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Can someone very simply explain to me how to compute the expected distance from the origin for a random walk in $1D, 2D$, and $3D$? I've seen several sources online stating that the expected distance is just $\sqrt{N}$ where $N$ is the number of time steps, but others say that the expected distance is $\sqrt{\frac{2N}{\pi}}$. Which one is it and is it the same regardless of dimension?

Thanks

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Maybe this will help: random walk –  Joel Cornett Jan 28 '12 at 1:29
    
What about this? mathworld.wolfram.com/RandomWalk1-Dimensional.html –  Diego Jan 28 '12 at 1:32
    
almost duplicate: math.stackexchange.com/questions/118889/… –  leonbloy Apr 16 '12 at 17:38
    
a question asked on 1/28 is an almost duplicate of a question asked on 3/11? –  Diego Apr 16 '12 at 17:49

1 Answer 1

up vote 8 down vote accepted

The expected value of the square of the absolute distance from the origin is $N$ (you are adding together $N$ independent random variables with mean $0$ and absolute magnitude $1$), and this is true in any dimension.

So those sources which are telling you $$\sqrt N$$ are giving you this as in some sense the "root mean square" distance from the starting point. It is not the expected value of the distance.

For a one dimensional random walk the expected absolute distance from the origin after $N$ steps is not easy to state explicitly, but as $N$ increases it becomes close to $$\sqrt{\dfrac{2N}{\pi}}.$$ So the sources which give you that are in a sense talking about a limit.

This changes for higher dimensions: if there are $d$ dimensions then the expected absolute distance from the origin after $N$ steps becomes close to $$\sqrt{\dfrac{2N}{d}} \dfrac{\Gamma(\frac{d+1}{2})}{\Gamma(\frac{d}{2})}$$ where $\Gamma$ is the Gamma function. As the number of dimensions increases, this get close to but is still below $\sqrt N$.

This is related to the means of the chi- and chi-squared distributions.

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There is a related question, where the OP asks for some reference for the formula from your post. As he does not have a privilege to post comments, I am pinging you instead of him. –  Martin Sleziak Sep 1 '13 at 17:50

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