3 ms·

Then try normalizing a random 1000-element vector. The average of the vector elements is around 0.027.

• akomtu 5y ago
  The fact that average of D is sqrt(2) is the easy part. But average doesnt mean that random D values would concentrate in that spot. D may be evenly distributed. So the question is what's the variance of D? Using your formulas above: D^2 = (1 - x1)^2 + (1 - x1^2) = 2 - 2x1 We know that x1 is 0 on average, but its distribution is restricted by the fact that we choose a random point on the n-sphere. So what's the probability of that random point falling into a thin stripe where abs(x1) < eps? And how does this probability behave for large n? This eps-stripe is basically the (n-1)-sphere of width eps, so the stripe's area A(n)=eps•S(n-1), and so our probability p=eps•S(n-1)/S(n). The "magical" property of n-spheres is that their area grows at first, reaches maximum in 7 dimensions and then falls off rapidly to zero. Using formulas from wikipedia, I get: p = eps•(n/4)^(n/2) for large n. In other words, the distribution of x1 approaches the look of the delta function at the n^n pace and for all practical matters, D=sqrt(2) with high precision for n > 10. Where