# Random walk

1. Apr 5, 2009

### superwolf

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. Apr 6, 2009

### lanedance

Hi Superwolf

$$3\sigma$$ (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

3. Apr 6, 2009

### superwolf

So L sqrt(N) is the standard deviation?

4. Apr 6, 2009

### lanedance

i would strat with the definition of variance and work from there
$$\sigma^2 = <(x - \bar{x})^2>$$
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