Geometric Brownian Motion Probability of hitting uper boundary

Let $(S_t)$ be a geometric Brownian Motion, i.e. $S_t = S_0 e^{\left(\mu-\frac{\sigma^2}{2}\right)t+ \sigma W_t }$. Let $\alpha>0$ and $\tau = \inf\{t>0 | S_t \geq \alpha\}$. Compute $P(\tau \leq t)$.

What I have done:

I know that for $X_t = W_t - \beta t$, $X_t^* = sup_{0\leq s \leq t} X_s$ and $\alpha \geq 0$

$$P(X_t^* \geq \alpha ) = 1- \Phi\left( \frac{\alpha + \beta t}{\sqrt{t}} \right) + e^{-2 \alpha \beta} \Phi\left( \frac{\beta t - \alpha}{\sqrt{t}} \right)$$ Furthermore, I know that $P(\tau \leq t) = P(S_t^* \geq \alpha )$. Hence I would have started with $$\log(S_t)= \log(S_0) + \left(\mu - \frac{\sigma^2}{ 2 }\right)t + \sigma W_t \geq \alpha$$ $$W_t \geq \frac{ \log\left(\frac{\alpha}{S_0}\right) - \left(\mu - \frac{\sigma^2}{ 2 }\right)t}{\sigma}$$ Now I would set $$\alpha^{'} = \frac{\log\left(\frac{\alpha}{S_0}\right)}{\sigma}$$ and $$\beta = -\frac{\left(\mu - \frac{\sigma^2}{ 2 }\right)}{\sigma}$$ and would use the formula for $P(X_t^* \geq \alpha^{'} )$ as stated above. Is this correct? I am not sure and I feel like I have done something wrong. Any hints or comment welcome -thanks!

• For deterministic $\theta_t$, $\mathbb{P}(W_t \geq \theta_t ) = 1 - \Phi(\theta_t)$. As you have already calculated, here $$\theta_t = \frac{1}{\alpha}\left(- \left(\mu - \frac{\sigma^2}{2}\right)t+\log \frac{\alpha}{S_0}\right).$$ – ChargeShivers May 6 '17 at 14:34
• @ChrisVarghese you mean $\frac{1}{\sigma}$ don't you? But I can/should use the formula given by $P(X_t^{*} \geq \alpha)$ with my values for the cosntant $\alpha$ and the drift $\beta$ since I can rewrite my $W_t$ as $X_t = W_t + \frac{\mu - \frac{\sigma^2}{2}}{\sigma}$ define $X_t^{*} = \sup_{0\leq s \leq t} W_s$ and then plug in? – Methamortix May 6 '17 at 14:58

$$P(\tau \leq t ) = P(S_s \geq \alpha, s \in [0,t]) = P(\log(S_0) + \left(\mu - \frac{\sigma^2}{ 2 }\right)s + \sigma W_s \geq \alpha, s \in [0,t])$$ $$=P\left( W_s \geq \frac{ \log\left(\frac{\alpha}{S_0}\right) - \left(\mu - \frac{\sigma^2}{ 2 }\right)s}{\sigma}, s \in [0,t] \right)$$
Now set $$\alpha^{'} = \frac{\log\left(\frac{\alpha}{S_0}\right)}{\sigma}$$ and $$\beta = -\frac{\left(\mu - \frac{\sigma^2}{ 2 }\right)}{\sigma}$$ Then you have $$P\left( W_s \geq \alpha^{'} + \beta s, s \in [0,t] \right)$$ Define $X_t = W_t - \beta t$ and $X_t^* = \sup_{0\leq s \leq t} X_s$ and use the formula $$P(X_t^* \geq \alpha^{'} ) = 1- \Phi\left( \frac{\alpha^{'} + \beta t}{\sqrt{t}} \right) + e^{-2 \alpha^{'} \beta} \Phi\left( \frac{\beta t - \alpha^{'} }{\sqrt{t}} \right)$$