Transition probabilities for the symmetric random walk on the integers

Question

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).

Answer 1

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

$$ p_t(0,n)=e^{-t} \sum_{k=0}^\infty \Big(\frac{t}2\Big)^{n+k+k}\frac{1}{(n+k)!k!}. $$