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I am considering the Inverse Gaussian distribution as the hitting time distribution for a Wiener process, $W(t)$, with drift parameter $\nu$ and variance parameter $\sigma$. Define the hitting time as the first time the process reaches the level $a > 0$:

$T = \min_{t>0}\{ W(t) \geq a\}.$

The hitting time, $T$, of the level $a$ is then Inverse Gaussian. Let $\mu = a/\nu$ and $\lambda = a^2/\sigma^2$:

$f_T(t) = \sqrt{\frac{\lambda}{2 \pi t^3}} \exp\Big\{- \frac{\lambda(t-\mu)^2}{2\mu^2t}\Big\}$

$P(T \leq t) = F(t) = \Phi\Big(\sqrt{\frac{\lambda}{t}}\Big(\frac{t}{\mu} - 1\Big)\Big) - \exp\Big(\frac{2\lambda}{\mu}\Big)\Phi\Big(- \sqrt{\frac{\lambda}{t}}\Big(\frac{t}{\mu} + 1\Big)\Big),$

where $\Phi(\cdot)$ is the standard normal distribution CDF.

I am looking to find the conditional Inverse Gaussian distribution, conditioned on the endpoint value. Say the endpoint value, $W(t)$ = $r < a$ is known, I want to find the probability that the threshold has been crossed:

$P(T \leq t \big| W(t) = r < a)$

What I have done leaves me with quite nasty expressions. Any help on approximations as well as insight on simplifications of the expressions is appreciated. What I have done is: for $r < a$

$P(T \leq t \big| W(t) = r) = \frac{P(T \leq t, W(t) = r)}{P(W(t) = r)} $

Then, for the sake of notation, consider the problem "discretely". I say that for each $0 < s < t$, the joint probability can be found by considering the hitting time $s$ and the increment from $s$ to $t$, $W(t) - W(s),$ which must be $r - a$: $ P(T = s) P(W(t) - W(s) = r - a)$, so I must sum over $s$. In continuous notation:

$P(T \leq t, W(t) = r) = \int_{0}^t f_T(s) f_{W(t) - W(s)}(r-a) ds$

Note that $W(t) - W(s)$ is normally distributed with mean $(t-s)\nu$ and variance $(t-s)\sigma^2$.

Finally in my problem I must sum over all $0 < r < a$ which leaves me with a double integral, to integrate out $W(t) = r$ on which it was conditioned in the beginning.

$\displaystyle\int_{r=0}^a\frac{\int_{0}^t f_T(s) f_{W(t) - W(s)}(r-a) ds}{f_{W(t) - W(0)}(r)} dr.$

How can I simplify this?

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