q^N$ be an equivalent martingale measure for the strictly positive numéraire $N$. Then $M_t=S_t/N_t$ is a martingale. Define the discounted European call option payoff $Y_t=(S_t-K{)}^{+}/N_t=(M_t-K/N_t{)}^{+}$. Because $N_{t+1}\ge N_t$ and $K>0$, we have $K/N_{t+1}\le K/N_t$. The conditional Jensen inequality therefore givesFor the middle step, the strike $K/N_t$ is fixed conditionally on ${\mathcal{F}}_t$; alternatively use $\mathbb{E}[X^{+}\mid {\mathcal{F}}_t]\ge \max \{\mathbb{E}[X\mid {\mathcal{F}}_t],0\}$. Thus $Y$ is a submartingale. Risk-neutral valuation in units of $N$ gives $C(T,K)=N_0{\mathbb{E}}^{Q^N}{Y}_T$, with the usual fixed initial prices. ConsequentlyThe source's word “increasing” means nondecreasing: a European call option can have zero payoff at successive maturities, so strict increase is not guaranteed. Integrability is understood in the claim class for which the stated replication prices exist.

Let $b$ and $h$ be the numbers of units of the numéraire and the stock in a replicating strategy. The two terminal payoffs of the European call option are $2$ and $0$, soSolving these simultaneous equations gives $b=-6/35$ and $h=8/35$. The initial replication cost isAs an independent pricing check, let $q$ be the risk-neutral probability of the state $(15,20)$ under the numéraire measure. The discounted stock must have initial value one and terminal mean one:Risk-neutral valuation then gives $10(3/7)(2/15)=4/7$. The physical probabilities $1/2,1/2$ are not the numéraire-measure probabilities.

The positive-part function is convex, so pathwise Jensen inequality givesTo compare the Asian option price with European call option prices at different dates, carry each earlier payoff forward using the numéraire. Purchase $1/T$ of each replicating strategy for maturity $t$, and, when its payoff is received, reinvest it in $N$ until $T$. This is a self-financing portfolio, with terminal wealthThe first inequality uses $N_T\ge N_t$ and nonnegative payoffs. Since the Asian option is replicable in the complete market, absence of arbitrage makes its replication cost no larger than this superhedging cost. ThereforeEquivalently, divide the first pathwise bound by $N_T$, use $(S_t-K{)}^{+}/N_T\le (S_t-K{)}^{+}/N_t$, and take expectations under the numéraire equivalent martingale measure. Merely averaging earlier payoffs without reinvestment would miss the difference in payment dates.

Backward induction makes every $U_t$ adapted and integrable: the conditional expectation of an integrable random variable is integrable, and $|\max (a,b)|\le |a|+|b|$. The Snell envelope recursion directly gives $U_t\ge \mathbb{E}[U_{t+1}\mid {\mathcal{F}}_t]$, so $U$ is a supermartingale.

If $Z$ is a submartingale, iterating its defining inequality gives $\mathbb{E}[Z_T\mid {\mathcal{F}}_t]\ge Z_t$. Starting at $U_T=Z_T$ and applying the tower property of conditional expectation backwards showsThis last conditional expectation is a true martingale. The conclusions areIn the second case waiting until $T$ attains the optimal stopping value. It is not generally true that $U_t=Z_t$: that further equality holds when $Z$ itself is a martingale.

For the intended random walk model, take $S_0$ to be deterministic, or independent of all the independent and identically distributed random variables $Y_j=S_j-S_{j-1}$. If $\nu$ is their common probability distribution, then $Y_{t+1}$ is independent of ${\mathcal{F}}_t=\sigma (S_0,Y_1,\dots ,Y_t)$. This supplies the Markov property that the backward dynamic programming argument needs. Define the transition operatorand the deterministic optimal stopping value function recursively byAt any state where the conditional law is evaluated, independence givesBackward induction, starting with $U_T=f(S_T)$, proves the intended representationIntegrability along the actual random walk makes the recursion finite at the states used by that process, up to null sets. If finite real-valued value functions on all of $\mathbb{R}$ are intended, a convenient sufficient convention is statewise integrability: $\mathbb{E}|f(s+Y_1+\cdots +Y_k)|<\infty$ for every $s$ and $0\le k\le T$. Indeed each stopping value is bounded in absolute value by the expectation of the sum of the absolute rewards. This convention makes the displayed recursion finite and measurable everywhere; the source only assumes integrability from the given starting process.

The printed independence of the increments alone does not suffice if $S_0$ can reveal future increments. Here is a bounded finite-state counterexample, also with a convex reward. Let $T=2$, let ${\epsilon }_1,{\epsilon }_2$ be independent fair signs, and setThe two increments are exactly ${\epsilon }_1,{\epsilon }_2$, hence independent and identically distributed. But ${\mathcal{F}}_1$ already knows both signs, so $U_1=\max \{S_1^{+},S_2^{+}\}$. On the two positive-probability histories with $S_1=0$, its values are respectively $0$ and $1$. Thus no deterministic $V(1,0)$ can work. Adding independence of $S_0$ from the increments repairs this missing Markov property hypothesis.

Use the random walk interpretation and optimal stopping value function recursion specified in part (b). If $h$ is convex, translating and integrating its convexity inequality gives, for $0<\theta <1$,Thus the transition operator preserves convex functions. Also the pointwise maximum of convex functions is convex: each of two convex functions at an intermediate point is bounded by the same convex combination of their pointwise maximum at the endpoints. Starting with $V(T,\cdot )=f$, backward induction intherefore proves the asserted convexity in the intended model:The statewise integrability convention in part (b) gives finite convex functions on all real states. The same inequality holds for the canonical extended value function wherever the expectations are well-defined. Integrability only along one started process need not make that canonical function finite at unused states: for $T=1$, $S_0=0$, $f(s)=e^{s^2}$ and increment density proportional to $e^{-y^2-|y|}$, $\mathbb{E}f(S_1)<\infty$, but $\mathbb{E}f(s+Y_1)=\infty$ for $s>1/2$. This concerns the canonical recursion away from visited states, rather than the almost-sure identity for $U$. Arbitrary off-state versions of $V$ need not be convex; the recursive version is the one meant here. No zero-mean assumption on the increments was used, and convexity alone does not make $f(S_t)$ a submartingale for an arbitrary drift.

Apply the Itô formula to ${\xi }_t=V(t,S_t)$. The backward equation cancels its drift, leavingBounded $a$ and $V_S$ make this Itô integral square-integrable on the finite horizon; in addition $V$ itself is bounded. Hence $\xi$ is a true martingale, not just a local martingale, and ${\xi }_T=g(S_T)$. Taking conditional expectations givesThe stock is understood to be adapted to the stated Brownian filtration, with its initial value fixed there. This is the Feynman-Kac formula in the zero-potential, zero-drift case.

Differentiate the backward equation for $V$ with respect to the stock state, and put $w=V_S$. The product derivative of $a^2/2$ is $a{a}^{\prime }$, givingThe assumed smoothness and bounded derivatives put $w$ in the uniqueness class for the equation defining $U$. Therefore $U=V_S$, and ${\pi }_t=V_S(t,S_t)$ is the option delta. Substituting in the Itô integral from part (a) and using $d{S}_t=a(S_t)d{W}_t$ gives the self-financing representationThe boundedness hypotheses justify both stochastic integrals. This derivation does not divide by $a$ and remains valid even at states with zero diffusion amplitude.

The positive density process is the stochastic exponentialSince $a^{\prime }$ is bounded, the Novikov condition holds; $Z$ is a true martingale with $\mathbb{E}{Z}_T=1$ and $Z_T>0$. Thus it defines the stated equivalent probability measure. The positive sign in the exponent means that the Girsanov theorem givesUnder this change of measure, applying the Itô formula to ${\pi }_t=U(t,S_t)$ produces precisely the drift in its backward equation, which vanishes:Bounded $a$ and $U_S$ make $\pi$ a true martingale under $\hat{\mathbb{P}}$. Its terminal value is $g^{\prime }(S_T)$. ConsequentlyThe sign can also be checked before changing measure: under the original measure $d\pi =-a{a}^{\prime }{U}_{S}dt+a{U}_{S}dW$, while the quadratic covariation term in $d(Z\pi )$ is $Za{a}^{\prime }{U}_{S}dt$, exactly cancelling its drift.

Take the original measure as the risk-neutral measure for this zero-interest model, or interpret $S$ as a discounted stock price. ${\xi }_t$ is the arbitrage-free value of the European contingent claim with payoff $g(S_T)$, and ${\pi }_t$ is its option delta and replicating stock holding. Part (b) supplies the self-financing portfolio:With the cash asset identically one, its value is ${\xi }_t$ and its gains satisfy $d{\xi }_t={\pi }_{t}d{S}_t$, so it exactly replicates $g(S_T)$. This delta hedge is admissible since the assumed nonnegative $V$ gives nonnegative wealth. In this continuous-path setting the holdings can be taken predictable.

The auxiliary change of measure in part (c) expresses the option delta as an expectation of terminal payoff sensitivity. It changes the stock drift to $a{a}^{\prime }$ and is generally not the original risk-neutral measure for the zero-interest stock market. Its role is to value sensitivity under the tilted state dynamics, while the actual replicating strategy trades under the original dynamics. No general claim of market completeness is needed, including when $a$ can vanish.