# Expectation of first-passage-time of a diffusion process with negative drift

Take the stochastic process $X_0 = 0$ and $X_t = \nu t + \sigma W_t$ where $W_t$ is standard Brownian motion and $\nu$ is a drift which may be negative (this is the key complexity to the question)

Let $\alpha > 0$ be a fixed level, then define the first passage time as the random variable: $T = \inf\{ 0 < t \mid X_t=\alpha \}$.

Then, given some $r > 0$, what is the following expectation? $$U \equiv \mathbb{E}\left(e^{-r T} \right)$$

A related question may be: what is the moment generating function for the $T$ stopping time? Note that I do not care about the pdf or cdf of $T$, just the above expectation.

Solving the $\nu \geq 0$ case: I believe this turns out to be easy as this follows the example in https://en.wikipedia.org/wiki/Inverse_Gaussian_distribution#Relationship_with_Brownian_motion . Hence, $T$ is distributed as an Inverse Gaussian: $T\sim IG(\tfrac\alpha\nu, \tfrac {\alpha^2} {\sigma^2})$ in that notation.

Moreover, the moment generating function for the $IG(\mu,\lambda)$ is $$M(t;\mu,\lambda) \equiv \exp\left[{\frac{\lambda}{\mu}\left(1-\sqrt{1-\frac{2\mu^2t}{\lambda}}\right)}\right]$$

Finally, note that the expectation I gave is exactly the definition of the MGF at $t=-r$, so we have our answer, $$\mathbb{E}\left(e^{-r T} \right) = M\left(-r;\tfrac\alpha\nu, \tfrac {\alpha^2} {\sigma^2}\right)$$

Solving the $\nu < 0$ case: If you look carefully at theorems such as in Karlin and Taylor "A First Course in Stochastic Processes" page 362 and theorem 5.3, or the wikipedia page above, they always assume that $\nu \geq 0$. As Karlin and Taylor puts it: "When $\nu < 0$, $T$ has a defective probability distribution, that is $T$ is infinite with positive probability".

However, to find $\mathbb{E}\left(e^{-r T}\right)$ we don't really need to find the probability distribution. Is the moment generating function (and hence the direct solution for $U$) still defined? Is the formula the same as above?

• For the less general case, see stats.stackexchange.com/questions/232363/… for more details and the connection to the IG distribution. – jlperla Aug 31 '16 at 4:12
• Is it possible for your model to consider the case when a is also less than zero when the drift coefficient is negative, as in that case first passage density is well defined and not defective. – Comic Book Guy Aug 31 '16 at 21:36
• Thanks. You are right that it works if the drift is towards the barrier. Sadly, I need it to move in the opposite direction – jlperla Aug 31 '16 at 21:41
• Please see Theorem 2 in Feller (who i believe coined the term defective density). math.ucsd.edu/~lni/math257/Tauberian-Theorem.pdf – Comic Book Guy Aug 31 '16 at 23:09
• @ComicBookGuy Thank you, very helpful. I don't quite understand the application of the theorem to my problem (if you mean the continuity theorem on page 431) since the limits are unclear for me. However, this may be part of the solution: There is no problem whatesoever in using laplace transforms/mgfs of distributions with defective densities – jlperla Sep 1 '16 at 3:45

The answer is $$E[e^{-rT}] = exp\left\{\frac{\mu+\sqrt{\mu^2+2\sigma^2r}}{\sigma^2} \alpha\right \}$$ if $\alpha<0$, and $$E[e^{-rT}] = exp\left\{\frac{\mu-\sqrt{\mu^2+2\sigma^2r}}{\sigma^2} \alpha \right\}$$ if $\alpha>0$

To establish these formulas one can follow the approach from Karlin and Taylor. Since Karlin and Taylor provide the derivations for the case $\alpha>0$, I derive the formula for the case $\alpha<0$.

Let $V_{\lambda}(t)=exp\{\lambda X(t) - q(\lambda)t \}$ where $q(\lambda) = \mu \lambda+\frac{1}{2}\sigma^2\lambda^2$. Note $V_{\lambda}(t)$ is a martingale.

Consider $V_{\lambda}(t \wedge T )$. Then $V_{\lambda}(t \wedge T )$ is also a martingale and is bounded for $\lambda<0$. We assume then that $\lambda<0$, and from the above observations we have, $$E[V_{\lambda}(t \wedge T )] = E[V(0)] = 1$$ Moreover $$\lim_{t \rightarrow \infty} V_{\lambda}(t \wedge T) = \left\{ \begin{array}{ll} exp\{\lambda T - q(\lambda) T & \text{ if } T<\infty \\ 0 & \text{ if } T=\infty \end{array} \right.$$

Therefore, $$\lim_{t \rightarrow \infty} E[V_{\lambda}(t \wedge T )]= E\left[\lim_{t \rightarrow \infty} V_{\lambda}(t \wedge T )\right] = e^{\lambda \alpha}E[e^{-q(\lambda)T}]$$

From the above derivations it follows that $$1=e^{\lambda \alpha}E[e^{-q(\lambda)T}]$$ or $$e^{-\lambda \alpha}=E[e^{-q(\lambda)T}]$$ Set $q(\lambda)=r$ and solve the resulting quadratic to obtain $$\lambda=\frac{-\mu \pm \sqrt{\mu^2+2\sigma^2 r}}{\sigma^2}$$ Since we require $\lambda<0$ the correct root is the negative one. Thus, we obtain $$E[e^{-rT}]=\exp \left\{ \frac{\mu + \sqrt{\mu^2+2\sigma^2 r}}{\sigma^2} \alpha \right\}$$

Finally, note that if $\mu>0$ then $\lim_{r \rightarrow \infty} E[e^{-rT}] \neq 1$ which implies that the associated distribution function is defective, while if $\mu>0$ then $\lim_{r \rightarrow \infty} E[e^{-rT}]=1$ and the associated distribution is not defective (Theorem 2 in Feller that Comic Book Guy referred to).

• Looks good to me. Any comments from others? – jlperla Sep 3 '16 at 22:24