Past exam of the martingales and financial mathematics course of the University of Oxford / 2009 / B (Mathematical Models of Financial Derivatives)

Apply Itô's formula to $\log X$ for positive initial value $x_0$. The drift correction is $-{\sigma }^2/2$, soThis is geometric Brownian motion. Under the physical probability measure, $W_T/\sqrt{T}$ has the standard normal distribution. For positive volatility,The drift here is $\mu$, because this is a physical bankruptcy probability. The risk-neutral measure will instead give drift $r$ for pricing. If $\sigma =0$, the value is deterministic and the default probability is the indicator of $x_0{e}^{\mu T}<m$; the usual positive-volatility convention is used in the remaining formulas.

In the Merton model, the total equity payoff is $(X_T-m{)}^{+}$, exactly a European call option on the company value with strike $m$. Total debt pays $\min (X_T,m)=X_T-(X_T-m{)}^{+}$. The first payoff has slope zero below $m$ and slope one above it; the second has slope one below $m$ and is constant above it.

LetThe Black-Scholes formula gives total equity value $C=x_0\Phi (d_1)-m{e}^{-rT}\Phi (d_2)$. Because the total company value is a traded asset, replication gives total debt value $x_0-C$. Divide by the numbers of securities to obtainThese are prices per security; the payoff diagram represents the two aggregate claims.

As insurance against credit default, the payment is the positive shortfall of recovery below face value:Thus the credit default swap in this question is one $m$th of a European put option on the firm value. Its price isEquivalently, adding this insurance to one corporate bond produces a sure unit payment, so its price is $e^{-rT}$ minus the bond price from part (b). This also fixes the positive orientation of the printed “difference”: insurance compensates face value minus recovery.

At maturity, one stock share is worth $(X_T-m{)}^{+}/n$. For a nonnegative strike $K$, the digital call option therefore pays one exactly onUnder the risk-neutral measure, $\log (X_T/x_0)$ is normal with mean $(r-{\sigma }^2/2)T$ and variance ${\sigma }^{2}T$. The risk-neutral pricing consequently givesThe effective company-value strike is $m+nK$, because both the outstanding debt and the number of shares matter. At $K=0$, the strict positive-share event remains $X_T>m$.

Write $\tau =T-t$. In the Black-Scholes formula, both $d_1$ and $d_2$ depend on $x,K$ only through $\log (x/K)$. Replacing them by $\lambda x,\lambda K$ leaves that ratio unchanged. Each prefactor in the put formula acquires the same factor $\lambda$, givingThis is degree-one homogeneity of the European put option price in its spot and strike together.

Differentiate the homogeneity identity from part (a) with respect to $\lambda$ at $\lambda =1$. The chain rule gives Euler homogeneous function theorem in the formUsing the supplied put option spot derivative $p_x=-\Phi (-d_1)$ and the given price,For positive strike, division by $K$ gives the put-price strike derivativeIt is nonnegative, as increasing the strike increases the terminal put payoff.

At $T_1$, with underlying spot $s$, the call option on a put option paysLet $D=K_2{e}^{-r(T_2-T_1)}$. The underlying put option value is strictly decreasing and convex in $s>0$, with limits $D$ at zero and zero at infinity. If $0<K_1<D$, there is a unique $s_{\ast }$ at which that put value equals $K_1$. Hence the compound option payoff is a decreasing convex curve from $D-K_1$ to zero on $[0,s_{\ast }]$, followed by a horizontal zero segment. There is an upward kink at $s_{\ast }$. If $K_1\ge D$, the payoff is identically zero.

For $K_1=0$, the holder obtains the underlying put at no cost, so the payoff is its full value at $T_1$. The discounted put price is a martingale under the risk-neutral measure. Using the tower property in the risk-neutral pricing,Explicitly, this is $K_2{e}^{-r(T_2-t)}\Phi (-d_2)-s\Phi (-d_1)$ with $d_1,d_2$ evaluated over $T_2-t$. The intermediate choice time disappears when its exercise cost is zero.

Write $p(s)=p(T,s;K_3,T_3)$ and $c(s)=c(T,s;K_3,T_3)$. Because the right can be left unexercised, the chooser with positive exercise cost has payoffLet $\tau =T_3-T$ and $D=K_3{e}^{-r\tau }$. Put-call parity gives $c(s)-p(s)=s-D$. Thus the put is selected for $s<D$ and the call for $s>D$; the two are equal at $s=D$. Before truncation at zero, the payoff is decreasing on the left and increasing on the right.

At the meeting point, the Black-Scholes formula givesIf $K<M$, the payoff has a positive minimum $M-K$ at $D$. If $K=M$, it touches zero there. If $M<K<D$, each side has a distinct exercise threshold and the payoff is zero between them. If $K\ge D$, even the left-end put value cannot cover the exercise cost, and only the right call branch can be positive. The illustration shows the middle case, including the competing underlying option values.

If $d_1>d_2$, buy $1/100$ of stock 1 and sell $1/100$ of stock 2. Initial cost is zero; the one-period payoff is $u_1-u_2>0$ in the up state and $d_1-d_2>0$ in the down state. This is an arbitrage. If terminal rather than time-one arbitrage is required, reinvest the positive receipts in either positive-price stock.

For $d_1<d_2$, define the one-period Arrow state pricesThe denominator is positive because $u_1/u_2>1>d_1/d_2$. Both state prices are strictly positive and satisfy ${\pi }_u{u}_i+{\pi }_d{d}_i=1$ for $i=1,2$. Thus every portfolio's present value equals the positive state-price-weighted value of its two future payoffs.

Use the same strictly positive pricing rule at every node of the two-period tree. A nonnegative terminal payoff that is positive in at least one state has strictly positive price by backward induction, so a zero-cost self-financing strategy cannot generate an arbitrage. Therefore the market is arbitrage free when $d_1<d_2$, despite having no initially supplied bond. This is a two-stock binomial market without a supplied bond.

For holdings $h_1,h_2$, a sure payment of 100 requiresSubtracting gives $2h_1+h_2=0$, so $h_1=-1$ and $h_2=2$. Thus sell one share of stock 1 and buy two shares of stock 2. Its initial cost is $100(-1+2)=100$.

This synthetic zero-coupon bond pays exactly its initial cost after one period, and thereforeAt later nodes the holdings can be rescaled to generate the same one-period risk-free return; a physical bond need not be listed separately.

With the rate $r=0$, the risk-neutral probability in a binomial market must satisfyThe first stock gives $0.8+0.6q=1$, hence $q=1/3$. The second gives $0.9+0.3q=1$, confirming the same answer. ThusTake the successive up/down choices independent under this measure, so both stock prices are martingales. The martingale transition condition applies from times zero to one and one to two; the displayed $t+1$ condition in the source should be read for these two transitions.

The exchange option has terminal payoffs determined by the two successive up/down moves:The two stocks span both one-step states, so this finite market is complete. The risk-neutral pricing uses the independent transition probabilities $q=1/3$ and $1-q=2/3$. Since the synthetic risk-free rate is zero,The price is approximately 7.56 in the currency units of the stock prices.