That's the probability density, not the cumulative density. —Dan
On Mar 18, 2016, at 5:14 AM, Mike Speciner <ms@alum.mit.edu> wrote:
p[1](x) = 1 ?
On 17-Mar-16 16:21, Eugene Salamin via math-fun wrote:
While I haven't worked this out for Gaussian random variables, it is easy to get the distribution of the product of n independent random variables uniform on [0,1]. It is
p[n](x) = (-log x)^(n-1) / (n-1)!
The singularity at 0 should be the same for a Gaussian, and indeed for x near 0 (and > 0)
K[0](x) ~ -log x + constant.
-- Gene
From: Dan Asimov <asimov@msri.org> To: Eugene Salamin <gene_salamin@yahoo.com>; math-fun <math-fun@mailman.xmission.com> Sent: Saturday, March 5, 2016 12:53 PM Subject: Re: [math-fun] How to generate a random variable with a BesselK[0] distribution Interesting (though I've never needed a Bessel distribution).
I wonder what the distribution is of the product of n i.i.d. N(0,1) random variables is.
—Dan
On Mar 5, 2016, at 12:25 PM, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Suppose, for some reason, you need to generate random variables whose normalized probability distribution is
P(z) = (1/pi) BesselK[0] ( |z| ).
Here's a clever way to do it without having to muck with Bessel functions. Let x and y be independent Gaussian random variables with zero mean and unit variance. Then z = xy has the desired distribution.
Furthermore, the sum of n such products, has distribution
P[n](z) = (1 / (sqrt(pi) Gamma(n/2)) (|z| / 2)^((n-1)/2) BesselK[((n-1)/2] ( |z| ).
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