Consider a one-dimensional random walk on the positive integers with:
(1) – A fully reflecting boundary at "x = 0".
(2) – Initiation of the walker at "x = 1".
(3) – A P[+1 Step] / P[RHS Step] probability for the walker less than or equal to the P[-1 Step] / P[LHS Step] probability.
What's the average distance of the walker from the origin and the probability distribution of the walker?
Best Answer
Added: As Douglas Zare points out, there is no limiting distribution. There are two different limits as time runs over even or odd values. The distribution I describe below is the average of these two, and hence my $\pi_n$ means the long run average probability that the walk finds itself in state $n$.
More formally, I mean that $$\pi_n=\lim_{t\to\infty} {1\over 2} \left(\mathbb{P}(X(t)=n)+\mathbb{P}(X(t+1)=n)\right)$$ where $X(t)$ is the position of the walk at time $t$.
Suppose that the walk has been running for a long time and has reached its equilibrium. We need the probability of a negative step (which I'll call $q$) to be strictly greater than probability of a positive step (which I'll call $p$), otherwise the equilibrium doesn't exist.
The steady state probabilities are $$\pi_n=\cases{{q-p\over 2q} & $n=0$\\[8pt] {q-p\over 2pq} \left({p\over q}\right)^n & $n\geq 1$}$$
Therefore, the average position of the walk (in the long run) is $$\sum_{n=0}^\infty n \pi_n ={1\over 2(q-p)}.$$
When $p$ and $q$ are close together, this number is very large. But as $q$ gets close to 1, the walk spends most of its time jumping back and forth between positions 0 and 1, and its average position is close to $1/2$.