Applying the Black–Scholes PDE to a Nonstandard Payoff

# Black-Scholes Equation with dividend


# Black-Scholes Equation with dividend












Consider a European option with payoff $$g(S_T) = S_T^{-5}e^{10S_T}$$ Assume that the interest rate is $r = .1$ and the underlying asset satisfies $S_0 = 2, \sigma = .2$, an pays dividend at continuous rate equal to $q(t,S_t) = qS_t$ and $q = .2$

a.) Write the Black-Scholes equation for this problem.

b.) Solve the problem analytically by the method of separation of variables. Plug into the equation a solution candidate of the form $e^{a\tau}S^{-5}e^{10S}$ and determine $a$.

Attempted solution for a.) The Black-Scholes model with dividend is given by the SDE $$dS_t = S_t(r - q(t,S_t))dt + \sigma S_t dB_t$$ and the Black-Scholes equation is given by $$\begin{cases} \partial_\tau V(\tau,S) &= \frac{\sigma^2 S^2}{2}\partial_{SS} V(\tau,S) + (r - q(t,S))S \partial_S V(\tau,S) - rV(\tau,S)\\ V(\tau,0) &= e^{-r\tau}g(0)\\ V(0,S) &= g(S) \end{cases}$$ thus with the parameters above we have $$\begin{cases} \partial_{\tau}V(\tau,S) &= \frac{(.2)^2(2)^2}{2}\partial_{S S}V(\tau,S) + (.1 - 2(.2))2V(\tau,S) - .1V(\tau,S)\\ V(\tau,0) &= e^{-.1\tau}g(0) = 0\\ V(0,S) &= g(2) = 2^{-5}e^{20} \end{cases}$$

A tad confused about b.). Any suggestions would be greatly appreciated. Also, if anyone can check part a.) solution that would be great.

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/25266

Let \begin{align*} V(t, S_t) = E\Big(e^{-r(T-t)} g(S_T)\mid \mathcal{F}_t \Big) \end{align*} be the risk-neutral value at time $t$ of the option payoff $g(S_T)$. Then $\{e^{-rt}V(t, S_t), 0 \le t \le T\}$ is a martingale. Consequently, \begin{align} -rV + \frac{\partial V}{\partial t} + (r-q)S\frac{\partial V}{\partial S_t}+\frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S_t^2} = 0,\tag{1} \end{align} which is the Black-Scholes equation for the solution.

For a solution of the form \begin{align*} V(t, S_t) = e^{a(T-t)}S_t^{-5}e^{10S_t}. \end{align*} Note that \begin{align*} \frac{\partial V}{\partial t} &= -a V, \\ \frac{\partial V}{\partial S_t} &= \Big(-\frac{5}{S_t}+10\Big) V,\ \mbox{ and}\\ \frac{\partial V^2}{\partial S_t^2} &= \frac{5}{S_t^2}V + \Big(-\frac{5}{S_t}+10\Big)^2 V\\ &=\Big(\frac{30}{S_t^2} - \frac{100}{S_t} + 100\Big) V. \end{align*} Substitute into Equation (1) and evaluate at $t=0$, \begin{align*} -r-a +(r-q)(-5+10S_0) + \frac{1}{2}\sigma^2 \big(30-100 S_0 + 100 S_0^2\big) = 0. \end{align*} That is \begin{align*} a &= -r+(r-q)(-5+10S_0) + \frac{1}{2}\sigma^2 \big(30-100 S_0 + 100 S_0^2\big)\\ &= -0.1+ (0.1-0.2)\times (-5+10\times 2) + 0.5 \times 0.2^2 \times (30-100 \times 2 + 100 \times 4)\\ &=3. \end{align*} Finally, the option value is given by \begin{align*} V(0, S_0) &= e^{a(T-0)}S_0^{-5}e^{10S_0}\\ &= \cdots \end{align*} Please fill-in here. What is $T$?