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Thread: Markov Chain problem

  1. #1
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    Markov Chain problem

    Given a markov chain with states {$\displaystyle e_1,e_2,e_5$} and the transition probability matrix

    $\displaystyle P = \begin{bmatrix}0.2 & 0.8 & 0\\1 & 0 & 0\\0.7 & 0 & 0.3\end{bmatrix}$

    which on of the following is false?
    a) $\displaystyle e_1$ communicates with $\displaystyle e_2$
    b) $\displaystyle e_1$ is accessible from $\displaystyle e_3$
    c) $\displaystyle e_3$ communicates with $\displaystyle e_1$
    d) $\displaystyle e_2$ is accessible from $\displaystyle e_3$
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  2. #2
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    Hello, Bucephalus!

    Given a Markov chain with states {$\displaystyle e_1,e_2,e_5$} and the transition probability matrix

    . . $\displaystyle P \;=\; \begin{bmatrix}0.2 & 0.8 & 0\\1 & 0 & 0\\0.7 & 0 & 0.3\end{bmatrix}$

    Which one of the following is false?

    a) $\displaystyle e_1$ communicates with $\displaystyle e_2$
    I assume that "communicates" is a two-way relationship.
    . . That is: .$\displaystyle e_1\to e_2$ .and .$\displaystyle e_2\to e_1$

    Since .$\displaystyle \begin{Bmatrix}P(e_1\to e_2) &=& 0.8 \\ P(e_2\to e_1) &=& 1\end{Bmatrix}$ . the statement is true.




    b) $\displaystyle e_1$ is accessible from $\displaystyle e_3$

    Since .$\displaystyle P(e_3\to e_1) \:=\: 0.7,$ .the statement is true.




    c) $\displaystyle e_3$ communicates with $\displaystyle e_1$

    We have: .$\displaystyle P(e_3\to e_1) \:=\:0.7$

    . . . .but: .$\displaystyle P(e_1\to e_3) \:=\:0$

    $\displaystyle e_1$ and $\displaystyle e_3$ do not communicate.

    This statement is false.




    d) $\displaystyle e_2$ is accessible from $\displaystyle e_3$

    Since .$\displaystyle \begin{Bmatrix}P(e_3\to e_1) &=& 0.7 \\
    P(e_1\to e_2) &=& 0.8 \end{Bmatrix}$

    . . then: .$\displaystyle P\left([e_3\to e_1] \wedge [e_1\to e_2]\right) \;=\;(0.7)(0.8) $

    Hence: .$\displaystyle P(e_3\to e_2) \:=\:0.56$ . . . The statement is true.

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  3. #3
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    Awesome reply

    Thanks for your help.
    David.
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