# Transition probabilities for the symmetric random walk on the integers

I found that most references for the symmetric random walk on the integers are for the discrete time case, i.e. the ones that gives us explicit transition probabilities. Now, I am looking at a random walk $X:[0,\infty) \rightarrow \mathbb{Z}$ with rates $q(i,i-1)=q(i,i+1)=\frac{1}{2}$ and want to know the transition probabilities $p_t(0,n)=?$, but I would also like to know if there are more general results.

The more general question could be: Does anybody know if there is a way to state the transition probabilities explicitly for arbitrary transition rates?

Theoretically, we only need to solve the infinite system of ODE's:

$\frac{dp_t(x,y)}{dt} = \frac{1}{2}\left(p_t(x-1,y)+p_t(x+1,y)\right)-p_t(x,y)$ for all pairs of $x,y \in \mathbb{Z}$ and initial condition $p_t(x,y)=\delta_{x,y}$ but this is not that simple (if one does not know the answer).

• The natural continuous analog of a RW is Brownian motion. – Christian Remling Feb 1 '16 at 1:38
• @ChristianRemling is this a hint or a general remark? if it is a hint, I have troubles to interpret it. – Acuriousmind Feb 1 '16 at 1:44
• It's a general comment. – Christian Remling Feb 1 '16 at 2:52
• "Solve an infinite system of ODEs" is the same as "compute the semigroup generated by an infinite matrix". Often the best place to start is diagonalizing the matrix, i.e. look for eigenfunctions or a spectral decomposition. – Nate Eldredge Feb 1 '16 at 3:15

I think this is as explicit as you're going to get: Of course $p_t(x,y)=p_t(0,y-x)$, so we may as well concentrate on that.
You can think of a continuous time random walk as writing out a list of instructions at various times along the real line (which represents time). The instructions are either $+1$ or $-1$; and they occur at points of independent Poisson processes with rate $\frac 12$ according to the convention above.
Now $p_t(0,n)$ is the probability that there $n$ more $+1$'s than $-1$'s up to time $t$ (I'll assume $n>0$ because the case $n<0$ is similar). That is it is the sum over $k$ of the probability that there are $n+k$ $+1$'s and $k$ $-1$'s up to time $t$. You have precise expressions for each of these since it's a Poisson process, so
$$p_t(0,n)=e^{-t} \sum_{k=0}^\infty \Big(\frac{t}2\Big)^{n+k+k}\frac{1}{(n+k)!k!}.$$