Let $X$ be a geometric Brownian motion $dX_t = \mu X_t dt + \sigma X_t dW_t, X_0 > 0$ and ${\cal F}$ its natural filtration. Let $\tau_a$ be the first hitting time of $a$ by $X$. How can we relate the Laplace transform of the hitting time of $X$ to the one of $W$ for which we know the expression?
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It might help to note that $\tau_a$ is the hitting time by $W$ of the line $t\mapsto [\log(a/x)+(\sigma^2/2-\mu)t]/\sigma$, where $x=X_0$. – John Dawkins May 07 '16 at 16:42
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Hi @JohnDawkins, thanks. I tried finding the corresponding hitting level for $W$ but then I was stuck. I was not aware of this result about the hitting time of a line by a Brownian motion. Just a correction, IMO, the line should be $t \mapsto [\log(a/x) - \mu t ] (2/\sigma^2)$ where $x = X_0$ – green diod May 07 '16 at 16:57
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But $X_t = X_0\exp(\sigma W_t-\sigma^2t/2+\mu t)$. – John Dawkins May 08 '16 at 16:20
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You're right, my mistake, too much stuff on my pad. – green diod May 08 '16 at 23:11