Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 211 (Advanced Financial Models)

For holdings $h\in {\mathbb{R}}^n$, let $V_0=h\cdot P_0$ and $V_1=h\cdot P_1$. Under the investment-consumption arbitrage convention used here, an arbitrage satisfiesThe initial consumption is $-V_0$, and the terminal consumption is $V_1$: there is a possible gain with no external funding and no negative consumption. A pure-investment arbitrage additionally requires $V_0=0$. This convention is important because the last part distinguishes initial consumption from pure investment.

A pricing kernel is a strictly positive random variable $R$ such that $R{P}_1^i$ is integrable andIt is a one-period state-price density, converting a terminal payoff into its initial price. Strict positivity means $R>0$ almost surely, not merely $R\ge 0$. When cash has price $1$ at both dates, this identity forces $\mathbb{E}R=1$, so $d\mathbb{Q}=Rd\mathbb{P}$ defines an equivalent martingale measure.

For any holdings $h$, the pricing kernel identity implies $V_0=\mathbb{E}[R{V}_1]$. If $V_1\ge 0$ almost surely, then $V_0\ge 0$, so an arbitrage cannot have $V_0<0$. If also $V_0=0$, then the nonnegative integrable random variable $R{V}_1$ has zero expectation and hence is zero almost surely. Since $R>0$ almost surely, $V_1=0$ almost surely, ruling out a positive terminal gain. Therefore a pricing kernel rules out arbitrage.

Write $q_{15},q_{12},q_9$ for the state probabilities under an equivalent martingale measure. Since cash is constant and the European put option pays only in the lowest state, the pricing equations areTheir unique solution is $q_9={\xi }_0/2$, $q_{15}={\xi }_0/2-2/3$, $q_{12}=5/3-{\xi }_0$. Every physical state has positive probability, so equivalence requires all three values to be strictly positive. ThusFor each price in this open interval the pricing kernel takes values $3q_{15},3q_{12},3q_9$ in the three equally likely states, proving absence of arbitrage.

The necessity, including the exclusion of endpoints, can also be checked directly. The three terminal payoff vectors of cash, the stock and the European put option form the invertible matrixEach unit state payoff therefore has a replicating strategy, whose initial cost is the corresponding $q$. A zero or negative $q$ supplies an arbitrage, so the endpoints are genuinely excluded.

Write $(a,b,c)$ for holdings of cash, the stock and the European put option, and let $\gamma =-V_0$ be initial consumption. At the given price, $a=-10b-c-\gamma$, and the three terminal payoffs, in descending order of the stock price, areNonnegative terminal payoffs require $c+\gamma \le 2b$ and $c-\gamma \ge b$. Together with $\gamma \ge 0$, these imply $b\ge 2\gamma \ge 0$. If $b=0$ they force $\gamma =c=a=0$, which is not an arbitrage. If $b>0$, the highest-state payoff is at least $3b>0$, so every such choice is an arbitrage. Consequently the complete set isThe pure-investment arbitrages are exactly the choices $\gamma =0$:For example, $(a,b,c)=(-11,1,1)$ costs zero and pays $(4,1,0)$.

Let ${\tau }_n\uparrow \infty$ be a localizing sequence for the discrete-time local martingale $X$. For $t\ge 1$, the increment of the stopped martingale isand $\{{\tau }_n\ge t\}\in {\mathcal{F}}_{t-1}$. Thus, for any $A\in {\mathcal{F}}_{t-1}$,The assumed integrability gives the dominating random variable $|X_t|+|X_{t-1}|$. The dominated convergence theorem now yields $\mathbb{E}[1_A(X_t-X_{t-1})]=0$, or $\mathbb{E}[X_t\mid {\mathcal{F}}_{t-1}]=X_{t-1}$. Iterating the tower property of conditional expectation proves an integrable discrete-time local martingale is a true martingale. The proof uses the discrete-time increment indicator; the conclusion does not extend to continuous time.

Under the standard local martingale definition, $X_0$ is integrable. For a nonnegative $X$, the Fatou lemma and a localizing sequence giveHence every $X_t$ is integrable, and part (a) applies. Therefore a nonnegative discrete-time local martingale is a true martingale. This is stronger than the continuous-time supermartingale conclusion.

Take an increasing localizing sequence ${\tau }_n$ for $X$ and define the stopping timesThe predictable process condition $K_{u+1}\in {\mathcal{F}}_u$ makes $R_n$ a stopping time. For each fixed finite time range, the finitely many coefficients $K_1,\dots ,K_t$ are finite almost surely; hence $T_n\uparrow \infty$ almost surely.

The stopped martingale transform can be writtenIts coefficient is ${\mathcal{F}}_{s-1}$-measurable and bounded in absolute value by $n$. Each summand is consequently integrable with zero conditional expectation given ${\mathcal{F}}_{s-1}$, since $X^{{\tau }_n}$ is a true martingale. Thus $M^{T_n}$ is a true martingale andThe original PDF fixes $M_0=0$; the TeX extraction omits this equality.

For fixed integers $s\le t$, $A_s=\{\tau =s\}$ belongs to ${\mathcal{F}}_s$ and $X_s=0$ there. The supermartingale property and nonnegativity implyTherefore $X_t=0$ almost surely on $A_s$. Taking the finite union over $s\le t$ gives $X_t=0$ on $\{\tau \le t\}$ almost surely; taking the countable intersection over all integer $t$ makes the assertion simultaneous at every time. Thus zero is absorbing for a nonnegative supermartingale in discrete time.

Assume the innovations are independent of the previous history and have finite exponential moments at every real argument, as required by the finite-valued affine definition. Conditioning on $X_{t-1}=x$ givesHence the autoregressive process of order one is an affine process, withHere and in the remaining parts, an affine process uses a time-homogeneous Markov process: the one-step transform is the same at every time. The displayed definition at time $1$ alone would not determine later transitions of an arbitrary time-inhomogeneous Markov process.

Use the time-homogeneous Markov property and define backward effective parameters byConditioning the last exponential factor on ${\mathcal{F}}_{t-1}$ replaces ${\theta }_t{X}_t$ by $A(q_t)X_{t-1}+B(q_t)$. Repeating the tower property of conditional expectation combines this with the preceding exponent, then with each earlier exponent. The last remaining conditional transform is at time $1$, soThese are finite because the one-step affine process transforms are finite at every real parameter. The argument establishes the entire joint affine transform, including when the coefficients ${\theta }_j$ have different signs.

Iterating the autoregressive process of order one givesFor $q_r={\sum }_{j=r}^t{\theta }_j{a}^{j-r}$, regrouping the sum of the exponents givesThe independent innovations factorize the moment-generating function. ConsequentlyFinite sums avoid a division by $a-1$, so these formulas include $a=0$ and $a=1$; zeroth powers in these sums are $1$.

A unit-face-value zero-coupon bond pays $1$ at maturity. The martingale deflator pricing identity isUse the affine process Markov property with respect to the market filtration; if that filtration contains extra predictive information, the natural Markov property alone would not suffice. Part (b), with every future coefficient equal to $1$, gives the exponential-affine form. More explicitly, the exponential-affine bond pricing recursion isIndeed, conditioning the first future step in an $(n+1)$-step horizon transforms $e^{(1+\alpha (n))X_{t+1}+\beta (n)}$ into $e^{A(1+\alpha (n))X_t+B(1+\alpha (n))+\beta (n)}$. ThereforeThis uses the true martingale deflator pricing identity; a merely local martingale deflator would not by itself justify replacing prices by conditional terminal expectations.

Since the initial value and increments are integers, $m=S_T-K$ is an integer. If $m\le -1$, all three positive parts in the second difference are zero. If $m\ge 1$, the second difference is $(m-1)-2m+(m+1)=0$, including $m=1$. At $m=0$ it equals $0-0+1=1$. ThusThe original PDF has the first term $(S_T-K-1{)}^{+}$; the TeX's $(S_T-K+K{)}^{+}$ is an extraction error. This discrete call-price curvature identity isolates an individual integer state.

For an integer $m$ and $d\in \{-1,0,1\}$, verify the single-step identityIf $m\ge 1$, the positive part is linear across the step and $f(m)=1$; if $m\le -1$, it remains zero and $f(m)=0$. At $m=0$, $d^{+}=(d+d^2)/2$, which is checked for the three permitted values of $d$.

Apply this with $m=S_{t-1}-K$ and $d=S_t-S_{t-1}$, and telescope over $t$. The result is the discrete Tanaka formulaThe first sum is a martingale transform; the second records the correction at the kink of the positive part. The algebraic identity itself needs only the integer values and permitted increments, not the martingale property.

Taking expectations in the single-step discrete Tanaka formula, the martingale transform term vanishes because $f$ is bounded and ${\mathcal{F}}_T$-measurable. HenceAt a state with $p_K=\mathbb{P}(S_T=K)>0$, the tower property of conditional expectation gives $\mathbb{E}[S_{T+1}\mid S_T=K]=K$, so the conditional second moment of the increment is ${\sigma }^2(T,K)$. Part (a) gives $p_K=C(T,K+1)-2C(T,K)+C(T,K-1)$. Combining these identities yields the discrete Dupire equationThe conditional variance needs a positive-probability conditioning state. The bound $|K-S_0|\le T$ does not guarantee $p_K>0$. If $p_K=0$, both the call-price curvature and the time increment above are zero; the equation can be extended by assigning an arbitrary finite value to ${\sigma }^2(T,K)$ at such a state, but its conditional variance is not determined there.

Let $p_K=C(T,K+1)-2C(T,K)+C(T,K-1)>0$ and $D_K=C(T+1,K)-C(T,K)$. Given $S_T=K$, the next value is one of $K-1,K,K+1$. The martingale property makes its upward and downward conditional probabilities equal; their sum is the conditional variance of the increment, ${\sigma }^2(T,K)=2D_K/p_K$. ThusThe discrete Dupire equation therefore recovers all positive-probability one-step conditional transitions from the European call option price surface. No Markov property is required: these are conditional probabilities given the current value, not necessarily given the whole past.

Recovery is impossible at a zero-probability state. For example, $S_t\equiv S_0$ satisfies every stated assumption, but at $T\ge 1$ a different integer $K$ can satisfy $|K-S_0|\le T$ while $p_K=0$. Its conditional transitions can be specified arbitrarily without altering $C$. Thus the literal all-states claim needs the positive-probability qualification.

For a fixed realization $\xi =a$ and $\mathrm{Re}z>1$, direct integration givesTo interchange the integral and expectation, write $x=\mathrm{Re}z>1$ and estimateThe Fubini's theorem now proves the Mellin transform of call prices identityThe required exponential moment is $\mathbb{E}{e}^{x\xi }<\infty$ at the chosen real part; the assertion for every $x>1$ presupposes these exponential moments for every such $x$.

Set $z=x_0+iy$. Since $|M(z)|\le \mathbb{E}{e}^{x_0\xi }$ and $|z(z-1){|}^{-1}$ is integrable over $y\in \mathbb{R}$, the proposed contour integral is absolutely convergent. The same bound justifies the Fubini's theorem when substituting $M(z)=\mathbb{E}{e}^{z\xi }$:The last equality uses the contour identity provided in the original PDF. ConsequentlyThis is contour inversion for call prices. The supplied TeX corrupts the permitted identity into a reciprocal and drops the imaginary units from its endpoints; the PDF has $(e^a-1{)}^{+}$ and the vertical contour used above.

Assume the usual positive initial stock price and strike, so the logarithm is defined, and interpret $U$ as a classical solution of the displayed partial differential equation. For a fixed complex $z$, put $F_t(z)=S_t^{z}U(t,{\sigma }_t,z)$, defining $S_t^z=\exp (z\log S_t)$. The Itô formula for the two Brownian motions with correlation coefficient $\rho$ givesThe partial differential equation cancels the entire drift. By the permission to treat the resulting local martingales as true martingales, and the terminal condition, $F_t(z)=\mathbb{E}[S_T^z\mid {\mathcal{F}}_t]$.

The bounded spot volatility ensures $\mathbb{E}{S}_T^{x_0}<\infty$: the Itô formula for a real power and localization bound its moment by $S_0^{x_0}\exp (x_0(x_0-1)\Vert \sigma {\Vert }_{\infty }^{2}T/2)$. Rewrite the proposed integrand asNow $|F_t(x_0+iy)|\le \mathbb{E}[S_T^{x_0}\mid {\mathcal{F}}_t]$, an integrable bound independent of $y$. Conditional Fubini's theorem and part (b) therefore implyThis route justifies the contour exchange without assuming bounds on $U$ uniform in all complex $z$. Cash is constant, $S$ is a true martingale under the stated allowance, and the displayed $C$ is a true martingale. The original measure is thus an equivalent martingale measure for all three assets. The fundamental theorem of asset pricing gives the market has no arbitrage under the usual admissible trading convention.

Use positive initial asset prices and the usual augmentation of the natural Brownian filtration. Define the market price of riskContinuity and strict positivity of $\sigma$ imply ${\int }_0^T{\lambda }_t^{2}dt<\infty$ on every finite horizon almost surely, because each path has a positive minimum of $\sigma$ there. Hence the strictly positive stochastic exponentialis defined and has $Y_0=1$ and $d{Y}_t=-Y_t(r_{t}dt+{\lambda }_{t}d{W}_t)$. The Itô product rule givesso it is a local martingale deflator.

For uniqueness, let $\tilde{Y}$ be any normalized strictly positive local martingale deflator. The Brownian martingale representation theorem makes $B\tilde{Y}$ a continuous local martingale with a Brownian motion integral representation. Dividing by $B$ therefore gives $d\tilde{Y}=-r\tilde{Y}dt+\eta dW$. The vanishing drift of $S\tilde{Y}$ requires $\tilde{Y}(\mu -r)+\eta \sigma =0$, so $\eta =-\tilde{Y}\lambda$. This scalar linear stochastic differential equation has exactly the exponential solution above, proving the normalized local martingale deflator is unique.

The assumptions do not make $\lambda$ deterministically bounded: $\sigma$ need not be bounded away from zero uniformly over outcomes. Thus a true martingale density or Novikov condition is not inferred here; the required conclusion is local.

Let ${\theta }_t$ and ${\eta }_t$ be the predictable holdings of the stock and the bank account. With no consumption, the self-financing portfolio has $X=\theta S+\eta B$ and $dX=\theta dS+\eta dB$. The Itô product rule, together with the dynamics from part (a), givesThe drift vanishes because $X=\theta S+\eta B$ and $\sigma \lambda =\mu -r$. Thus $XY$ is a local martingale. It is nonnegative by the assumed nonnegative wealth and strict positivity of $Y$, and henceFor example, this last fact follows directly from Conditional Fatou lemma applied to a localizing sequence; the initial capital is assumed finite. As usual, holdings must be integrable against the asset semimartingales for the self-financing portfolio equation to be defined.

Fix the maturity horizon. Since $YB$ is a nonnegative local martingale and $r$ is bounded, $\mathbb{E}{Y}_T<\infty$: $B_T\ge B_0{e}^{-\Vert r{\Vert }_{\infty }T}$ and $\mathbb{E}(Y_T{B}_T)\le B_0$. The bounded nonnegative payoff thus makes $Y_T{\xi }_T$ integrable. SetThe Brownian martingale representation theorem yields $N_t=N_0+{\int }_0^t{\zeta }_{s}d{W}_s$, using its locally square-integrable version for an integrable terminal random variable. The Itô formula for $N/Y$ givesChoose the replicating strategyIts self-financing portfolio equation has exactly the displayed drift and diffusion, because $\mu -r=\sigma \lambda$. All coefficients are locally integrable after stopping; the holdings are predictable in the augmented natural Brownian filtration. This construction has $X_t\ge 0$ and $X_T={\xi }_T$, so it is an admissible replicating strategy under the question's nonnegative-wealth convention.

For any other nonnegative self-financing portfolio $\hat{X}$ replicating the same payoff, part (b) implies ${\hat{X}}_0\ge \mathbb{E}[Y_T{\xi }_T]$. The constructed strategy attains equality, since $Y_0=1$ and $N_0=\mathbb{E}[Y_T{\xi }_T]$ in the augmented natural Brownian filtration. HenceThis is deflator-based claim replication; it does not require upgrading the local martingale deflator to a true martingale density.

With constant coefficients, $\lambda =(\mu -r)/\sigma$ is constant, so the stochastic exponential $Z_t=e^{rt}{Y}_t$ is a true martingale. Under its equivalent martingale measure $\mathbb{Q}$, the stock follows the Black-Scholes model $d{S}_t=S_t(rdt+\sigma d{W}_t^{\mathbb{Q}})$. The Gaussian exponential moment gives the value of the power optionThe option delta is ${\partial }_{s}V=V/(2s)$, so the replicating strategy holdsHalf the wealth value is in the stock and half in the bank account, with continuous rebalancing. To check the self-financing portfolio property under the original measure, the Itô formula gives $d{V}_t=\frac{1}{2}(r+\mu )V_{t}dt+\frac{1}{2}\sigma V_{t}d{W}_t={\theta }_{t}d{S}_t+{\eta }_{t}d{B}_t$. Also $V_T=\sqrt{S_T}$ and $V_t>0$.

The payoff is unbounded, so the bounded-payoff restriction of part (c) is not invoked automatically; the Black-Scholes model has the necessary finite Gaussian moments, and the explicit strategy attains $V_0=\mathbb{E}[Y_T\sqrt{S_T}]$. The supermartingale bound from part (b) therefore proves its minimality. The cost is independent of the physical drift $\mu$.