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Random walk confusion.

  1. Aug 25, 2012 #1
    I read about the random walk the other day. The simplest 2D form, where you start at zero and move up or down one unit at random, both are as likely.

    To get an the average distance from zero after N steps, the following argument was used: The distance after one step is 1. If after some steps, the distance is D, then with the next step the distance is either D-1 or D+1. The squares of the distances after the next step are either D^2-2D+1 or D^2+2D+1. Since both are equally likely, the change in distance is just the average of them. Adding them up and dividing by two, to take the arithmetic mean. the new distance squared is D^2+1. Since the square of the distance is 1 after the first step and increases by one every step, therefore the distance after N steps is the square root of N, on average when both jumps happen as often.

    Considering the case after one step, since both jumps are as likely, to zero and two, the same argument says that the average distance is the square root of two. But why is that, since jumping to 0 or 2 is just as likely, they happen as often and therefore the average distance should be one.

    Instead shouldn't it be said that since the distance is either 0 or 2, we should take the average of them and not of the squares to get the average distance?

    What is the reasoning for using the squares? If I square 9 and 11 and take the square root of their average, I don't get 10.
     
  2. jcsd
  3. Aug 25, 2012 #2

    uart

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    High chingle. You are confusing the "square root of the average squared distance" (aka rms distance) with the average distance.

    - The rms distance is sqrt(N)

    - That average distance is always zero.

    BTW. The average "distance" was one in the final example you gave above, but this was merely because you started from one. The average displacement was still zero, as is always the case.
     
  4. Aug 25, 2012 #3

    Mute

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    Do you mean a 1d random walk? I don't see what is two-dimensional in the kind of walk you describe, just going up or down.
     
  5. Aug 25, 2012 #4

    mathman

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    The argument is for the root mean square distance after N steps, not the average.
     
  6. Aug 26, 2012 #5
    Yes, you are all right, I was confused, it is the average of the squared distances, not the distances themselves. Yes I should have written 1d random walk.

    What confused me is for example the last paragraph there:

    http://www.mit.edu/~kardar/teaching/projects/chemotaxis(AndreaSchmidt)/random.htm

    If I take the square root of the average squared distances, it's not the average distance I should expect to be at? After 25 steps, I wouldn't expect to be 5 steps away from the beginning on the average, or would I?

    For example the case with two steps, out of four tries, I should be at 2 once, at -2 once and at 0 twice. Distance is 2 half of the times (counting -2 also as a distance 2 away from 0), so average distance is one, meaning that the rms distance is not necessarily related to the average distance?
     
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