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RANDOM STEPS IN A DIMENSION


We are interested in the average length of the vector X sum of N unit vectors randomly chosen in D dimensions, ie, unformly distributed on the unit N sphere and independent.


In one dimension, D=1, the identities of Quiz 23, can be used to compute the average length of the vector, and its average square length,

   E[X] = const sqrt(N)

E[X^2] = N


Similar relations hold in any dimension:

lim E[Xk] = a(k,D) Nk/2.
Exercise: even powers


The relation E[X^2] = N is easy to prove: the N unit vectors are "decoupled" in the sum X2.


Prove (or disprove) the relations for higher even powers:
E[X4] = N2 ( 1 + 2/D )
E[X6] = N3 ( 1 + 6/D + 8/D2 )


Exercise
a(k+2,D) = (D+k)/D a(k,D).


Given that a(2,D)=1 for every D, this recursion solves the problem once the values for k=1 and k=2 are known.


What are the values a(1,D) ?





Hint. The distribution of |X| is the Rayleigh distribution. From this the recursion relation is easily proved, and the coefficient a(k,D) can be computed. For example, for even k we have
   D =   2      3      4      5      6
k=-2          3/1    4/2    5/3    6/4
   0     1      1      1      1      1
   2     1      1      1      1      1
   4   4/2    5/3    6/4    7/5    8/6
   6  24/4   35/9   48/16  63/25  80/36


The values for k=1 are (integral from 0 to infinity)

a(1,D) = (D/2)(-1/2) 2/Gamma(D/2) Int xD exp(-x2) dx


The first few values are ( P = sqrt(pi) )

   D =  2           3              4               5               6
k= 1   P/2  sqrt(2/3)*2/P  sqrt(2/4)*3P/8  sqrt(2/5)*8/3P  sqrt(2/6)*15P/16


Marco Corvi - 2016