Timeline for How to pick a random direction in n-dimensional space
Current License: CC BY-SA 2.5
8 events
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| May 28, 2010 at 17:21 | comment | added | Kevin Walker | Yes, as I noted in the 2nd paragraph this is inefficient for large $n$, but in applications $n$ is not always large. I've employed this method when $n=3$. In this case I suspect it's faster than using Gaussian distributions. | |
| May 28, 2010 at 13:42 | comment | added | Nate Eldredge | Indeed, you should expect $(n/2)!(\pi/2)^{-n/2}$ trials before success. For $n=10$ it's about 1000, and for $n=100$... let's just say, too many. | |
| May 28, 2010 at 10:56 | comment | added | Peter Shor | It sounds right. It also sounds truly inefficient ... in high dimensions you will have to sample a ridiculously large number of points to get one that works. | |
| May 27, 2010 at 21:32 | comment | added | Kevin Walker | @Ben: Oops, you're right. That's what I meant but apparently not what I typed. Fixed now. @Nate: Does it sound right after the correction? I'm just choosing a random point in the unit ball and rescaling. | |
| May 27, 2010 at 21:27 | history | edited | Kevin Walker | CC BY-SA 2.5 |
added 6 characters in body
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| May 27, 2010 at 20:08 | comment | added | Ben Wieland | Those $\sum x_i$ are supposed to be $\sum x_i^2$. | |
| May 27, 2010 at 16:53 | comment | added | Nate Eldredge | That doesn't sound right. Can you give or refer to a proof? Anyway, I think the usual approach of "generate $n$ iid standard Gaussian and divide the resulting vector by its norm" is simpler to program and faster to run. | |
| May 27, 2010 at 14:09 | history | answered | Kevin Walker | CC BY-SA 2.5 |