# Proof of R = s√N

This proof of *R* = *s√N* describes an algebraic approach to the random walk problem, sometimes known as the drunkard’s walk.

*If the drunk takes a large number of strides (N) each of the same length (s) in succession, but in random directions, what is the resultant distance of travel (R)?*

Obviously this will vary from one batch of *N* strides to another and may often be zero or as large as *s* x *N*. We want the average distance from start to finish, averaged over many batches of *N*strides. Observe a walk of *N* strides and find the resultant travel distance (*R*) from start to finish. Observe a large number of such walks starting afresh each time and find the average value of *R*for all those walks. Because it leads to a simple result, you can find the average value of *R*^{2} and take the square root, obtaining a root mean square (R.M.S.) average. You can show that this average should approach the value *s* √*N*. The two-dimensional proof is shown below. The three-dimensional one is similar.

The diagram shows a few strides of a random walk. Choose a set of perpendicular co-ordinates, such as x and y, then resolve stride number 1 into components *x _{1}* and

*y*, stride number 2 into

_{1}*x*and

_{2}*y*, and so on. Then the resultant of that walk,

_{2}*R*, has an

*x*-component (*x _{1} *+

*x*+....

_{2}*x*)

_{N}
and a *y*-component (*y _{1}* +

*y*+ ....

_{2}*y*)

_{N}and

R2 = (x1 + x2 + .... xN)2 + (y1 + y2 + .... yN2)

= x1 2 + x2 2 + etc +2x1x2 + 2x1x3 + etc

+ y1 2 + y2 2 + etc + 2y1y2 + 2y1y3 + etc

= s12 + s2 2 + etc + ZERO

= s2N

The cross terms such as 2x1x2, add up to zero in averaging over many walks, because those terms are as often negative as positive, and they range similarly from 0 to 2s2. Similarly for the y cross terms.

Then the average value of R = s√N.

The proof is better if you use trigonometry and resolve each stride, s, into horizontal and vertical components scos θ and s sin θ. Then the cross terms in the expression for R2 take the form 2s2 cos (θ 1 - θ 2), etc, and you can argue that the cosines are as often positive as negative.