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Sagot :
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]
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]
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