Q.9.6 - A First Course in Probability

Question

Compute the limiting probabilities for the model of Problem 9.4. 

Step-by-Step Solution

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 1: 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.

Step 2: 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 

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

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.