Results 1 to 2 of 2

Math Help - Markov Chain HELP!!!!

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    2

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,806
    Thanks
    697
    Hello, stephy7878!

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

    Find the fixed probability vector.

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

    We have: . (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: . \begin{array}{ccc}\frac{3}{8}x + \frac{3}{8}y + \frac{3}{8}z &=& x \\ \\[-4mm]<br />
\frac{5}{8}x + \frac{3}{8}y + \frac{1}{2}z &=& y \\ \\[-4mm]<br />
\qquad \frac{1}{4}y + \frac{1}{8}z &=& z \end{array}

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


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

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

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

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

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


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


    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Markov Chain of random variables from a primitive markov chain
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: October 19th 2011, 08:12 AM
  2. Markov Chain Help
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: June 28th 2010, 07:37 AM
  3. Markov Chain
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: February 12th 2010, 07:23 AM
  4. Markov Chain
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: December 12th 2009, 04:52 PM
  5. Replies: 2
    Last Post: October 28th 2008, 06:32 PM

Search Tags


/mathhelpforum @mathhelpforum