Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 38 (Advanced Financial Models)

Choose a localizing sequence of discrete-time stopping times ${\tau }_j\uparrow \infty$ for $X$. For fixed integer $t$, every stopped value is bounded in absolute value by the finite sumThat sum is integrable under the hypothesis. Since $X^{{\tau }_j}$ is a martingale,The dominated convergence theorem, including its conditional version, now removes the stopping. Thus $\mathbb{E}[X_t\mid {\mathcal{F}}_{t-1}]=X_{t-1}$. The process is adapted and integrable by hypothesis, so $X$ is a true discrete-time martingale. This is the integrable discrete-time local martingale is a martingale criterion. The finite sum dominating stopped values is the crucial discrete-time feature.

The stopped processes are nonnegative martingales. Their expectations equal the finite deterministic value $X_0$. The Fatou lemma gives, for each fixed integer $t$,Thus every $X_t$ is integrable. Apply the preceding discrete-time criterion to conclude $X$ is a martingale, not merely a supermartingale. This is the nonnegative discrete-time local martingale is a martingale result; its conclusion does not extend to arbitrary continuous-time nonnegative local martingales.

Predictability means $K_s$ is ${\mathcal{F}}_{s-1}$-measurable. If $|K_s|\le L$, each product $K_s(M_s-M_{s-1})$ is integrable, and the finite sum defining $Y_t$ is integrable. Pulling the bounded predictable factor out of the conditional expectation givesTherefore the bounded predictable martingale transform $Y$ is a martingale starting at zero.

Stop just before a large predictable coefficient would be used. Setwith the infimum of the empty set equal to infinity. Because $K_{t+1}$ is ${\mathcal{F}}_t$-measurable, ${\sigma }_j$ is a stopping time. The increment of the stopped martingale transform isIts coefficient is predictable and bounded by $j$: the first coefficient exceeding $j$ occurs at the step after stopping, and is never included. Part (c) makes $Y^{{\sigma }_j}$ a martingale. Since the finitely many $K_s$ on any fixed finite horizon are finite almost surely, ${\sigma }_j\uparrow \infty$ almost surely. ThusThis is predictable-coefficient localization of a martingale transform. Stopping after taking the large increment would not give the required bound.

First propagate terminal nonnegativity backwards; it is not necessary to assume nonnegative wealth at intermediate dates. Suppose $Y_t\ge 0$ and defineThese events increase to the whole space up to a null set. On $A_j$, both the old value and the coefficient are bounded, so $1_{A_j}{Y}_t$ is integrable andThe left side is nonnegative; hence $Y_{t-1}\ge 0$ on every $A_j$, and therefore almost surely. Starting from $t=T$, induction gives $Y_t\ge 0$ for every $0\le t\le T$.

The process stopped at $T$ is now a nonnegative discrete-time local martingale. Part (b) makes it a martingale with initial value zero. Consequently $\mathbb{E}{Y}_T=0$, and a nonnegative random variable with zero expectation vanishes almost surely:This is the terminal nonnegativity criterion for a finite-horizon martingale transform. A finite deterministic horizon is essential to the backward induction.

A trading strategy chooses a vector $H_t$ of holdings for the period $(t-1,t]$, with $H_t$ measurable with respect to ${\mathcal{F}}_{t-1}$. Its end-of-period wealth is $X_t=H_t\cdot P_t$. Rebalancing at date $t$ is self-financing whennew holdings cost exactly the value released by the old holdings. With fixed initial capital $x$, the equivalent gains identity isA European contingent claim is a maturity-$T$ payoff $\xi$ measurable with respect to ${\mathcal{F}}_T$. It is attainable if there exists a predictable self-financing strategy and an initial capital $x$ for which $X_T=\xi$ almost surely. Such a strategy is a replicating strategy, and $x$ is its initial replication cost. Holdings are understood only up to the maturity being replicated.

On a finite sample space, all real-valued holdings over the finite interval $0,\dots ,T$ are bounded after null states are discarded. The gains identity is therefore a bounded predictable martingale transform of the vector martingale $P$, summed over its coordinates. It follows that $X$ is a martingale, soHere the replication cost is a prescribed deterministic initial capital, as in the definition of attainability. The stronger intermediate identity is $X_t=\mathbb{E}[\xi \mid {\mathcal{F}}_t]$. If initial capital is instead allowed to be ${\mathcal{F}}_0$-measurable and random, the corresponding statement is $X_0=\mathbb{E}[\xi \mid {\mathcal{F}}_0]$; its unconditional expectation still equals $\mathbb{E}\xi$.

Let ${\pi }_t$ be the predictable stock holding during $(t-1,t]$. Cash has constant price, so self-financing givesPut $\mathcal{G}={\mathcal{F}}_{t-1}$. The tower property of conditional expectation and ${\mathcal{F}}_t$-measurability of $S_t$ giveSince $X_{t-1}$, $S_{t-1}$ and ${\pi }_t$ are $\mathcal{G}$-measurable, substituting the gains identity yieldsThe denominator is positive on every positive-probability parent atom. If it were zero on such an atom, $S_t$ would be constant there, and the martingale property would force that constant to equal $S_{t-1}$, contradicting the nonzero-increment assumption. HenceThis is conditional covariance hedge ratio. It uses attainability; a regression coefficient alone would not replicate a general unattainable payoff. After maturity one may liquidate into cash, so the same formula gives zero for later dates wherever its denominator remains nonzero. On a finite sample space a martingale cannot have nonzero increments forever; the stated nondegeneracy is naturally a finite-maturity assumption.

Work conditionally on a positive-probability atom of ${\mathcal{F}}_{T-1}$. Let $Z$ have the conditional distribution of $S_T$ there, and let $Z^{\prime }$ be an independent copy of that conditional law. The finite sample space makes all expectations finite. ThenStrict increase of $g$ makes the integrand nonnegative, and strictly positive whenever $Z\ne Z^{\prime }$. The conditional variance is positive by part (c), so the conditional distribution is nondegenerate and $\mathbb{P}(Z\ne Z^{\prime })>0$. Thus the numerator of the hedge ratio is strictly positive on every such atom. The denominator is also positive, givingThis is strict positive covariance with an increasing payoff. The independent copy is taken from the conditional distribution, not from an unrelated unconditional distribution.

Write the discount factor as $D_t=B_t^{-1}=\exp (-{\int }_0^t{r}_{s}ds)$. Splitting the time integral at $t$ givesThe random variable $D_T$ lies in $(0,1]$ because the short rate is nonnegative and continuous on the finite maturity interval. A process of conditional expectations of an integrable terminal variable is a martingale, by the tower property of conditional expectation. Therefore

The terminal density is strictly positive and has expectation one under the usual deterministic initial bond-price convention. Its density process isFor $0\le s\le t\le T$, the Bayes formula for conditional expectation under a change of measure givesThe same calculation at $s=0$ gives the finite expectation $1/P(0,T)$, so this is a true martingale, not just a formal conditional identity. Thus the continuous-time bank account measured in units of the maturity-$T$ bond is a ${\mathbb{Q}}_T$-martingale. This is the forward measure change of numéraire. If the initial bond price were random rather than given, integrability of its reciprocal would need to be included for this true-martingale assertion.

To avoid confusing the continuous-time bank account with the coefficient of $r_t$, denote the latter by $b(\tau )$ and the other coefficient by $a(\tau )$, where $\tau =T-t$. Apply the Itô formula to $D_t[a(\tau )+b(\tau )r_t]$. Since the expression is affine in $r$, its second rate derivative is zero. Its drift isThe quadratic terms cancel. The remaining expression isIt vanishes when $a^{\prime }=0$ and $b^{\prime }=-a-b$. For a unit bond payoff choose terminal conditions $a(0)=1$, $b(0)=0$, givingThe resulting local martingale is $D_t[1-(1-e^{-\tau })r_t]$. The allowed bound $0\le r_t\le 1$ puts it between zero and one, so the bounded local martingale criterion makes it a true martingale. At maturity it equals $D_T$. Comparing with part (a) therefore givesIn particular $e^{-(T-t)}\le P(t,T)\le 1$, and $P(T,T)=1$. This is linear bond pricing in a bounded short-rate diffusion; choosing zero coefficients would produce a local martingale but would not price the required terminal payoff.

Use the same affine drift cancellation with terminal conditions $a(0)=0$, $b(0)=1$. It gives $a(\tau )=0$, $b(\tau )=e^{-\tau }$, soThe process is bounded, which justifies the conditional expectation identity. The Bayes formula for conditional expectation for the forward measure now givesThe denominator is positive; the ratio lies in $[0,1]$ and equals $r_T$ at maturity. This is the forward-measure terminal rate in a linear bond model.

Buy one lower-strike European call option and sell one higher-strike European call option. The initial cost is $C(K_i)-C(K_{i+1})<0$, so the strategy releases strictly positive cash. Its terminal payoff isfor every stock price. Thus it is an arbitrage: a positive initial receipt accompanies a nonnegative terminal obligation. One may consume the receipt immediately, or hold it in cash to make a zero-initial-capital strategy with strictly positive terminal wealth. This is the vertical-spread arbitrage for increasing call prices.

Buy half a call at each neighboring strike and sell one call at the middle strike. Its cost isFor fixed terminal stock price $s$, the function $K\mapsto (s-K{)}^{+}$ is convex. Since $K_i$ is the midpoint, the payoffis nonnegative for every $s$. More explicitly, it is zero outside $[K_{i-1},K_{i+1}]$, equals $(s-K_{i-1})/2$ on $[K_{i-1},K_i]$, and equals $(K_{i+1}-s)/2$ on $[K_i,K_{i+1}]$. The negative cost and nonnegative payoff produce an arbitrage. This is the butterfly-spread arbitrage for nonconvex call prices.

Use positive strikes, the natural real-power domain of this price curve. For any such $K$, the stock-minus-call payoff isalmost surely, since $S_1>0$. Therefore the call price must be strictly below the stock price $S_0=1$: if $C(K)\ge 1$, buying stock and selling the call has nonpositive initial cost and strictly positive terminal payoff. Also a negative call price is an immediate arbitrage by buying the call.

For $p<0$, $1+K^p>K^p$ and raising to the negative power $1/p$ reverses the inequality, giving $(1+K^p{)}^{1/p}<K$ and $C(K)<0$. The expression is undefined at $p=0$. For $0<p<1$, strict concavity of the power implies $(1+K{)}^p<1+K^p$, whenceAt $p=1$, $C(K)=1$. Every defined case with $p\le 1$ therefore violates the necessary no-arbitrage bounds. ConsequentlyThis argument does not require a dense family of strikes; even one positive-strike call gives the contradiction. The strict stock-minus-call payoff explains why the borderline $p=1$ is also excluded.

Use zero-interest cash as the one-period numéraire, consistent with the stated expectation-price formula. For $p>1$, differentiate the proposed call-price curve twice. The first derivative and the candidate density areThis is the power call-curve pricing density. It is strictly positive. Its integral is $C^{\prime }(\infty )-C^{\prime }(0+)=1$, and its survival function is $-C^{\prime }(u)$. Since $C(0+)=1$ and $C(\infty )=0$,Likewise, integrating the survival function from $K$ onwards givesThus a market whose terminal stock has this law under an equivalent martingale measure prices the stock at one, every proposed call at $C(K)$, and any integrable claim $g(S_1)$ at $\int g(u)f_p(u)du$. The finite-market fundamental theorem of asset pricing says that an equivalent measure pricing every traded discounted payoff by expectation excludes arbitrage. This proves the intended conclusion when such an equivalent pricing law is part of the model. For example, take the canonical terminal state space $(0,\infty )$ with stock equal to its coordinate and physical law equivalent to the positive density $f_p$.

There is, however, a genuine insufficiency in the literal finite-strike formulation: a finite list of call prices and no-arbitrage alone do not force this pricing law, nor even a continuous terminal distribution. Here is an explicit counterexample. Take $p=2$, one strike $K=1$, and two terminal stock valuesGive the lower stock value the remaining strictly positive probability and take this as the physical measure too. Direct calculation gives $\mathbb{E}{S}_1=1$ andso the stock/cash/call market is arbitrage-free. Now letThis bounded nonnegative function is zero at both actual stock values, so $g(S_1)=0$ almost surely. Yet $\int g(u)f_2(u)du>0$. Charging that positive amount for the identically zero payoff creates an arbitrage by selling it. In fact no Lebesgue probability density can price every claim correctly on this two-state market.

Therefore the displayed $f_p$ is the intended continuous pricing density, but the promised no-arbitrage extension requires an equivalent pricing measure with this terminal law; a full call curve identifies that law if such a measure exists, but it does not follow from the printed finite-strike hypotheses alone. This is the finite-strike nonidentification of a pricing density. The counterexample and the corrected sufficient hypothesis account for the literal and intended readings separately.

Multiply the linear stochastic differential equation by $e^{-at}$. The Itô product rule givesThe integrand is deterministic, so the Itô integral has a centered normal distribution. Its variance, by the Itô isometry, isHenceAt $a=0$ the correct continuous-limit formula is $N(z,b^{2}t)$. A zero variance, for example when $b=0$, denotes the deterministic distribution. For $a<0$ the variance is still positive because both numerator and denominator in its quotient are negative. This is the explicit Ornstein-Uhlenbeck solution, allowing either sign of the linear drift coefficient.

Choose the market price of risk ${\lambda }_t=(\mu -r{S}_t)/\sigma$. The Girsanov theorem, with the hypotheses allowed in the question, gives an equivalent martingale measure $\mathbb{Q}$ under whichThus the stock under the risk-neutral measure is a linear Gaussian diffusion. In particular, for $\tau =T-t$ its conditional mean and standard deviation arePut $\varphi (x)=(2\pi {)}^{-1/2}{e}^{-x^2/2}$ and let $\Phi$ be the standard normal distribution function. For $N\sim N(0,1)$,since ${\int }_{-d}^{\infty }x\varphi (x)dx=\varphi (d)$. Discounting this expectation gives the call price in an arithmetic stock model with interest. A particularly convenient expression isAt maturity define $C(T,s)=(s-K{)}^{+}$. This value is nonnegative because it is a discounted expectation of a nonnegative payoff.

For $t<T$, hold ${\pi }_t=C_s(t,S_t)$ shares and hold ${\beta }_t=[C(t,S_t)-{\pi }_t{S}_t]/B_t$ units of the continuous-time bank account. The pricing function solvesThe Itô formula under the physical measure therefore givesThis proves self-financing and terminal replication, with wealth always $C(t,S_t)\ge 0$. The coefficients are locally smooth before maturity, and the strategy extends to maturity through its continuous wealth limit and the square-integrable discounted payoff representation.

To see minimality, any other nonnegative self-financing portfolio replicating the payoff has discounted wealth a nonnegative local martingale under $\mathbb{Q}$, hence a supermartingale. Its initial capital $x$ must satisfy $x\ge {\mathbb{E}}^{\mathbb{Q}}[e^{-rT}(S_T-K{)}^{+}]=C(0,S_0)$. The strategy constructed above attains equality. ThusThe additive physical diffusion may take negative stock values; the formula and nonnegative replicating wealth remain valid. Replacing it by a multiplicative Black–Scholes diffusion would give the wrong price and hedge.

Differentiate the Gaussian price with respect to $s$. The terms involving derivatives of $d$ cancel because ${\varphi }^{\prime }(d)=-d\varphi (d)$ and $s-K{e}^{-r\tau }=\nu d$. Thus the delta hedge isFor every $t<T$, $\nu >0$ and $S_t$ is finite, so $0<{\pi }_t<1$. At maturity its limiting value is the payoff derivative $1_{\{S_T>K\}}$ except at the kink, an event of probability zero under the equivalent Gaussian law. Consequently the stock holding is always nonnegative and never exceeds one. The initial drift $\mu$ does not enter this hedge; it is removed by the change to the risk-neutral measure.

The mark-to-market value of holdings $H_t$ in the asset-price vector is the dot product $X_t=H_t\cdot P_t$. A self-financing strategy pays for every rebalancing from within the portfolio. Over a short interval, the gains on the currently held assets are $H_t\cdot d{P}_t$, while consumption removes $c_{t}dt$ units of wealth. ThereforeThe holdings are predictable and integrable against the price semimartingale; the consumption rate is nonnegative and suitably measurable. This is the continuous-time self-financing-with-consumption convention.

Use the state-price density as a positive local martingale deflator, so each component of $YP$ is a local martingale. The Itô product rule and the wealth equation giveThe consumption term has finite variation, and the quadratic covariation of a stochastic integral satisfies $d[Y,X]=H\cdot d[Y,P]$. Also $X=H\cdot P$. Regrouping the terms therefore givesThe differential before $YP$ is necessary: the first term is a stochastic gain, not the level of the deflated portfolio. It is missing in the printed display. This is the deflated wealth equation with consumption.

The integral against the local-martingale vector $YP$ is a local martingale, provided the predictable holdings are stochastically integrable. Integrating part (b) givesBoth $Y_t{X}_t$ and the cumulative deflated consumption are nonnegative. Hence $M_t\ge -Y_0{X}_0$. With the usual finite deterministic initial capital, the shifted processis a nonnegative local martingale, and is therefore a supermartingale. For completeness, a localizing sequence turns it into true martingales; conditional Fatou lemma for their nonnegative stopped values gives the supermartingale inequality and ordinary Fatou gives integrability at each time. Subtracting the initial constant provesThis is supermartingale control of deflated consumption gains. It uses $c\ge 0$ as well as $X\ge 0$; an unrestricted stochastic integral is not necessarily a true supermartingale.

Part (c) gives $\mathbb{E}{M}_t\le 0$. Since $Y_t{X}_t\ge 0$, its integrated identity impliesfor every finite $t$. The cumulative consumption increases with $t$, so the monotone convergence theorem yieldsThis is the infinite-horizon state-price budget constraint. No terminal-wealth convergence or vanishing assumption is needed for this inequality.

Concavity gives the supporting-tangent inequalityMultiply by the discount factor $e^{-bt}$ and use the first-order condition $e^{-bt}{U}^{\prime }(c_t)=Y_t$. This gives the pointwise marginal-utility verification of optimal consumption inequalityApply the budget inequality from part (d) to the competing admissible strategy, and use equality for the proposed one:The two weighted consumption integrals are finite, so their difference is integrable. Under the usual positive discount-rate assumption $b>0$, the utility integrals are also integrable: $U(0)\le U(x)\le L$ for a finite upper bound $L$, and ${\int }_0^{\infty }{e}^{-bt}dt=1/b$. Integrating the tangent inequality and taking expectations therefore givesEconomically, both consumers face the same state-price budget, and the candidate spends it exactly where its discounted marginal utility equals the state price. This is utility duality with martingale deflators in its consumption form.

A positive $b$, or another hypothesis making the infinite-horizon objectives well defined and permitting this integration, is needed. The printed question does not specify the sign of $b$. Bounded utility alone does not ensure that an undiscounted infinite time integral exists: a bounded integrand can have both infinite positive and negative parts. Thus the conclusion is established under the standard discount convention $b>0$, and also whenever the displayed objectives satisfy the stated integrability conditions; without either convention the literal infinite-horizon comparison need not be a defined mathematical expression.

This can occur within an admissible financial model, not just for an abstract bounded integrand. Take $U(x)=1-2e^{-x}$ and $b=0$. Choose a deterministic smooth nonnegative $c$ with successive unit-length plateaus alternating between zero and the integer $n$, and connect them over intervals whose lengths have finite sum. Then ${\int }_0^{\infty }2c_t{e}^{-c_t}dt<\infty$: the high-plateau contributions sum to ${\sum }_{n\ge 1}2n{e}^{-n}$ and the transition integrand is bounded by $2/e$. SetThe bank account is a positive deterministic Itô process and $Y_t{B}_t=Y_0$, so $Y$ is a state-price density. Holding $X_t/B_t$ bank units finances consumption with nonnegative wealth and exact budget equality. Also $U^{\prime }(c_t)=Y_t$. Nevertheless every zero plateau contributes $-1$ to the utility integral, while each sufficiently high plateau contributes a fixed positive amount. Both its negative and positive parts are infinite, so the undiscounted objective is undefined. This establishes why the missing discount or objective-integrability hypothesis is substantive.