Chapter 9, Problem 1RE

Given the transition matrix $P$ and initial-state matrix $S_0$ shown below, find $S_1$ and $S_2$, and explain what each represents:

$\begin{array}{l}\begin{array}{cc}A& B\end{array}\\ P=\begin{array}{c}A\\ B\end{array}\left[\begin{array}{cc}.6& .4\\ .2& .8\end{array}\right]S_0=\left[\begin{array}{cc}.3& .7\end{array}\right]\end{array}$

To determine

To determine: The first state transition matrix, $S_1$, and the second state transition matrix, $S_2$, of the given matrix.

Answer to Problem 1RE

The first state transition matrix is given by $S_1=\left[\text{0}\text{.32 0}\text{.68}\right]$. This represents that after one trial, the probability of being; in state $A$ is $0.32$ ; and in $B$ is $0.68$.

The second state transition matrix is given by $S_2=\left[\text{0}\text{.328 }0.\text{672}\right]$. This represents that after two trials, the probability of being; in state $A$ is $0.328$ ; and in state $B$ is $0.672$.

Explanation of Solution

:

Consider the given information,

Here, $P$ represents the transition matrix, $\begin{array}{l}AB\\ p=\begin{array}{c}A\\ B\end{array}\left[\begin{array}{cc}0.6& 0.4\\ 0.2& 0.8\end{array}\right]\end{array}$, and $S_0$ is the $\text{initial-state}$ matrix, $S_0=\left[0.3\text{ 0}.7\right]$.

The first state transition matrix is given by $S_1=S_{0}P$.

Thus,

$\begin{array}{c}{S}_1=S_{0}P\\ =\left[\text{0}\text{.3 0}\text{.7}\right]\left[\begin{array}{cc}0.6& 0.4\\ 0.2& 0.8\end{array}\right]\\ =\left[\left(\text{0}\text{.3}\right)\left(\text{0}\text{.6}\right)\text{+}\left(\text{0}\text{.7}\right)\left(\text{0}\text{.2}\right)\left(0.3\right)\left(0.4\right)+\left(0.7\right)\left(0.8\right)\right]\\ =\left[\text{0}\text{.32 0}\text{.68}\right]\end{array}$

Hence, the first state transition matrix is given by $S_1=\left[\text{0}\text{.32 0}\text{.68}\right]$.

The first entry in the first row represents the probability of being in state $A$ after one trial, and the second entry in the first row represents the probability of being in state $B$ after one trial.

The second state transition matrix is given by $S_2=S_{1}P$.

$\begin{array}{c}{S}_2=S_{1}P\\ =\left[\text{0}\text{.32 0}\text{.68}\right]\left[\begin{array}{cc}0.6& 0.4\\ 0.2& 0.8\end{array}\right]\\ =\left[\left(\text{0}\text{.32}\right)\left(\text{0}\text{.6}\right)\text{+}\left(\text{0}\text{.68}\right)\left(\text{0}\text{.2}\right)\left(0.32\right)\left(0.4\right)+\left(\text{0}\text{.68}\right)\left(0.8\right)\right]\\ =\left[\text{0}\text{.328 0}\text{.672}\right]\end{array}$

Hence, the second state transition matrix is $S_2=\left[\text{0}\text{.328 0}\text{.672}\right]$.

The first entry in the first row represents the probability of being in state $A$ after two trials, and the second entry represents the probability of being in state $B$ after two trials.

Therefore, the first state transition matrix is given by $S_1=\left[\text{0}\text{.32 0}\text{.68}\right]$. This represents that after one trial, probability of being; in state $A$ is $0.32$ ; and in $B$ is $0.68$.

The second state transition matrix is given by $S_2=\left[\text{0}\text{.328 }0.\text{672}\right]$. This represents that after two trials, probability of being; in state $A$ is $0.328$ ; and in state $B$ is $0.672$.