Question

Compute the limiting probabilities for the model of Problem 9.4.

Short answer

The limiting probabilities is$\mathrm{\pi }=\left[\begin{array}{c}\frac{1}{20}\\ \frac{9}{20}\\ \frac{9}{20}\\ \frac{1}{20}\end{array}\right]$

Step-by-step solution

Given Information 

We have given that 3 white and 3 black balls are distributed in two urns in which each urn contains 3 balls.

We need to find the limiting probability.

Simplify

To find the distribution $\mathrm{\pi }=\underset{\mathrm{n}\to \infty }{\lim }{\mathrm{P}}^{\mathrm{n}}{\mathrm{s}}_0$for every starting distribution $s_0$, we are solving

$\mathrm{\pi }=\mathrm{\pi P}$

Which yield the equalities

localid="1648053603504" $$\begin{gathered} \mathrm{\pi }=\frac{1}{9}{\mathrm{\pi }}_1 \\ {\mathrm{\pi }}_1={\mathrm{\pi }}_0+\frac{4}{9}{\mathrm{\pi }}_2 \\ {\mathrm{\pi }}_2=+\frac{4}{9}{\mathrm{\pi }}_1+\frac{4}{9}{\mathrm{\pi }}_2+{\mathrm{\pi }}_3 \\ {\mathrm{\pi }}_3=+\frac{1}{9}{\mathrm{\pi }}_2 \end{gathered}$$

Solving this system and using $\sum _i{P}_i=1$we end up with distribution

$\mathrm{\pi }=\left[\begin{array}{c}\frac{1}{20}\\ \frac{9}{20}\\ \frac{9}{20}\\ \frac{1}{20}\end{array}\right]$

This is the required stationary and limit distribution.