To solve the problem, we need to determine the stationary distribution [tex]\( S \)[/tex] and then use [tex]\( S \)[/tex] to construct the limiting matrix [tex]\( \overline{P} \)[/tex]. Here's a step-by-step solution:
1. Understand the Stationary Distribution:
For the stationary distribution [tex]\( S \)[/tex] of a transition matrix [tex]\( P \)[/tex], we need:
[tex]\[
S \cdot P = S
\][/tex]
Additionally, the elements of [tex]\( S \)[/tex] must sum to 1.
2. Set Up the Equations:
Let [tex]\( S = \begin{bmatrix} s_1 & s_2 \end{bmatrix} \)[/tex]. The requirements are:
[tex]\[
\begin{bmatrix} s_1 & s_2 \end{bmatrix} \cdot \begin{bmatrix} 0.3 & 0.7 \\ 0.3 & 0.7 \end{bmatrix} = \begin{bmatrix} s_1 & s_2 \end{bmatrix}
\][/tex]
Which leads to the system of linear equations:
[tex]\[
\begin{cases}
0.3s_1 + 0.3s_2 = s_1 \\
0.7s_1 + 0.7s_2 = s_2 \\
s_1 + s_2 = 1
\end{cases}
\][/tex]
3. Simplify the Equations:
Simplifying the first two equations, we get:
[tex]\[
\begin{cases}
0.3s_1 + 0.3s_2 = s_1 \\
0.7s_1 + 0.7s_2 = s_2
\end{cases}
\][/tex]
This can be rearranged to:
[tex]\[
\begin{cases}
-0.7s_1 + 0.3s_2 = 0 \\
0.7s_1 - 0.3s_2 = 0
\end{cases}
\][/tex]
4. Solve the System:
Since the equation system is dependent, we rely on the constraint [tex]\( s_1 + s_2 = 1 \)[/tex]. Solving one of the simplified equations:
[tex]\[
-0.7s_1 + 0.3s_2 = 0 \implies 0.3s_2 = 0.7s_1 \implies s_2 = \frac{7}{3} s_1
\][/tex]
Substituting into [tex]\( s_1 + s_2 = 1 \)[/tex]:
[tex]\[
s_1 + \frac{7}{3} s_1 = 1
\][/tex]
[tex]\[
\frac{10}{3}s_1 = 1 \implies s_1 = \frac{3}{10} = 0.3
\][/tex]
Thus,
[tex]\[
s_2 = 1 - s_1 = 1 - 0.3 = 0.7
\][/tex]
5. Stationary Distribution:
[tex]\[
S = \begin{bmatrix} 0.3 & 0.7 \end{bmatrix}
\][/tex]
6. Construct the Limiting Matrix:
The limiting matrix [tex]\( \overline{P} \)[/tex] has rows equal to the stationary distribution [tex]\( S \)[/tex]:
[tex]\[
\overline{P} = \begin{bmatrix}
0.3 & 0.7 \\
0.3 & 0.7
\end{bmatrix}
\][/tex]
Hence, the stationary matrix [tex]\( S \)[/tex] and the limiting matrix [tex]\( \overline{P} \)[/tex] are:
[tex]\[
S = \begin{bmatrix} 0.3 & 0.7 \end{bmatrix}
\][/tex]
[tex]\[
\overline{P} = \begin{bmatrix}
0.3 & 0.7 \\
0.3 & 0.7
\end{bmatrix}
\][/tex]
