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Question about probability with absorption

  1. Apr 23, 2012 #1
    say we have integers 0-10. We start with N and the probability that N grows by 1 is .69. The probability that N decreases by 1 is .31.
    Thus obtaining N+1 after one time step is .69 and similar obtaining N-1 is .31.
    Once N reaches 0 or 10 it is absorbed and cant move from there.
    My question is what is the probability that N is eventually at 10 if we start at N=2? also what is the probability of absorption if we start at N=2?
    Last edited: Apr 23, 2012
  2. jcsd
  3. Apr 23, 2012 #2
    Unless I'm missing something, the probability of absorption is 1 no matter where you start. The probability that N=10 is 0.69 since there is a 0.31 probability that N=1 on the first step.
  4. Apr 23, 2012 #3
    im sorry i meant to ask
    Starting from N=2, what is the expected number of steps before absorption?
  5. Apr 23, 2012 #4
    i changed the problem a bit to make more sense
  6. Apr 23, 2012 #5
    Google for "gambler's ruin problem"
  7. Apr 23, 2012 #6


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    Hey chuy52506.

    You will need to setup a markov chain and calculate the steady-state distribution for the long-term probabilities for the probability of getting a 1 vs getting a 10. If there is a non-zero chance of always getting to N=10, then the long-term probability will always be 1 since N=10 is an absorbing state, and judging from your post, I think this is the case.

    If you want to figure out for a finite number of transitions, then construct your transition matrix M, find M^n for n transitions and then multiply this matrix by your initial probability vector which will have a 1 entry in the second element (this is a column vector) and when you multiply M by this column vector you will get probabilities of being in each state from N=1 to N=10 as expressed by the column.

    If you want to prove long-term absorption (when n->infinity) then you will need to use the definition of absorption in markovian systems.
  8. Apr 24, 2012 #7
    By the definition of the Markov property, the probability of a future state is dependent only on the present state. The probability of moving toward the upper absorbing state is always 0.69 and the toward the lower absorbing state is is always 0.31. I assume the process continues to termination (absorption) with p=1. Therefore the probability the process terminates at N=11 is 0.69 and at N=1; 0.31. If the process terminates at N=11 and the process starts at N=2, then the marker must visit N=10 with probability 0.69 .

    (We are not talking about consecutive moves in one direction. If that were true, the probability of 9 consecutive moves from N=2 to N=11 would be [itex] 0.69^9 = 0.035[/itex].)
    Last edited: Apr 24, 2012
  9. Apr 24, 2012 #8


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    I actually missed the criteria that it gets stuck at 0 which screwed up my analysis.

    So yeah this is going to be what micromass said: i.e. a two-absorption state system.
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