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Thread: Random walk with 1/(1+x) probability

  1. #1
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    Random walk with 1/(1+x) probability

    You are walking along the x axis and you start at x = 0. At each time step you move to x + 1 with probability p = \frac{1}{1+x} and to x-1 with probability p = \frac{x}{1+x}. What is the expected number of steps until you return to 0 for the first time?
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    Re: Random walk with 1/(1+x) probability

    Should I post a solution or are people trying this? o.o
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    Re: Random walk with 1/(1+x) probability

    I looked at it for 10 min and decided it was more than I wanted to spend time on.
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    Re: Random walk with 1/(1+x) probability

    number of steps 1 with probability 0

    number of steps 2 with probability 1/2!

    number of steps <2n-1> with probability 0

    number of steps <2n> with probability 1/(n+1)!

    Does this make sense?
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    Re: Random walk with 1/(1+x) probability

    If n is number of steps then
    distance = (1+x)[n/(1+x)] + (1-x)[nx/(1+x)] = 0
    gives n(1+x^2) = 0
    n = 0 since 1+x^2 is not zero
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