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Homework Help: Statistics - Discrete Markov Chains

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    I'm not really sure if this belongs here or in the precalculus mathematics section. I had to take calculus before taking this class so I'm putting it here.

    I'm confused about part (b). I don't really understand how I'm supposed to find [itex]P(X_{n+2}=0|X_{n}=2)[/itex] because I don't know what state n+1 is. Thanks for any help.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 30, 2013 #2
  4. Apr 30, 2013 #3
    [itex]P(X_{n+2}=0) = P(X_{n+2}=0|X_{n}=2)P(X_{n}=2) + P(X_{n+2}=0|X_{n}=1)P(X_{n}=1) + P(X_{n+2}=0|X_{n}=0)P(X_{n}=0)[/itex]
    [itex]P(X_{n+2}=0|X_{n}=2)P(X_{n}=2) = P(X_{n+2}=0) - P(X_{n+2}=0|X_{n}=1)P(X_{n}=1) - P(X_{n+2}=0|X_{n}=0)P(X_{n}=0)[/itex]

    I'm not sure how this helps.
  5. Apr 30, 2013 #4
    You're using the wrong events.

    Do this

    P(X_{n+2} = 0~\vert~X_n=2)
    & = & P(X_{n+2} = 0~\vert~X_n=2,~X_{n+1}=0)P(X_{n+1}=0~\vert~X_n=2)\\
    & & + P(X_{n+2} = 0~\vert~X_n=2,~X_{n+1}=1)P(X_{n+1}=1~\vert~X_n=2)\\
    & & + P(X_{n+2} = 0~\vert~X_n=2,~X_{n+1}=2)P(X_{n+1}=2~\vert~X_n=2)
  6. Apr 30, 2013 #5

    Ray Vickson

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    If you had been told that ##X_0 = 2## would you have been able to work out the probability that ##X_2 = 0?## Have you really never seen how to get multi-step transition probabilities?

    Note: I am waiting for answers to these questions before offering more help.
    Last edited by a moderator: May 6, 2017
  7. May 2, 2013 #6
    Well for b) I got .21 and believed that I solved the problem correctly. I don't know exactly what c is even asking me. Find P(X_1 = 0). What exactly are the alphas? Like what do they represent? Alpha 1 = probability x equals zero is .25.

    I have indeed never seen how to get multi-step transition probabilities =( but i believe i figured it out correctly and got the answer.

    Thanks for your help guys.
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