# Stationary distribution and limiting distribution of a random walk

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!

• Given your transition matrix, do you also mean than the walk can also move from $N$ to $0$ as if $0=N+1$? – Saad Oct 8 '18 at 13:15
• yes. the walk can get from N to 0 and from 0 to N – stedmoaoa Oct 8 '18 at 13:18

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.

• I do not quite understand the argument since invariant under the map. Can you be more specific? – stedmoaoa Oct 8 '18 at 13:39
• @stedmoaoa, sure I added a bit more; you can also just look at the direct argument instead, but the soft argument is the more intuitive reason. – Marcus M Oct 8 '18 at 15:59