This may be a duplicate, but I’ve done some searching and I can’t find exactly this problem setting, probably due to not knowing the right terminology for how to refer to the transition matrix.
I’m considering absorbing random walks ${X_t}$ which have symmetric transition probabilities inside a window of size $k$. For example, on ${0, 1, 2, dots, 100}$, with $k=3$ if $X_t = 5$
begin{align*}
P[X_{t+1} = 4] &= P[X_{t+1} = 6]\
P[X_{t+1} = 3] &= P[X_{t+1} = 7]\
P[X_{t+1} = 2] &= P[X_{t+1} = 8]\
end{align*}
When the walk is within $k$ of the endpoint, the allowable transitions are truncated on both sides, e.g. if $X_t = 1$, $P[X_t = 3] = 0$.
I wanted to figure out the probability of the walk being absorbed at the upper endpoint, so I did some numerical experiments with walks on the integers from 0-100.
I calculated $T^n$ for the transition matrix $T$ and a large $n$. I then looked at $(T^n)_{i, 100}$ to see the probability of being absorbed at 100. It appears that the value is always equal to $i/100$, regardless of how I set the transition distribution or window-size.
The transition matrix for a walk on the integers in $[0, N]$ has the following properties:
- $T$ is (row) stochastic
- $T$ is a band matrix with upper and lower band size $k$
- $T_{i, j} = 0$ if $|i – j| > max (i, N – i)$
- $T_{i,i + r} = T_{i, i – r}$ (this is the main property I don’t know the name of)
- $T_{0, 0} = T_{1, 1} = 1$
I would like a proof or counterexample of the following statement:
begin{equation*}
limlimits_{n to infty} (T^n)_{i, N} = i/N
end{equation*}
If the statement above is true, is it still true if the transition distributions are no longer left-right symmetric, but it’s still the case that $E[X_{t + 1}] = X_t$?
