Question

A transition probability matrix is said to be doubly

stochastic if

$\sum _{i=0}^M{P}_{ij}=1$

for all states j = 0, 1, ... , M. Show that such a Markov chain is ergodic, then

j = 1/(M + 1), j = 0, 1, ... , M.

Short answer

It is proved that a Markov chain is ergodic, then ${\mathrm{\pi }}_j=\frac{1}{M+1}$for$j=0,1,...,M$

Step-by-step solution

Given Information

We have given that the transition probability matrix is doubly stochastic if

$\sum _{i=0}^M{P}_{ij}=1$

for all states$j=0,1,...,M$.

Simplify

As the chain ergodic and the transition matrix is doubly stochastic, there exists a unique stationary distribution $\mathrm{\pi }$. Now, we just have to check that is that localid="1648139523807" ${\mathrm{\pi }}_i=\frac{1}{\mathrm{m}+!}$solution of the system of the equation $\mathrm{\pi }=\mathrm{\pi P}$i.e., is it true

${\mathrm{\pi }}_j=\sum _{\mathrm{i}}{\mathrm{p}}_{\mathrm{ij}}{\mathrm{\pi }}_{\mathrm{j}}$${\mathrm{\pi }}_j=\sum _{\mathrm{i}}{\mathrm{p}}_{{\mathrm{i}}_{\mathrm{j}}}{\mathrm{\pi }}_{\mathrm{j}}$

but, we havelocalid="1648139404599" $\sum _{i}pij=1$, which means

$\frac{1}{M+1}=\frac{1}{M+1}.\sum _{i}pij=\frac{1}{M+1}.1=\frac{1}{M+1}$

So, we have proved that the stationary distribution is localid="1651480780070" ${\mathrm{\pi }}_j=\frac{1}{\mathrm{m}+1}$.