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:
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) ?
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)
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