user366312
Gold Member
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 2
 Problem Statement

If ##(X_n)_{n≥0}## is a Markov chain on ##S = \{1, 2, 3\}## with initial distribution ##α = (1/2, 1/2, 0)## and transition matrix
## \begin{bmatrix} 1/2&0&1/2\\ 0&1/2&1/2\\ 1/2&1/2&0 \end{bmatrix},##
then ##P(X_2 = 2) = ?## and ##E(X_2)=?##.
 Relevant Equations
 Markov Chain
My solution:
##X_1 = \begin{bmatrix} 1/2&1/2&0 \end{bmatrix} \begin{bmatrix} 1/2&0&1/2\\ 0&1/2&1/2\\ 1/2&1/2&0 \end{bmatrix} = \begin{bmatrix} 1/4&1/4&1/2 \end{bmatrix}##
##X_2 = \begin{bmatrix} 1/4&1/4&1/2 \end{bmatrix} \begin{bmatrix} 1/2&0&1/2\\ 0&1/2&1/2\\ 1/2&1/2&0 \end{bmatrix} = \begin{bmatrix} 3/8&3/8&1/4 \end{bmatrix}##
So, ##P(X_2=2) = 3/8##
##E(X_2=2) = 1 * 3/8 + 2 * 3/8 + 3 * 1/4 = 15/8##
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Is this solution correct?
Why or why not?
##X_1 = \begin{bmatrix} 1/2&1/2&0 \end{bmatrix} \begin{bmatrix} 1/2&0&1/2\\ 0&1/2&1/2\\ 1/2&1/2&0 \end{bmatrix} = \begin{bmatrix} 1/4&1/4&1/2 \end{bmatrix}##
##X_2 = \begin{bmatrix} 1/4&1/4&1/2 \end{bmatrix} \begin{bmatrix} 1/2&0&1/2\\ 0&1/2&1/2\\ 1/2&1/2&0 \end{bmatrix} = \begin{bmatrix} 3/8&3/8&1/4 \end{bmatrix}##
So, ##P(X_2=2) = 3/8##
##E(X_2=2) = 1 * 3/8 + 2 * 3/8 + 3 * 1/4 = 15/8##
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Is this solution correct?
Why or why not?
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