# Optimal Choice of exceeding time

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 $P(S_T>a)$.

• 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. – aajajim Nov 21 '13 at 21:44
• I might have chosen a wrong notation. To use the relfection principle I need to take the supremum of $S_0+\sigma B_t$ – TI Jones Nov 21 '13 at 21:50
• 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. – TI Jones Nov 21 '13 at 21:52
• Ohhh sorry, you're right. I was a bit faster on reading your question :)! I will think about it! – aajajim Nov 21 '13 at 22:38

## 1 Answer

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

• 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 $V'(a)=0$? – TI Jones Nov 24 '13 at 14:29
• Maybe, I would try differentiating the expression as it is and shove it into Mathematica :) The integral appears quite messy :/ – jensa Nov 24 '13 at 14:56