The transition probability matrix of Markov Chain is

$$ \left[\begin{array} r 0.5 & 0.4 & 0.1 \\ 0.3 & 0.4 & 0.3 \\ 0.2 & 0.3 & 0.5 \end{array}\right] $$

Find the limiting probabilities.

Mumbai University > Electronics and Telecommunication > Sem5 > Random Signal Analysis

Marks: 10M

Year: May 2015, Dec 2014

To find the limiting probabilities i.e $\lim_{n→∞}⁡P^n $

Let the limiting probabilities be $π=[π_1 \ π_2 \ π_3 ]$. Then we have $πP=π$ such that $∑π_i=1$

$$ ∴[π_1 \ π_2 \ π_3 ]\left[\begin{array} r 0.5 & 0.4 & 0.1 \\ 0.3 & 0.4 & 0.3 \\ 0.2 & 0.3 & 0.5 \end{array}\right]=[π_1 \ π_2 \ π_3 ] $$

$$0.5π_1+0.3π_2+0.2π_3=π_1 --- (1)$$

$$0.4π_1+0.4π_2+0.3π_3=π_2 --- (2)$$

$$0.1π_1+0.3π_2+0.5π_3=π_3 --- (3)$$

We know $∑π_i=1$

$$π_1+ π_2+ π_3=1 --- (4)$$

i.e $π_3=1-π_1-π_2$

Substituting the above value in equation (1) and (2)

$0.5π_1+0.3π_2+0.2(1-π_1-π_2)=π_1$

$∴ - 0.7π_1+0.1π_2=-0.2 ---- (5)$

$0.4π_1+0.4π_2+0.3(1-π_1-π_2)=π_2$

$∴ 0.1π_1-0.9π_2=-0.3 --- (6)$

Multiply 9 with equation (5) and adding equation (5) & (6) we get,

$$-6.2π_1=-2.1$$

$$∴ π_1=0.3333$$

$$∴ π_2=0.3703$$

$$∴ π_3=1-π_1-π_2$$

$$π_3=1-0.3333-0.3703$$

$$π_3=0.2964$$

$$∴ π_1=0.3333 \ \ π_2=0.3703 \ \ π_3=0.2964$$