Past exam of the mathematical models of financial derivatives course of the University of Oxford / 2023 / 2 / b

Write $\tau =T-t>0$ and $g_n(s)=(K-s{)}_{+}^n$. The European power put formula isHere $Y>0$ has a lognormal distribution. Since $g_n$ is decreasing and is a convex function, its discounted expected value is also decreasing and convex as a function of $S$. For $S>0$ it is smooth and strictly positive: the normal distribution assigns positive probability to $SY<K$ whatever the current positive spot price. Its endpoints areThe dominated convergence theorem justifies the second limit, because $0\le (K-SY{)}_{+}^n\le K^n$. For a quantitative sketch, the binomial theorem and truncated lognormal moment givewhere $\Phi$ is the standard normal distribution function. This finite sum is useful for drawing the curve; its positive expectation form makes its qualitative shape clearer.

The solid curves use $K=1$, $r=0.05$, $\sigma =0.30$ and $\tau =1$; the dashed curves show their terminal financial payoffs for comparison. Each price remains positive at $S=K$ and tends smoothly towards zero beyond it. At small $S$, discounting gives a price below the undiscounted financial payoff; European exercise is permitted only at expiry.

Assume $r>0$, put $b=S^{\star }\in (0,K)$ and let ${\tau }_b=\inf \{u\ge 0:S_u\le b\}$. In the continuation region $S>b$, the time-independent Black-Scholes equation is a Cauchy-Euler equation. Substituting $V=S^m$ givesSet $\gamma =2r/{\sigma }^2>0$. The two powers are $S$ and $S^{-\gamma }$. The boundary condition $V(S)\to 0$ as $S\to \infty$ removes the growing solution; value matching at $b$ sets the coefficient. Hence the fixed-trigger value isFor $0\le S\le b$ the policy exercises immediately, giving $V_b(S)=(K-S{)}^n$. The quoted continuation expression alone does not apply below the exercise boundary. Equivalently, the discounted boundary-hitting representation gives ${\mathbb{E}}_S[e^{-r{\tau }_b};{\tau }_b<\infty ]=(S/b{)}^{-\gamma }$, which multiplies the exercise financial payoff.

For $S>b$, maximize the coefficient $H(b)=b^{\gamma }(K-b{)}^n$ of $S^{-\gamma }$. Its logarithmic derivative isIt changes sign once, at $b=\gamma K/(\gamma +n)$. The perpetual power put option therefore has candidate valueThe same boundary follows from smooth fit, $-\gamma (K-b{)}^n/b=-n(K-b{)}^{n-1}$. To establish optimality among all stopping times, rather than only fixed barriers, we verify the candidate. Let $\mathcal{L}=rS{\partial }_S+{\sigma }^2{S}^2{\partial }_{SS}/2$ be the infinitesimal generator. The maximum property of $H$ gives $V_n\ge g_n$ everywhere, and $(\mathcal{L}-r)V_n=0$ for $S>b$. For $S\le b$, set $y=S/(K-S)\le \gamma /n$. ThenThus the discounted candidate is a nonnegative supermartingale. The smooth fit removes the derivative-jump local time of a semimartingale term in the Itô formula. The optional stopping theorem and localization bound every exercise policy's discounted expected value by $V_n$. Immediate exercise attains it below $b$; the first hitting time of $b$ attains it above $b$. The terminal remainder vanishes since $0\le e^{-ru}{V}_n(S_u)\le K^n{e}^{-ru}\to 0$. This proves the displayed value.

For $n=1$ it reduces to the ordinary perpetual put option. Larger $n$ lowers the optimal trigger, because the more strongly curved financial payoff makes further downward moves more valuable. If $r=0$, under the risk-neutral measure the stock tends to zero almost surely, so the supremum is $K^n$, approached as the exercise time tends to infinity and not attained at a finite time when $S>0$. If $r<0$, the values at deterministic exercise times grow without bound; the perpetual value is infinite. These cases do not admit the positive-rate formula above.