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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.

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    $\begingroup$ Those are not difficult questions. Try to work them out yourself will be better for you, in particular, for your PhD thesis later on. If your text book is not clear, try some other books as well, for example, the book "Stochastic Calculus for Finance II" by Shreve, which is pretty readable. $\endgroup$ – Gordon Apr 4 '16 at 18:55
  • $\begingroup$ Ok, thank you I will $\endgroup$ – Wolfy Apr 4 '16 at 19:38
  • $\begingroup$ For (b), both $\tau$ and $T$ appeared in $e^{a\tau}S^{-5}e^{10S_T}$. Is it a typo? Why do you need the $\tau$? $\endgroup$ – Gordon Apr 4 '16 at 20:32
  • $\begingroup$ It could be a typo but I am not sure $\endgroup$ – Wolfy Apr 4 '16 at 20:43
  • $\begingroup$ I think its safe to assume it is a typo $\endgroup$ – Wolfy Apr 4 '16 at 20:48
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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 $T$?

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  • $\begingroup$ Thank you for your patience I will ask my professor what $T$ is and get back to you $\endgroup$ – Wolfy Apr 6 '16 at 18:06
  • $\begingroup$ I will leave the specific calculations for you to fill-in. $\endgroup$ – Gordon Apr 6 '16 at 18:07
  • $\begingroup$ So, is my part a.) of the solution incorrect? $\endgroup$ – Wolfy Apr 6 '16 at 23:30

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