# Barrier Derivative Pricing

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$. What is the present value of this derivative? ## 1 Answer As is often the case, there are generally two solution strategies here. 1. (Probabilistic) You explicitly solve for the expected discount factor at the first passage time$\nu$of$S$to the level$B$under the risk-neutral probability measure$\mathbb{P}^*$, i.e. $$V_0 = \mathbb{E}_{\mathbb{P}^*} \left[ e^{-r \nu} \right].$$ 2. (Differential Equation) The option value$V$satisfies the ODE $$\frac{1}{2} \sigma^2 S^2 \frac{\mathrm{d} V^2}{\mathrm{d} S^2} + r S \frac{\mathrm{d} V}{\mathrm{d} S} - r V = 0$$ 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 $$S^2 \frac{\mathrm{d} V^2}{\mathrm{d} S^2} + \lambda S \frac{\mathrm{d} V}{\mathrm{d} S} - \lambda V = 0,$$ where$\lambda = 2r / \sigma^2$. This equation is of the Euler-Cauchy type and we thus try the solution $$V(S) = S^\beta$$ and get $$\beta (\beta - 1) S^\beta + \beta \lambda S^\beta - \lambda S^\beta = 0.$$ This equation holds for all values of$S$if $$\beta^2 + \beta (\lambda - 1) - \lambda = 0.$$ Solving for$\beta$yields $$\beta_\pm = \frac{1}{\sigma^2} \left( -\left( r - \frac{1}{2} \sigma^2 \right) \pm \left( r + \frac{1}{2} \sigma^2 \right) \right)$$ and we notice that$\beta_+ = 1$and$\beta_- = -\lambda$. The general solution to the ODE is given by $$V(S) = c_- S^{-\lambda} + c_+ S,$$ 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, $$V(S) = \left( \frac{S}{B} \right)^{-\lambda}.$$ References Wilmott, Paul (2006) Paul Wilmott on Quantitative Finance, Vol. 1: Wiley, 2nd edition. • 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\$ ? – chengcj Mar 1 '17 at 21:58
• As the option is perpetual, its price cannot depend on time but only on the spot. Any time is like any other. – LocalVolatility Mar 1 '17 at 22:06