Intuition question. You draw two vectors from an n-dimensional isotropic Gaussian where n>>1. What is the typical angle between them?
Prikaži ovu nit
1/ Do any of you posting your musings here know how to establish that the answer is pi/2? It is not done by simulations or by looking it up in mathoverflow. It is only established through a mathematical proof. Figure it out. It involves solid angles.
-
-
2/ If you do not know it, learn it. Be it because yo want to understand DL models with many weights or brain circuits with many neurons, you will need an intuition for the properties of high dimensional spaces.
-
3/ And you do not need to be a mathematician. I am not - I am just a physicist. But enough of a scientist to know that whether a `fact' is true is not decide by voting!!!
- Još 3 druga odgovora
Novi razgovor -
-
-
I assume most of my readers would go through the distribution of dot product path. Or do you mean how to derive the ab=||a||||b||cos \theta equation in the first place?
Hvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi
-
-
-
The mean dot product is zero and that fact is trivial. The next question would be whether you expect concentration of measure.
-
norms grow linearly with n. variance of sum of independent variables also grows linearly with n. And there is your concentration of measure.
- Još 1 odgovor
Novi razgovor -
-
-
Basic concentration inequalities give not just the proof but bounds on how close you get to pi/2 for finite sized spaces :)
Hvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi
-
-
-
I would tip around the idea of inner product and its relation to the angle of two random vectors, and what happens when the dimension increases. Fix the first one to [1,0,0,0...] if you like... :)
Hvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi
-
Čini se da učitavanje traje već neko vrijeme.
Twitter je možda preopterećen ili ima kratkotrajnih poteškoća u radu. Pokušajte ponovno ili potražite dodatne informacije u odjeljku Status Twittera.