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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!

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

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  • $\begingroup$ I do not quite understand the argument since invariant under the map. Can you be more specific? $\endgroup$ – stedmoaoa Oct 8 '18 at 13:39
  • $\begingroup$ @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. $\endgroup$ – Marcus M Oct 8 '18 at 15:59

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