# Limit Behavior of Transition Probability Matrix

So the problem is: Let $$P = \begin{bmatrix}0.2 &0.8&0&0&0&0&0 \\0.7 & 0.3&0&0&0&0&0\\0&0&0.3&0.5&0.2&0&0\\0&0&0.6&0&0.4&0&0\\0&0&0&0.4&0.6&0&0\\0&0.1&0.1&0.2&0.2&0.3&0.1\\0.1&0.1&0.1&0&0.1&0.2&0.4 \end{bmatrix}$$

Find $\lim_{n\to \infty}P^{n}$.

So I am thinking maybe I should apply the conclusion I got from the question before this one which is, Let since $j$ is recurrent, non-null and aperiodic, then for any $i∈E$

$$\lim_{n\to \infty}P^{n}(i, j)= F(i,j)(\pi_j)$$ in which $F(i,j) = P_i(T_1<\infty$), where $T_1$ is the first hitting time of $j$.

But I am not exactly sure how to work this problem out. Appreacie any help!

There are three classes of states in this Markov chain:

\begin{align} C_1 &= \{1,2\}\\ C_2 &= \{3,4,5\}\\ C_3 &= \{6,7\}. \end{align}

$C_1$ and $C_2$ are recurrent, while $C_3$ is transient.

Conditioned on $X_0\in C_1$, we find the limiting distribution to be $$\pi_1=\left(\frac7{15},\frac8{15}\right)$$

Conditioned on $X_0\in C_2$, we find the limiting distribution to be $$\pi_2=\left(\frac6{23},\frac7{23},\frac{10}{23}\right).$$

Let $q_{ij}$ be the probability of absorption into state $j$ conditioned on $X_0=i$, for $i=6,7$ and $j=1,2$. Then we have the system of equations \begin{align} q_{61} &= \frac1{10} + \frac3{10}q_{61} +\frac1{10}q_{71}\\ q_{62} &= \frac12 + \frac3{10}q_{62}+ \frac1{10}q_{72}\\ q_{71} &= \frac15 + \frac25 q_{71}+ \frac15 q_{61}\\ q_{72} &= \frac15 + \frac25 q_{72}+ \frac15 q_{62},\\ \end{align} which yields $$q_{61} = \frac15,\ q_{62}=\frac45,\ q_{71}=\frac25,\ q_{72}=\frac35.$$ It follows that \begin{align} \lim_{n\to\infty}P^n&= \begin{bmatrix} \pi_1(1) & \pi_1(2)&0&0&0&0&0\\ \pi_1(1) & \pi_1(2)&0&0&0&0&0\\ 0&0&\pi_2(3) & \pi_2(4) & \pi_2(5) & 0 & 0\\ 0&0&\pi_2(3) & \pi_2(4) & \pi_2(5) & 0 & 0\\ 0&0&\pi_2(3) & \pi_2(4) & \pi_2(5) & 0 & 0\\ q_{61}\pi_1(1) & q_{61}\pi_1(2) & q_{62}\pi_2(3) & q_{62}\pi_2(4) & q_{62}\pi_2(5) & 0 & 0\\ q_{71}\pi_1(1) & q_{71}\pi_1(2) & q_{72}\pi_2(3) & q_{72}\pi_2(4) & q_{72}\pi_2(5) & 0 & 0\\ \end{bmatrix}\\\\ &=\begin{bmatrix} \frac{7}{15} & \frac{8}{15} & 0 & 0 & 0 & 0 & 0 \\ \frac{7}{15} & \frac{8}{15} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{6}{23} & \frac{7}{23} & \frac{10}{23} & 0 & 0 \\ 0 & 0 & \frac{6}{23} & \frac{7}{23} & \frac{10}{23} & 0 & 0 \\ 0 & 0 & \frac{6}{23} & \frac{7}{23} & \frac{10}{23} & 0 & 0 \\ \frac{7}{75} & \frac{8}{75} & \frac{24}{115} & \frac{28}{115} & \frac{8}{23} & 0 & 0 \\ \frac{14}{75} & \frac{16}{75} & \frac{18}{115} & \frac{21}{115} & \frac{6}{23} & 0 & 0 \\ \end{bmatrix}. \end{align}

• Thanks a lot! One question: In computing the $q_{61},q_{62}, q_{71},$ and $q_{71}$, I don't see how to construct the equations. For example, in computing $q_{61}$, where do the coeffcients $\frac1{10}, \frac3{10}$ and $\frac1{10}$ come from? – Kior Oct 12 '16 at 10:32