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?
4
-
$\begingroup$ 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 $x=X_0$. $\endgroup$– John DawkinsMay 7, 2016 at 16:42
-
$\begingroup$ 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 $x = X_0$ $\endgroup$– green diodMay 7, 2016 at 16:57
-
$\begingroup$ But $X_t = X_0\exp(\sigma W_t-\sigma^2t/2+\mu t)$. $\endgroup$– John DawkinsMay 8, 2016 at 16:20
-
$\begingroup$ You're right, my mistake, too much stuff on my pad. $\endgroup$– green diodMay 8, 2016 at 23:11
Add a comment
|