# When uniformly sampling points from a high dimensional ball, intuitively, why are most of the points at the crust?

This is related to the curse of dimensionality. Many proofs can be found showing that the volume is mostly at the crust, but I don't quite understand the intuition.

Consider a $$d$$ dimensional unit ball - centered at the $$d$$ dimensional origin and having radius $$1$$. We can draw the points uniformly by drawing from $$x_i \sim N(0, 1)$$ for all $$d$$ components in the vector, and then normalize said vector. (Ref: https://stats.stackexchange.com/questions/7977/how-to-generate-uniformly-distributed-points-on-the-surface-of-the-3-d-unit-sphe)

This is telling me that, on average, each component of the vector will be close to $$0$$ - so wouldn't the points mostly be concentrated at the center of the sphere, not the crust?

• A $d$-dimensional ball of radius $r$ has volume proportional to $r^d$. If $d$ is large, and $r$ is even a little less than one, then $r^d$ is tiny. Commented Sep 8, 2022 at 6:47

Imagine the $$d$$-ball to be made up of infinitely many concentric $$(d-1)$$-spheres. The surface of each of these is propositional to $$r^{d-1}$$. So if you take an infinitesimal or actual range of radii, the volume in the shell corresponding to that range will be proportional to that radius power, too.
For the $$1$$-ball (line segment) you get uniform weight: equal density at all radii. For the $$2$$-ball you already have density increasing linearly with radius. The higher the dimension, the higher the exponent and the stronger the concentration of density at the high end of the radius.
Consider a $$d$$ dimensional unit ball - centered at the $$d$$ dimensional origin and having radius 1. We can draw the points uniformly by drawing from $$x_i\sim N(0,1)$$ for all $$d$$ components in the vector, and then normalize said vector.