Is this geometrical 'derivation' of Brownian motion legitimate? Here's a simple 'derivation' of the Brownian motion law that after N steps of unit distance 1, the total distance from the origin will be sqrt(N) on average. It's certainly not rigorous, but I'm wondering if people think it's reasonable, or possibly even a commonly known.


*

*An object takes one step from the origin, so is at a distance 1: d = 1 for N=1.

*On average, the next step will be neither exactly toward or exactly away from the origin, so you compromise and say it steps along a direction that's perpendicular to the vector connecting the origin to its present location - sort of half way between walking backwards and walking forwards. By Pythagoras's theorem, the average distance will then be d = sqrt(1^2+1^2) = sqrt(2) for N=2.

*Likewise for N=3, stepping in a normal direction gives d = sqrt( sqrt(2)^2 + 1^2 ) = sqrt(3), so in general d = sqrt(N).
This seems to work in dimensions 2 or higher.
 A: As a heuristic description, this is exactly right and correctly captures the essence of the subject. To turn it from a heuristic "derivation" into an actual derivation, you just need to make precise the notion that the two vectors are perpendicular on average. The precise meaning of "perpendicular on average" that's useful in this context is that the dot product is zero on average. That is, if $\vec r_n$ is the position vector after $n$ steps, and $\vec s_n$ is the vector representing the $n$th step, then
$$
\langle \vec r_n\cdot\vec s_{n+1}\rangle=0.
$$
The angle brackets mean an ensemble average -- that is, an average over many trials.
This statement is true -- the easiest way to prove it is that the probability distribution for $\vec s_{n+1}$ is symmetric about 0, so positive and negative contributions to the dot product occur equally. And it's sufficient to prove the standard formula. Since $\vec r_{n+1}=\vec r_n+\vec s_{n+1}$, 
$$
r_{n+1}^2=r_n^2+s_{n+1}^2+2\vec r_n\cdot\vec s_{n+1}.
$$
In the ensemble average, the last term vanishes, so $r^2$ increases, on average, by $s^2$ (i.e., by 1 for unit steps) on each step.
