Suppose you hold a share from company $Z$ whose vaue at time $t$ is $S_0+\sigma B_t$ where $B_t$ is Brownian Motion and $\sigma$ denotes some volatility. Now lets assume that company $Z$ may go bankrupt at some expoentially-distributed random variabl $T$ with mean $1/\lambda$. Now you plan to sell your share at the first time $H$ that the price exceeds $a$, i.e $H=\inf\{t: S_0+\sigma B_t>a\}$. If $H<T$ the vaue to you is $a\exp(-rH)$, otherwise it is worth nothing.

Do you have any idea how I can derive the optimal choice of $a$ ?

My intuitive way to solve this exercise is to first check what is the probability that $H<T$, i.e $P(H<T)$ Now I think I can use the Reflection Principle, so I define $S_t=\sup (S_0+\sigma B_t)$, then the Principle states that $P(S_T>a)=P(H<T)$. I think the solution to the problem is to find the maximal $a$ such that $P(S_T>a)$, but I do not know how to compute $P(S_T>a)$.

  • $\begingroup$ are you sure about your definition of $S_{t}$? Actually, $S_{t}=S_{0} + \sigma B_t = S_0 + \sigma \sqrt{t} G$, Where $G$ is a Gaussian standardized random variable. So if you are just looking for $P(S_T > a)$, then i can tell that its $P(S_T > a) = P(S_0 + \sigma \sqrt{T} G > a) = P(G > \frac{-(S_0-a)}{\sigma \sqrt{t}}) = P(G < \frac{(S_0-a)}{\sigma \sqrt{t}})=\Phi(\frac{(S_0-a)}{\sigma \sqrt{t}})$n where $\Phi(x)$ is Cumulative Distribution Function of the Gaussian random variable. $\endgroup$ – aajajim Nov 21 '13 at 21:44
  • $\begingroup$ I might have chosen a wrong notation. To use the relfection principle I need to take the supremum of $S_0+\sigma B_t$ $\endgroup$ – TI Jones Nov 21 '13 at 21:50
  • $\begingroup$ I think there is a mistake in your calculation, I have to use the fact that $T$ is eponentially distributed with mean $1/\lambda$ somewhere. $\endgroup$ – TI Jones Nov 21 '13 at 21:52
  • $\begingroup$ Ohhh sorry, you're right. I was a bit faster on reading your question :)! I will think about it! $\endgroup$ – aajajim Nov 21 '13 at 22:38

It's been quite a while since I did this stuff, but I'll add my input. Please correct me if appropriate.

$\{H < T\} = \{ \sup_{0\leq s \leq T} (S_{0} + \sigma B_{s}) > a \} = \{\sup_{0 \leq s \leq T} B_s > \frac{a-S_0}{\sigma}\}$.

Set $\mu := \frac{a-S_0}{\sigma}$ and $M_{T} := \sup_{0 \leq s \leq T} B_{s}$.

Then, $P(\{H < T\} = P(\{M_T > \mu \}) = 2\left(1 - \Phi\left(\frac{\mu}{\sqrt{T}}\right)\right)$.

Seek to maximize $V(a) := E\{ae^{-rH}1_{\{H < T\}}\} = a \int_{0}^{\infty}\lambda e^{-\lambda x} \int_{0}^{x}e^{-ry} \frac{d}{d\xi}\left(2 \Phi\left(\frac{\mu}{\sqrt{\xi}}\right)-1\right)(y) \, dy \, dx$.

  • $\begingroup$ Thanks, you think it is possible to get a closed formula for the maximal $a$? I mean differentiating $V(a)$ w.r.t $a$ and finding the $a$ s.t $V'(a)=0$? $\endgroup$ – TI Jones Nov 24 '13 at 14:29
  • $\begingroup$ Maybe, I would try differentiating the expression as it is and shove it into Mathematica :) The integral appears quite messy :/ $\endgroup$ – jensa Nov 24 '13 at 14:56

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.