First-passage probability with absorbing boundary at origin (No Laplace)

Question

I have the following problem which I would like to solve without using Laplace transform. Can you possibly help or provide pointers?

What is the first-passage probability, and mean first-passage time for a random walk to reach the origin if it starts from distance $M$ away from the origin?

I would then like to derive the same set of equations as above, but assuming there is an absorbing boundary at the origin. Any advice greatly appreciated.

Answer 1

I think the mean it is easy once $S_n -M$ is sum of $n$ independent random variables. Then, if $T \in L^1$ than it would satisfy Wald's identity, but $E[S_{T} -M] = E[T]E[X_1]$ gives you a contradiction if $M\ne0$ and the random walk is symmetric. Then $T\notin L^1$.

If the random walk has a drift to the right, then the SLLN gives you the same result.

Besides that, I have no idea. =)

Answer 2

The mean first passage time diverges, while the first passage time pdf is \begin{equation} f(t)=\frac{M}{\sqrt{4\pi D t^3}}e^{-M^2/4Dt} \end{equation} Now, as Redner states in his book First passage processes (beware, a lot of typos!) the relation between pdf's $p(x,t)$ and first passage pdf's $F(x,t)$ may be expressed as follows: \begin{equation} p(x,t)=\delta_{x,0} \delta_{t,0}+\sum_{t'=0}^{t}F(x,t')p(0,t-t') \end{equation} which may be inverted through Laplace transform.

Now, if you assume absorbing boundary conditions at $x=0$, then you just need to put an image at $-M$ (if theres a drift, then the image is a bit more complicated since), i.e., \begin{equation} p(x,t)=\frac{1}{\sqrt{4 \pi D t}}\left(e^{-(x-M)^2/4Dt}-e^{-(x+M)^2/4Dt} \right) \end{equation} Then, by flux considerations, the first passage time pdf is obtained from \begin{equation} f(t)=D \frac{\partial p(x,t)}{\partial x} \vert_{x=0} \end{equation}