Respuesta :
Answer:
The steady state proportion for the U (uninvolved) fraction is 0.4.
Step-by-step explanation:
This can be modeled as a Markov chain, with two states:
U: uninvolved
M: matched
The transitions probability matrix is:
[tex]\begin{pmatrix} &U&M\\U&0.85&0.15\\M&0.10&0.90\end{pmatrix}[/tex]
The steady state is that satisfies this product of matrixs:
[tex][\pi] \cdot [P]=[\pi][/tex]
being π the matrix of steady-state proportions and P the transition matrix.
If we multiply, we have:
[tex](\pi_U,\pi_M)*\begin{pmatrix}0.85&0.15\\0.10&0.90\end{pmatrix}=(\pi_U,\pi_M)[/tex]
Now we have to solve this equations
[tex]0.85\pi_U+0.10\pi_M=\pi_U\\\\0.15\pi_U+0.90\pi_M=\pi_M[/tex]
We choose one of the equations and solve:
[tex]0.85\pi_U+0.10\pi_M=\pi_U\\\\\pi_M=((1-0.85)/0.10)\pi_U=1.5\pi_U\\\\\\\pi_M+\pi_U=1\\\\1.5\pi_U+\pi_U=1\\\\\pi_U=1/2.5=0.4 \\\\ \pi_M=1.5\pi_U=1.5*0.4=0.6[/tex]
Then, the steady state proportion for the U (uninvolved) fraction is 0.4.
