Solution for american perpetual put

Question

I have been attempting an exercise in which I have to determine the value of an american perpetual put, $P$ in terms of the asset value $S$. The solution to the exercise says:

When $S>S_f$ (the optimal exercise boundary) P will satisfy

$\frac{1}{2}{\sigma}^2 S^2 \frac{\partial^2 P}{\partial S^2}-rP+rS\frac{\partial P}{\partial S}=0$ (standard BS pde)

where $r$ is risk free interest rate and $\sigma$ is the volatility of $S$

I understand that part but then it says to assume a solution of the form $P=S^{\lambda}$ where $\lambda$ is a real number.

Can anyone explain the idea behind this assumption?

Answer 1

To elaborate on the explanation provided by @Alex, the reasoning is because when we look at the PDE we notice that the $S$ terms appear in pairs with the $\dfrac{\partial}{\partial S}$, i.e. $S\dfrac{\partial}{\partial S}$ and $S^2\dfrac{\partial^2}{\partial S^2}$. What this says it that if we were to try a polynomial function of $S$ then after applying these operators then the exponent of $S$ would not change, i.e. $S\dfrac{\partial S^n}{\partial S} \to nS^n$ and similarly $S^2\dfrac{\partial^2 S^n}{\partial S^2} \to n(n-1)S^n$. This means that after trying this ansatz in the PDE it would cancel leaving only a polynomial equation for $n$, which if we can show has solutions, justifies the initial ansatz.

The overall reasoning behind this is that trying to solve PDEs in general is tedious, so always look for shortcuts specific to the problem. A good mentality to solving a PDE problem is that solving the actual PDE in general is relatively easy, the relatively hard bit is applying the boundary conditions! Hence whenever solving a PDE we should keep in mind how are our boundary conditions expressed? If they are a polynomial function of $S$ then we should try a polynomial function as our ansatz. If we have a wave boundary condition then try a wave ansatz, etc.

e.g.

The following $$ \dfrac{\sigma^2}{2} \frac{\partial^2 P}{\partial S^2}-rP+r\frac{\partial P}{\partial S}=0 $$ would typically motivate a solution $P = \exp(\mu S)$ and then we would solve for $\mu$.

A more general justification can be found in most courses in Linear Algebra and I would recommend seeing examples of the Sturm-Liouville problem. The general reasoning though is to change the basis into one that is easy to solve, e.g. we could solve your original PDE in terms of $\sin(S)$ and $\cos(S)$, or Legendre polynomials, Bessel functions, etc., but the solutions would not be anywhere as tidy. But you only learn which to try by drawing on experience, and even after years of experience and clever justifications it can frequently resort to trial and error.

I hope this helps.

Answer 2

With respect I think that this issue was associated Martingale properties AND dominated convergence theorem.(May be Wrong)

Let $L\in(0,K)$ a fixed price, we can consider the following choices for the exercise of a put option with strike price $K$:

• If $S_t\le K$, then we exercise contract at time $t$, and were delighted.
• O.W. we should wait until the first hitting time(of course we can play PS4 until the first hitting time!!) $$\tau_L=\inf \{u\ge t:S_u\le L \}$$ In case $S_T\le L$, the payoff will be $(K-S_t)^{+}$ since $K>L\ge S_t$. In case $S_t > L$, the price of the option will $$P_L(t,S_t)=E^{\mathbb{Q}}\left[e^{-r(\tau_L-t)}(K-S_{\tau_{L}})^+|S_t\right]$$ You know that the underlying asset price is written as $$S_t=S_0e^{rt+W^{\mathbb{Q}}_t-\frac{1}{2}\sigma^2\,t}\,\,(1)$$ We claim $$P_L(t,S_t=x)=\left\{ \begin{align} & K-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,0<x\le L \\ & (K-L){{\left( \frac{x}{L} \right)}^{-\frac{r}{2{{\sigma }^{2}}}}}\,,\,\,\operatorname{O}.W. \\ \end{align} \right. $$ As you see the starting date $t$ does not matter when pricing perpetual options hence $P_L(t, x)$ is actually independent of $t\in\mathbb{R}^+$, and the pricing of the perpetual put option can be performed by taking $t = 0$ and in the sequel we will work under $P_L(t,S_t)=P_L(S_t).$

Proof

Without loss of generality we take $t=0$

1. The result is obvious for $S_0=x\le L<K$ since in this case we have $\tau_L=t$ and $S_{\tau_L}=S_0=x$ and we have $$P_L(x)=E^{\mathbb{Q}}\left[e^{-r(t-t)}(K-x)^+|x\right]=K-x$$
2. (Fasten seat belt) Noe we should consider difficult case, i.e. $S_0=x>L$ .We know $$P_L(x)=E^{\mathbb{Q}}\left[e^{-r(\tau_L-0)}(K-S_{\tau_L})^+|S_0=x\right]=E^{\mathbb{Q}}\left[e^{-r\tau_L}(K-L)|S_0=x\right]$$ thus we just calculate $E^{\mathbb{Q}}\left[e^{-r\tau_L}|S_0=x\right]$. For this, we note that from $(1)$ for all $\lambda\in\mathbb{R}$ the process $\{Y_{\lambda}(t)\}_{t\ge 0}$ as defined $$Y_{\lambda}(t):={\left(\frac{S_t}{S_0}\right)}^{\lambda} e^{-t(r\lambda-\lambda(1-\lambda)\sigma^2/2)}=e^{\lambda\sigma W^{\mathbb{Q}}_t-\frac{1}{2}\lambda^2\sigma^2\,t}$$ is a martingale under the risk-neutral probability. Choosing $\lambda$ such that $$r=r\lambda-\frac{1}{2}\lambda(1-\lambda)\sigma^2$$ then $\lambda=1$ or $\lambda=-\frac{-2r}{\sigma^2}$. Choosing the negative solution, we have $\mathbb Q(\tau_L<\infty)=1$ and the bound
   $$0\le Y_{\lambda}(t)={\left(\frac{S_t}{S_0}\right)}^{\lambda} e^{-t(r\lambda-\lambda(1-\lambda)\sigma^2/2)}={\left(\frac{S_t}{S_0}\right)}^{-\frac{2r}{\sigma^2}} e^{-tr}\le {\left(\frac{L}{S_0}\right)}^{-\frac{2r}{\sigma^2}} $$ since $r>0$ and $\lambda=-\frac{-2r}{\sigma^2}$. This yields $${\left(\frac{L}{S_0}\right)}^{{-\frac{2r}{\sigma^2}}}E^{\mathbb{Q}}\left[e^{-r \tau_L}\right]=E^{\mathbb{Q}}\left[\left(\frac{S_{\tau_L}}{S_0}\right)^{-\frac{2r}{\sigma^2}}e^{-r\tau_L}\mathbb{1}_{\{\tau_L<\infty \}}\right]=E^{\mathbb{Q}}[Y_{\lambda}(t)\mathbb{1}_{\{\tau_L<\infty \}}]$$ as a result $${\left(\frac{L}{S_0}\right)}^{{-\frac{2r}{\sigma^2}}}E^{\mathbb{Q}}\left[e^{-r \tau_L}\right]=E^{\mathbb{Q}}\left[\underset{t\to \infty }{\mathop{\lim }} Y_{\lambda}(\tau_L\wedge t)\right]=\underset{t\to \infty }{\mathop{\lim }}E^{\mathbb{Q}}\left[Y_{\lambda}(\tau_L\wedge t)\right]=E^{\mathbb{Q}}\left[Y_{\lambda}(0)\right]$$ Indeed we used the dominated convergence theorem,we have $${\left(\frac{L}{S_0}\right)}^{{-\frac{2r}{\sigma^2}}}E^{\mathbb{Q}}\left[e^{-r \tau_L}\right]=E^{\mathbb{Q}}\left[Y_{\lambda}(0)\right]=1$$ In the oher words $$E^{\mathbb{Q}}\left[e^{-r\tau_L}|S_0=x\right]={\left(\frac{x}{L}\right)}^{-\frac{r}{2\sigma^2}}\,\,\,\,\,,\,\,\,\,x>L$$