Morning thought: 2x2 diagrams work so well because a random pair of vectors in a high dimensional space are orthogonal with probability ~1 @vgr
-
-
This Tweet is unavailable.
-
If they’re sparse product elements would all be zero with high probability. It might even be true without a sparsity condition via central limit theorem. Ie likely as many product elements being positive as negative, and normally distributed in [-1, 1]
End of conversation
-
-
-
For vectors drawn randomly from {-1,1}^N, the expected angle between them is 1/N.https://mathoverflow.net/questions/248466/why-are-two-random-vectors-in-mathbb-rn-approximately-orthogonal-for-large …
-
Right but why are you drawing things from the vertices of a high d unit cube, that is one spiky and not symmetric object I refuse to believe in coordinate axes in concept space, thought is covariant
End of conversation
New conversation -
-
-
Well-known result. Two ways to get there: 1) if the elements are iid and centered, apply LLN to have the inner product converge to zero 2) most of the area of a hypersphere is near the equatorhttps://www.johndcook.com/blog/2017/07/13/concentration_of_measure/ …
-
1) It doesn't converge because the product of the nth elements isn't within arbitrarily small epsilon of anything for any n. 2) Rotating a vector on an n-sphere into a new dimension and onto an n+1-sphere changes previous elements so the elements are not iid.
End of conversation
New conversation -
-
-
Update: don't need a sparsity condition (which makes sense since it would be weird to have this result be dependent on the coordinate system)
Thanks. Twitter will use this to make your timeline better. UndoUndo
-
-
-
It's roughly Gaussian; mean 0, stddev=dim^{-1/2}. Some intuition: 1. Mean 0: symmetry. 2. Mean square = 1/d: if v1,...,vd form orthogonal basis then sum_i (vi,w)^2 =1 for any unit vector w. Now use symmetry (added d terms). 3. Gaussian: dot product is a sum of many small things
-
Rigorously: WLOG let the first vector be v=(1,0,0,...); only 2nd must be random. Generate w's coords with d i.i.d. Gaussians, mean 0, stddev d^{-1/2}. w has uniformly random direction, norm about 1 (law of large numbers). Then (v,w) is approx. what you want and exactly Gaussian
End of conversation
New conversation -
-
-
i'm going to plug this random 2x2 generator i made again https://rothos.github.io/2x2/
- End of conversation
New conversation -
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.
