5 ms·
I'm not certain I know what you mean, but it doesn't seem to work. If you mean to take epsilon neighborhoods in Euclidean norm of the axes, then I'm skeptical t
by tgb 5y ago
I'm not certain I know what you mean, but it doesn't seem to work. If you mean to take epsilon neighborhoods in Euclidean norm of the axes, then I'm skeptical that that contains most of the mass of the sphere, since simulating random points on a thousand dimensional sphere doesn't seem to give points near these axes.
If instead you mean most of the mass of the ball is in the points which have all but one coordinate within epsilon of zero, then the intuition doesn't follow. It's equally true that you get most of the mass considering just points with all coordinates within epsilon. And for that you get that the mass of the ball is concentrated in an epsilon cube at the origin, which excludes exactly the spikes that you were basing the intuition on. The weird thing in this thought experiment is the epsilon cube not the sphere. For example the epsilon cube contains points much further than epsilon from the origin and so it's maybe not surprising that it contains most of the sphere.
- momenti 5y agohttps://news.ycombinator.com/item?id=3995615 https://news.ycombinator.com/item?id=3995615
- tgb 5y agoPerhaps you can be more specific since I don't see the claim here being discussed there.
- trhway 5y ago>since simulating random points on a thousand dimensional sphere doesn't seem to give points near these axes. that is kind of circular argument as "random" really depends on the density measure underlying the chosen sampling distribution.
- tgb 5y agoIt's no more ill defined than volume itself (and less so since it doesn't need an arbitrary scale).
- eru 5y agoI don't understand your 'less so' comment. Could you please elaborate?
- tgb 5y agoIf you want to define volume in D-dimensional space then you need to choose what the unit volume is, for example that a unit hyper cube has unit volume one. (That choice is relevant for these discussions and you could equally have chosen a unit ball to have volume one since you're defining D-dimensional volume you get to choose.) But a probability distribution is always normalized to integrate to one, which means either choice of volume definition will give you the same uniform probability distribution on the ball. So there's an extra arbitrary constant to choose for volume that isn't important in the probability distribution even though you define the probability density in terms of the volume.
- eru 5y agoOK. Though you could leave your volume definition parameterized, instead of picking a specific scale, couldn't you? I do agree that volume and probability distribution are essentially the same. (That's also why at my university measure theory and probability theory were handled in the same course.)
- kgwgk 5y agoIf only there was some kind of natural density measure to use in an Euclidean space R^D…