Probability first step conditioning.

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  1. JBKProductions's Avatar
    • Overlord in Training
    • Posts: 2,105
    Probability first step conditioning.
    Let  \alpha_k = P(Absorption\ into\ 1|X_0=k), P_{11} = Probability of going from 1 to 1. Let's assume there are 4 states {1, 2, 3, 4}. I just want to know by first step analysis why is(for example)  \alpha_3 = P_{33}\alpha_3 + P_{34}\alpha_4+P_{32}\alpha_2+P_  {31}\alpha_1 ? I don't understand this at all.
  2. nuodai's Avatar
    • PS Helper
    • TSR Legend
    Re: Probability first step conditioning.
    I take it this is a Markov chain of some variety? (i.e. future depends only on present state, not past states.)

    Suppose you start at position 3. Then:

    (a) You can go to position 1 with probability P_{31}, in which case you're absorbed to position 1 with probability \alpha_1
    (b) You can go to position 2 with probability P_{32}, in which case you're absorbed to position 1 with probability \alpha_2
    (c) You can stay at position 3 with probability P_{33}, in which case you're absorbed to position 1 with probability \alpha_3
    (d) You can go to position 4 with probability P_{34}, in which case you're absorbed to position 1 with probability \alpha_4

    Combining this information we get \alpha_3 = P_{31}\alpha_1 + P_{32}\alpha_2 + P_{33}\alpha_3 + P_{34}\alpha_4

    Since it's a Markov chain, the reason why all the probabilities "reset" is because when you know your second position it doesn't matter where you started off.
  3. JBKProductions's Avatar
    • Overlord in Training
    • Posts: 2,105
    Re: Probability first step conditioning.
    Thanks!
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