4
$\begingroup$

Assume constant interest rate $r$ and a stock with current price at $S_0$ that pays no dividend (assume $S_t\ge0$). When the stock price hits the barrier $B$ (where $B<S_0$) you receive \$$1$ and the derivative would terminate. This derivative doesn't have a maturity date. $S_t$ follows geometric Brownian motion with constant volatility $\sigma$.

What is the present value of this derivative?

$\endgroup$
1
$\begingroup$

As is often the case, there are generally two solution strategies here.

  1. (Probabilistic) You explicitly solve for the expected discount factor at the first passage time $\nu$ of $S$ to the level $B$ under the risk-neutral probability measure $\mathbb{P}^*$, i.e. \begin{equation} V_0 = \mathbb{E}_{\mathbb{P}^*} \left[ e^{-r \nu} \right]. \end{equation}

  2. (Differential Equation) The option value $V$ satisfies the ODE \begin{equation} \frac{1}{2} \sigma^2 S^2 \frac{\mathrm{d} V^2}{\mathrm{d} S^2} + r S \frac{\mathrm{d} V}{\mathrm{d} S} - r V = 0 \end{equation} subject to the contract-specific boundary conditions.

I will outline the second approach here and refer to e.g. Chapter 9 in Wilmott (2006) for further details. See e.g. this blog post for a solution to the finite-maturity American digital call option valuation problem using the first approach. In order to obtain the solution for the perpetual case, simply take the limit $T \rightarrow \infty$. The solution to the put is fully analogous.

ODE Approach

The ODE can be rearranged to

\begin{equation} S^2 \frac{\mathrm{d} V^2}{\mathrm{d} S^2} + \lambda S \frac{\mathrm{d} V}{\mathrm{d} S} - \lambda V = 0, \end{equation}

where $\lambda = 2r / \sigma^2$. This equation is of the Euler-Cauchy type and we thus try the solution

\begin{equation} V(S) = S^\beta \end{equation}

and get

\begin{equation} \beta (\beta - 1) S^\beta + \beta \lambda S^\beta - \lambda S^\beta = 0. \end{equation}

This equation holds for all values of $S$ if

\begin{equation} \beta^2 + \beta (\lambda - 1) - \lambda = 0. \end{equation}

Solving for $\beta$ yields

\begin{equation} \beta_\pm = \frac{1}{\sigma^2} \left( -\left( r - \frac{1}{2} \sigma^2 \right) \pm \left( r + \frac{1}{2} \sigma^2 \right) \right) \end{equation}

and we notice that $\beta_+ = 1$ and $\beta_- = -\lambda$. The general solution to the ODE is given by

\begin{equation} V(S) = c_- S^{-\lambda} + c_+ S, \end{equation}

where $c_\pm$ depend on the boundary conditions of the contract. In case of a put option we have the upper boundary condition $\lim_{S \rightarrow \infty} V(S) = 0$, which implied that $c_+ = 0$. The value matching condition at the lower boundary is $U(B) = 1$ and we thus obtain $c_- = B^\lambda$. Consequently,

\begin{equation} V(S) = \left( \frac{S}{B} \right)^{-\lambda}. \end{equation}

References

Wilmott, Paul (2006) Paul Wilmott on Quantitative Finance, Vol. 1: Wiley, 2nd edition.

$\endgroup$
  • $\begingroup$ Thank you very much for helping. Mind if I ask how come the differential equation is only dependant on $S$ and not dependant on $t$ ? $\endgroup$ – chengcj Mar 1 '17 at 21:58
  • $\begingroup$ As the option is perpetual, its price cannot depend on time but only on the spot. Any time is like any other. $\endgroup$ – LocalVolatility Mar 1 '17 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.