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

MATH152 April 2017

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Question B 02

For each matrix below left, interpret the matrix as a probability transition matrix, and match it to its corresponding equilibrium probability on the right. One mark each. Note that answers in the list (I)-(VII) may be a match more than once or not at all. (A) ${\begin{bmatrix}1/3&2/3\\2/3&1/3\end{bmatrix}},$ (B) ${\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}},$ (C) ${\begin{bmatrix}1/2&1/2\\2/3&1/3\end{bmatrix}}$, (D) ${\begin{bmatrix}9/10&0\\1/10&1\end{bmatrix}}$, (E) ${\begin{bmatrix}1&0\\0&1\end{bmatrix}}$, (I) ${\begin{bmatrix}0\\1\end{bmatrix}}$, (II) ${\begin{bmatrix}1\\1\end{bmatrix}}$, (III) ${\begin{bmatrix}1/2\\1/2\end{bmatrix}}$, (IV) ${\begin{bmatrix}1/3\\2/3\end{bmatrix}}$, (V) ${\begin{bmatrix}9/10\\1/10\end{bmatrix}}$, (VI) not a probability transition matrix, (VII) equilibrium probability not in the list, (VIII) no equilibrium probability

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Hint
Let $p_{i,j}$ be the probability of moving to state $i$ if you are in state $j$. Then the transition matrix matrix is defined by $P=[p_{i,j}]$. When $P$ has a single eigenvector for the eigenvalue $1$, we call the probability eigenvector as the equilibrium probability.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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: $\color {blue}{\text{A-(III), B-(III), C-(VI), D-(I), E-(VIII)}}$