Earlier quoted context omitted.
If you consider a point on the sphere it means choosing a bunch of xi such that: x1^2 + x2^2 + … + xn^2 = 1. Suppose wlog you pick (1,0,0,…,0). Then the distance from your point to a random point is: D = (x1-1)^2 + x2^2 + … + xn^2 And from the first equation we know: x1^2 = 1 - x2^2 - x3^2 - … - xn^2 Intuitionistically, your point will be far from a random point if x1 is close to zero, and x1 will be close to zero be…
Let me hijack your explanation starting from this point: D = (x₁-1)² + (x₂² + … + xₙ²) Since all the xₙ² sum to 1, as the dimensionality grows (∑xₙ²→1 as n→∞) each individual xₙ will converge towards 0. Since x₁ is almost 0, therefore the (x₁-1)² term will be almost 1. Since we know that ∑xₙ²=1, and that x₁² is almost 0, then we also know that ∑xₙ² - x₁² is almost 1, which is the 2nd half of the above expression for…
I'm not sure it will. x1 is chosen randomly in the -1..1 interval. I dont see how the million other dimensions would force it to stick to 0. Those N other dimensions shrink the stddev(xi) by sqrt(N), though.