Science:Math Exam Resources/Courses/MATH152/April 2017/Question B 02/Solution 1

First, we need to know that equilibrium probability is defined as a probability vector, i.e., the sum of its components is 1. Exclude ${\text{(II)}}$ from answer.

Second, from the definition of the probability transition matrix, we have $\sum _{i=1}^{2}P_{i,j}=1.\,$. (Total probability should be 1) So ${\text{(C)}}$ is not a transition probability matrix. (i.e., ${\text{(VI)}}$.)

Recall that when $P$ has a single eigenvector for the eigenvalue $1$, we call the probability eigenvector as the equilibrium probability. Since we only have $2\times 2$ transition matrices, the maximum number of eigenvectors corresponding to the eigenvalue $1$ is $2$ and in that case actually $P{\mathbf {x}}={\mathbf {x}}$ for any ${\mathbf {x}}$. i.e., $P=I$. Therefore ${\text{(E)}}$ has no equilibrium probability.

Since the remained probability transition matrices are not identity matrix and we only have limited number of options for equilibrium probabilities, instead of finding eigen-vectors corresponding to the eigenvalue $1$, we find equilibrium probabilities by trials and errors,

${\text{(A)}}$: ${\begin{bmatrix}1/3&2/3\\2/3&1/3\end{bmatrix}}\cdot {\begin{bmatrix}0\\1\end{bmatrix}}\neq {\begin{bmatrix}0\\1\end{bmatrix}}$.

${\begin{bmatrix}1/3&2/3\\2/3&1/3\end{bmatrix}}\cdot {\begin{bmatrix}1/2\\1/2\end{bmatrix}}={\begin{bmatrix}1/2\\1/2\end{bmatrix}}$.

Thus corresponding equilibrium probability of ${\text{(A)}}$ is ${\text{(III)}}$

${\text{(B)}}$: ${\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}}\cdot {\begin{bmatrix}0\\1\end{bmatrix}}\neq {\begin{bmatrix}0\\1\end{bmatrix}}$.

${\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}}\cdot {\begin{bmatrix}1/2\\1/2\end{bmatrix}}={\begin{bmatrix}1/2\\1/2\end{bmatrix}}$.

Thus corresponding equilibrium probability of ${\text{(B)}}$ is ${\text{(III)}}$

${\text{(D)}}$: ${\begin{bmatrix}9/10&0\\1/10&1\end{bmatrix}}\cdot {\begin{bmatrix}0\\1\end{bmatrix}}={\begin{bmatrix}0\\1\end{bmatrix}}$.

Thus corresponding equilibrium probability of ${\text{(D)}}$ is ${\text{(I)}}$

Answer: $\color {blue}{\text{A-(III), B-(III), C-(VI), D-(I), E-(VIII)}}$