Past exam of the martingales and financial mathematics course of the University of Oxford / 2007 / B (Mathematical Models of Financial Derivatives)

Let ${\Delta }_t$ be the number of stock units and ${\beta }_t$ the number of bank-account units, with predictable holdings satisfying the integrability conditions for the price integrals. The portfolio value is $V_t={\Delta }_t{S}_t+{\beta }_t{B}_t$. It is self-financing ifThus changes in value come only from gains on the assets held. Rebalancing transfers money between the stock and bank account; it does not inject or withdraw cash. The definition is an identity of stochastic integrals and automatically includes the bookkeeping needed when holdings themselves vary randomly.

Write the smooth claim value as $V(t,S)$. Itô's formula givesA replicating self-financing strategy with stock holding $\Delta =V_S$ and bank value $V-S{V}_S$ has gainsThe Brownian terms then agree. Equating the drift terms cancels $\mu S{V}_S$ and givesThe terminal condition enforces replication of the contingent claim payoff. Subject to the usual admissibility and growth conditions for the replicating price, this is the Black-Scholes equation. Its independence of the physical drift is a consequence of hedge cancellation, not an assumption that the drift equals the risk-free rate.

For deterministic square-integrable volatility, apply Itô's formula to the exponential to obtainThe stochastic integral is centered Gaussian with variance ${\int }_0^t\sigma (u{)}^{2}du$.

Under the risk-neutral measure the drift is $r$. Conditional on $S_t=s$, put $\tau =T-t$ and $v={\int }_t^T\sigma (u{)}^{2}du$. ThenIntegrating the discounted call option against this normal distribution yieldsThus the total remaining variance replaces ${\sigma }^2\tau$ throughout the ordinary formula; the corresponding effective constant volatility is $\sqrt{v/\tau }$. If $v=0$, use the limiting deterministic value $(s-K{e}^{-r\tau }{)}_{+}$.

Over the stated year, the true total variance isThe second trader uses $v_{\mathrm{offered}}=(3\alpha /2{)}^2=9{\alpha }^2/4$. Their difference iswhen $\alpha \ne 0$. A nondegenerate European call option has positive option vega, equivalently a price strictly increasing in total variance. The offered price is therefore below the model value, so buy the option in the stated comparison.

If the deterministic volatility path is indeed the true one and the frictionless Black-Scholes trading assumptions apply, buy the cheap option and short its replicating portfolio computed with that true path. The initial cash received is the positive price difference, while the two terminal payoffs cancel exactly. Investing the initial surplus yields a strictly positive terminal gain with no net claim liability, an arbitrage within this model. A mere uncertain volatility belief would not by itself provide that replication guarantee. At $\alpha =0$ both variances coincide and this price difference disappears.

Assume the SDE has the required solution and $f$ is sufficiently smooth. Itô's formula along that solution givesThe PDE cancels the first term. Integrating from $t$ to $T$ givesIf this integral is a true martingale, for example if its integrand has finite expected integrated square, its conditional expectation at time $t$ is zero. For the diffusion started at $x$ at time $t$, this yieldsThe notation in the question is the same Markov conditional expectation. The martingale or corresponding uniform-integrability growth hypothesis is part of the usual Feynman-Kac formula justification: a local martingale alone does not automatically have zero expected increment. Standard smooth solutions with the customary integrability bounds satisfy this requirement.

A stock holding receives both its price change and dividend, so its total gains are $dS+\delta Sdt$. The replicating portfolio therefore satisfiesItô's formula and the hedge $\Delta =(V_{\delta }{)}_S$ cancel the Brownian risk. Comparing drifts gives

To verify the claimed transformation, put $x=S{e}^{-\delta (T-t)}$ and set $U(t,S)=V_0(t,x)$. Then $x_t=\delta x$, $S{U}_S=x(V_0{)}_x$ and $S^2{U}_{SS}=x^2(V_0{)}_{xx}$. The dividend operator applied to $U$ becomesby the zero-dividend PDE. At maturity $x=S$, so the contingent claim payoff also agrees. Uniqueness in the pricing class proves

Differentiate the dividend PDE with respect to $S$. The diffusion term contributes ${\sigma }^{2}S{V}_{SS}+\frac{1}{2}{\sigma }^2{S}^2{V}_{SSS}$, and the stock-drift term contributes $(r-\delta )(V_S+S{V}_{SS})$. With $\Delta =V_S$, the coefficient of $\Delta$ is $(r-\delta )-r=-\delta$. ThereforeThe terminal condition is $\Delta (T,S)={\Psi }^{\prime }(S)$ wherever the contingent claim payoff is differentiable, with an appropriate weak or limiting interpretation at contingent claim payoff kinks. In the next part the squared positive-part contingent claim payoff is continuously differentiable, so its terminal option delta has no jump.

The contingent claim payoff derivative is ${\Psi }^{\prime }(s)=2(s-K{)}_{+}$. Differentiating the risk-neutral valuation is justified by the finite lognormal distribution, and $S_T$ is proportional to the starting stock value $s$. ThusFor a lognormal terminal stock, completing the square in its Gaussian exponent givesUse $p=2$ and $p=1$, and let $d_1=[\log (s/K)+(r+{\sigma }^2/2)\tau ]/(\sigma \sqrt{\tau })$. The result isAs $\tau \downarrow 0$ this tends to $2(s-K{)}_{+}$, including value zero at $s=K$, as required. The extra ${\sigma }^2$ in the option delta PDE's stock drift gives the same shifted Gaussian arguments. This is delta of a squared positive-part call.

Use the usual nondegenerate positive-rate case $r>0$, $\sigma >0$, $K>0$. Define the Black-Scholes equationThe American value dominates immediate exercise. Its discounted value is a supermartingale under the risk-neutral measure, so $\mathcal{L}V\le 0$, with equality where it is optimal to continue. The obstacle problem isEquivalently $\max (\mathcal{L}V,g-V)=0$.

For the put, write the exercise boundary as $S=b(t)$. On $S>b(t)$ solve $\mathcal{L}V=0$; on $0<S\le b(t)$ set $V=K-S$. Value matching and smooth pasting requireThe remaining conditions are $V(t,0)=K$, $V(t,S)\to 0$ as $S\to \infty$, and the terminal contingent claim payoff, with $b(T-)=K$ in this positive-rate setting. The boundary is an unknown to be found with the price, which is what makes this a free boundary problem.

put-call parity and nonnegativity of the European put option giveFor $r\ge 0$ this is at least $S-K$, and the call value is also nonnegative, so $C^E$ dominates the immediate call exercise contingent claim payoff. Its discounted price is a martingale. At any admissible exercise time, optional stopping theorem therefore bounds the expected discounted call option by the European call option price. Exercising at maturity attains that price. Hence American and European calls coincide for this non-dividend model with nonnegative rates.

For the strict put comparison assume $r>0$, $\sigma >0$, $S>0$ and $t<T$. Choose $u$ strictly between $t$ and $T$. A European put option at $u$ has value at most $K{e}^{-r(T-u)}$. On the eventimmediate exercise gives $K-S_u>K{e}^{-r(T-u)}\ge P^E(u,S_u)$. This event has positive probability because the lognormal stock density is positive on $(0,\infty )$. Exercise there at time $u$, and otherwise hold to maturity. This strategy strictly improves the European put option expected value, provingunder these positive-rate, pre-maturity assumptions.

The source's strict statement needs those qualifications. At maturity the prices agree, and at $r=0$ the put option is a convex function of the martingale stock and therefore is a submartingale, so waiting to maturity is optimal and the American and European puts coincide. Negative rates also invalidate the nonnegative-rate call argument. These are mathematical assumptions of the comparison, rather than changes to the printed question.

LetThe homogeneous function of the Black-Scholes formula gives its at-the-money value at the start date as $S_{T_1}c(T-T_1)$. Before that date, risk-neutral valuation and the discounted stock martingale giveAfter the start date the strike price is the realized value $S_{T_1}$, so the complete answer isAt maturity the second line is $(S_T-S_{T_1}{)}_{+}$. After entry the price depends on the historical strike price as well as the current stock, so that history cannot be discarded when giving the answer for all times.

At an entry date the holder chooses between entering the at-the-money call immediately and retaining the right to choose at the next date. The Snell envelope therefore gives, under risk-neutral valuation,with terminal value zero if entry has never occurred.

homogeneous function makes $B^{FSC}(T_i,s)=s{b}_i$. The discounted stock expected value is $s$, so the recursion reduces toIn the positive-rate case in the hint, put $\gamma =r/\sigma$ and $\beta =\sigma /2$. Then $c(u)$ is exactly its increasing function. The largest remaining maturity is $T-T_i$, so $b_i=c(T-T_i)$: immediate entry is optimal at every date while unentered.

At time zero this yieldsThus the Bermudan choice of start date adds no value over the ordinary at-the-money European call option in this setting. The maximum formula remains the general recursion; using monotonicity to select immediate entry uses the rate assumption supplied with the hint. This is immediate entry for a Bermudan forward-start call.

Set $x=B^2/S$ and $q=(S/B{)}^{\alpha }$, so $W=qV(t,x)$. Direct differentiation givesSubstitute into the Black-Scholes equation and divide by $q$:To recover the same operator on $V$, its first-derivative coefficient must be $r$. This givesFor this choice the last coefficient is $-r$, since $\frac{1}{2}{\sigma }^2(\alpha -1)+r=0$. Hence $\mathcal{L}W=q\mathcal{L}V=0$, proving the required invariance. The transformation is the stock-inversion reflection for the Black-Scholes equation.

A down-and-out option with no rebate pays $(S_T-K{)}_{+}$ only if the stock has stayed above the lower barrier $B$ up to maturity; if the barrier is hit, it expires worthless. For an unknocked-out contract currently at $S>B$, apply part (a) to the ordinary European call option price:Both terms solve the PDE. At $S=B$ they cancel, enforcing the absorbing boundary. At maturity and $S>B$, the reflected stock value is $B^2/S<B\le K$, so the reflected call option is zero and the required terminal contingent claim payoff remains. Standard growth and uniqueness conditions for the killed pricing problem identify this solution as the option value.

If the contract has already hit its barrier, its value is zero regardless of a later hypothetical stock recovery. Thus the formula is conditional on survival to the current time; the barrier history is part of the contract state.

At maturity debt receives the smaller of face value and firm assets:Thus it is equivalent to a zero-coupon payment $F$ minus a European put option on the firm value. Its time-$t$ price isEquity receives the residual $(V_T-F{)}_{+}$, hence is a European call option:put-call parity verifies $E_t+D_t=V_t$, consistent with the stated balance-sheet identity. As with the replication arguments elsewhere, the physical drift $\mu$ does not enter these option prices.

If default liquidates the firm as soon as its value crosses below $F$, equity is extinguished at that event. If default never occurs before maturity, equity receives $(V_T-F{)}_{+}$. It is therefore a no-rebate down-and-out option on firm value, with strike price and barrier both equal to $F$.

For a firm that has survived to $t$ and has $V_t>F$, part (b) givesIf default has already occurred, $E_t=0$. At maturity a surviving firm has equity $(V_T-F{)}_{+}$.

The source uses a strict crossing $V<F$. With nonzero volatility, continuity and immediate two-sided Brownian fluctuations at a first barrier hit make the pre-maturity absorbing value at $V=F$ zero as well: starting at the barrier, the process crosses below it immediately almost surely. At maturity the contingent claim payoff at equality is also zero. Thus this convention has the same pricing boundary as the usual barrier-hit convention in this nondegenerate model.