Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
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]
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
