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Random walk

  1. Apr 5, 2009 #1
    My textbook simply states that

    "For a simple 1D walk with step size L:
    <x^2> = NL^2

    So after N steps 99.7% of the particles will be closer then 3L sqrt(N) from the centre"

    How does it get from the first to the latter?
     
  2. jcsd
  3. Apr 6, 2009 #2

    lanedance

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    Hi Superwolf

    [tex] 3\sigma [/tex] (3 standard deviations) is 99.7% confidence limit for a normal distribution

    So the 99.7 and <x^2> seem to point to using variance of a normal distribution

    Which shouldn't be too had to get too asuming we know the total length is normally distributed...

    Taken from wiki:
    http://en.wikipedia.org/wiki/Central_limit_theorem

    So each step can be modelled as a binomial distribution with outcomes (L,-L) and 0.5 chance of success, and the sum giving the average length is then approximated by a normal distributions at large N
     
  4. Apr 6, 2009 #3
    So L sqrt(N) is the standard deviation?
     
  5. Apr 6, 2009 #4

    lanedance

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    i would strat with the definition of variance and work from there
    [tex] \sigma^2 = <(x - \bar{x})^2> [/tex]
    where the <> is expectation

    it should be a simple matter to get the standard deviation from there, and will probably end up as L sqrt(N), if you convince youself the mean is zero
     
    Last edited: Apr 6, 2009
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