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I have a question concerning the distribution of the first hitting time of Brownian Motion $\tau_x = \inf_{t\geq 0}\{W_t=x\}$, where $W_t$ is Brownian motion. Using some calculus, I found out that the density of $\tau_x$ should be $$f(t)=\frac{x}{\sqrt{2\pi}t^{\frac{3}{2}}}\exp(-\frac{x^2}{2t})$$

In some scripts in the internet I have seen that this seems to be right. However, the task was to proove that $\tau_x$ has an inverse gaussian distribution, e.g. the density should be of the form

$$f(x)=(\frac{\lambda}{2\pi x^3})^{\frac{1}{2}}\exp(-\frac{\lambda(x-\mu)^2}{2\mu^2 x})$$ and I think it's impossible to get the upper density into that form since I miss a $t$ in the numerator of my exponent. Can anyone give me a reason?

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  • $\begingroup$ your density is $f(x)$ right? or $f(t)$? $\endgroup$ – Seyhmus Güngören Jun 20 '14 at 12:34
  • $\begingroup$ It's $f(t)$ :-/ $\endgroup$ – AlexConfused Jun 20 '14 at 12:42
  • $\begingroup$ You may want to re-read the question and make sure it isn't asking for this en.wikipedia.org/wiki/Inverse-gamma_distribution $\endgroup$ – Chris Janjigian Jun 20 '14 at 12:45
  • $\begingroup$ okay just was to make sure $\endgroup$ – Seyhmus Güngören Jun 20 '14 at 12:47
  • $\begingroup$ Thank you, Chris Janjigian. The article says that a random variable with inverse gamma distribution with parameters $(\frac{1}{2},\frac{c}{2})$ is the same (in distribution) like a levy $(0,c)$ distributed random variable. I'm gonna check if this solves it. $\endgroup$ – AlexConfused Jun 20 '14 at 12:51

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