Past exam of the martingales and financial mathematics course of the University of Oxford / 2009 / B / 5

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.