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Thread: Markov Chain HELP!!!!

  1. #1
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    Markov Chain HELP!!!!

    Hello,
    Not sure where to put this question and let me know if it needs to be put somewhere else... I'm learning about Markov Chain Matrices and I am a little confused on how to take a transition matrix and finding the fixed probability vector??? Here is the problem.. let me know if you can help me out..

    This is a transition matrix

    0.375 0.625 0
    0.375 0.375 0.25
    0.375 0.5 0.125

    Find the fixed probability vector. I'm not really sure how to start this I'm really confused. Any ideas?


    The result should be a 1x3 Martix
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  2. #2
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    Hello, stephy7878!

    Given the transition matrix: .$\displaystyle A \;=\;\begin{pmatrix}\frac{3}{8} & \frac{5}{8} & 0 \\ \\[-4mm]
    \frac{3}{8} & \frac{3}{8} & \frac{1}{4} \\ \\[-4mm]
    \frac{3}{8} & \frac{1}{2} & \frac{1}{8} \end{pmatrix} $

    Find the fixed probability vector.

    We want a row vector: .$\displaystyle v \:=\:(x,y,z)\:\text{ so that }\:v\!\cdot\! A \:=\:v$

    We have: .$\displaystyle (x,y,z)\,\begin{pmatrix}\frac{3}{8}&\frac{5}{8}&0 \\ \\[-4mm] \frac{3}{8} & \frac{3}{8} & \frac{1}{4} \\ \\[-4mm] \frac{3}{8} & \frac{1}{2} & \frac{1}{8}\end{pmatrix} \;=\;(x,y,z)$

    Multiply: .$\displaystyle \begin{array}{ccc}\frac{3}{8}x + \frac{3}{8}y + \frac{3}{8}z &=& x \\ \\[-4mm]
    \frac{5}{8}x + \frac{3}{8}y + \frac{1}{2}z &=& y \\ \\[-4mm]
    \qquad \frac{1}{4}y + \frac{1}{8}z &=& z \end{array}$

    The equations simplify to: .$\displaystyle \begin{array}{cccc}\text{-}5x + 3y + 3z &=& 0 & {\color{blue}[1]} \\
    5x - 5y + 4z &=& 0 & {\color{blue}[2]} \\
    \qquad 2y - 7z &=& 0 & {\color{blue}[3]} \end{array}$


    Add [1] and [2]: .$\displaystyle -2y + 7z \:=\:0$ . . . whch is equivalent to [3].

    So we need another equaton.
    Here it is! . . . $\displaystyle x + y + z \;=\;1\;\;{\color{blue}[4]}$ .
    . . . This is true for all fixed probability vectors.

    From [3], we have: .$\displaystyle y \:=\:\tfrac{7}{2}z$

    Substitute into [2]: .$\displaystyle 5x - 5\left(\tfrac{7}{2}z\right) + 4z \:=\:0 \quad\Rightarrow\quad x \:=\:\tfrac{27}{10}z$

    Substitute into [4]: .$\displaystyle \tfrac{27}{10}z + \tfrac{7}{2}z + z \:=\:1 \quad\Rightarrow\quad z \:=\:\frac{5}{36}
    $
    . . . Then: .$\displaystyle x \:=\:\tfrac{27}{10}\left(\tfrac{5}{36}\right) \quad\Rightarrow\quad x \:=\:\frac{3}{8}$
    . . And: .$\displaystyle y \:=\:\tfrac{7}{2}\left(\tfrac{5}{36}\right) \quad\Rightarrow\quad y \:=\:\frac{35}{72}$


    Therefore, the fixed probabiity vector is: .$\displaystyle v \;=\;\left(\frac{3}{8},\:\frac{35}{72},\:\frac{5}{ 36}\right) $


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