Consider the random walk on $S={0, . . . , N}$, defined as follows. We are given $p, q, r >0$ with$ p+q+r= 1$. The walk increases by 1 with probability $p$, decreases by 1 with probability $q$, and stays put with probabity $r$.What is its stationary and limiting distribution?
I writed out the transition matrix.
$$
\begin{pmatrix}
r & p & \cdots &⋯ &⋯&q\\
q & r & p &\cdots&\cdots&\cdots \\
0 & q & r &p &\vdots&\vdots\\
0 & 0 & q &r &p &⋯\\
p & 0 & 0 &0 &q & r \\
\end{pmatrix}
$$
the stationary matrix satisfies $$\pi=\pi p$$
then it satisfies $$p_{i,j} \pi_i = p_{j,i} \pi_j$$
$$p\pi_1 = q\pi_2, p\pi_2 = q\pi_3, p\pi_3 = q\pi_4…$$
the stationary situation then becomes
$$
\begin{pmatrix}
1 & \frac{p}{q}\pi_1 & (\frac{p}{q})^2\pi_1 & (\frac{p}{q})^3\pi_1 &…&(\frac{p}{q})^N\pi_1\\
\end{pmatrix}
$$
We can then get the distribution because it adds up to 1.
Is my method correct?
Then, for the limiting distribution, I think there is no limiting distribution because it seems not to be converging to a specific state. Is it true? Thank you!
Best Answer
The stationary measure will be the uniform measure, which you could see by a soft argument: the transition matrix is irreducible and aperiodic, and thus has a stationary measure; furthermore, it is invariant under the map $k \mapsto k+1$, implying that each state must get the same measure.
You could also show this manually: let $\mu_j = \frac{1}{N+1}$ for each $j \in \{0,1,\ldots,N\}$. Then $$(P \mu)_j = p \mu_{j-1} + r \mu_{j} + q \mu_{j+1} = \frac{p + r + q}{N+1} = \frac{1}{N+1} = \mu_j$$
where the indices are $\mod{N+1}$.
EDIT: some further comment on the invariance thing; let $A$ be the map that cycles the states, i.e. $A (x_0 ,\ldots, x_{N})^T = (x_N,x_0,x_1,\ldots,x_{N-1})^T\,.$
Then note that $PA = P$. If you set $\mu$ to be the stationary measure of $P$ (i.e. $\mu P = \mu$), then it must be invariant under $A$: $$\mu PA = \mu P \implies \mu A = \mu\,.$$ Iteratively applying $A$ shows that $\mu$ must be uniform.