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Random Walk question

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    I need to determine the standard deviation from the mean position value <x> in a random walk scenario. We are given that the probability of taking a step to the right is p and the probability of taking a step to left is q. Please note that p and q need not be equal. The starting point is x = 0 and we are only moving in one dimension.

    2. Relevant equations

    3. The attempt at a solution
    I would just like to see if this solution seems valid. In a random walk, after taking N steps one's position is x = nR - nL where nR is the number of steps taken to the right and nL is the number of steps taken to the left. Both nR and nL of course follow binomial probability distributions. Using the fact that the variance = <x2> - <x>2, I went ahead and calculated these two expectation values for x2 and x.

    It would seem to me that the expectation value of x is
    <x> = <nR> - <nL>, thus using the fact that nR and nL follow binomial probability distributions, I get <x> = Np - Nq = N(2p-1). This formula seems reasonable to me as if p = 0.5, then <x> = 0 as one would expect.

    Now for <x2>, I simplifed nR - nL to 2nR - N. Therefore x2 = 4nR2 - 4NnR + N2. Thus,
    <x2> = <4NR2> - <4NnR> + <N2>

    Using the formulas for the binomial distribution, I get
    <x2> = 4(Np)2 - 4Np2 + 4Np - 4pN2 +N2.

    Putting all this together and taking the square root to get the standard deviation, I get
    [tex]\sigma[/tex] = 2[tex]\sqrt{Npq}[/tex].

    This all seems reasonable to me but I am not really an expert on probability and statistics so I was hoping to get some thoughts on this solution. Any help is very much appreciated.
  2. jcsd
  3. Sep 24, 2009 #2


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    Homework Helper

    I'm not an expert either, but I'm wondering about your separation of <x>. I believe that nR and nL are strongly correlated (r=-1), as demonstated by your treatment of <x2>. Doesn't this prevent you from separating <x> as you have done? I think that you should use x=2nR-N for <x> as well.
  4. Sep 24, 2009 #3
    I understand your concern but I checked and taking the expected value of <2nR -N> gives you the exact same answer. I actually didn't separate them when I did my homework and in the process of computing <x> I could see that <x> = <nR> - <nL> is the same as splitting them apart and I just used this simplifying fact when I made this post. Thank you very much for your comment though. I appreciate your taking the time to post your thoughts.
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