Proof of R = s√N
Teaching Guidance
for 14-16
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 (see below) 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 x1 and y1, stride number 2 into x2 and y2, and so on.
Then the resultant of that walk, R, has an
- x-component (x1+x2+...xN)
- y-component (y1+y2+...yN)
and
R 2 = (x1 + x2 + ... xN) 2 + (y1 + y2 + ... yN) 2
= x1 2 + x2 2 + etc. + 2x1x2 + 2x1x3 + etc.
+ y1 2 + y2 2 + etc. + 2y1y2 + 2y1y3 + etc.
= s1 2 + s2 2 + etc. + 0
= s 2N
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 2s 2. 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 scosθ and ssinθ. Then the cross terms in the expression for R 2 take the form 2s 2cos(θ1-θ2), etc, and you can argue that the cosines are as often positive as negative.
