- dear list,

The following may be slightly off topic (in which case I apologize), but

maybe somebody can help me:

I want to generate a set of points in a plane, {(xi,yi)}, such that the

points are distributed according to a chosen probability distribution

density f, i.e., the infinitesimal rectangle [x,x+dx] X [y,y+dy] shall

contain f(x,y)*dx*dy points. How do I do this ??

In other words, I am looking for a generalization of the 1-dim method,

where this is done with F*(z), where F=cumul(f) and F* = inverse of F, and

z = random out of [0,1] (or any appropriate interval if f is not

normalized).

I tried to do it the same way, just using complex numbers, but this does

not seem to work, and could also not be generalized for n>2 dimensions.

thanx for any hint,

p

(PS. sorry if it is trivial.)

--------------------------------------------------------------------------------------

Peter Bossew

Institute of Physics and Biophysics, University of Salzburg, Austria

home: Georg Sigl-Gasse 13/11, A-1090 Vienna, Austria

ph. +43-1-3177627, e-mail: p.bossew@... , peter.bossew@...

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* Support to the list is provided at http://www.ai-geostats.org - On 10-Mar-03 Peter Bossew wrote:
> I want to generate a set of points in a plane, {(xi,yi)}, such

Hi Peter,

> that the points are distributed according to a chosen probability

> distribution density f, i.e., the infinitesimal rectangle

> [x,x+dx] X [y,y+dy] shall contain f(x,y)*dx*dy points.

> How do I do this ??

>

> In other words, I am looking for a generalization of the 1-dim method,

> where this is done with F*(z), where F=cumul(f) and F* = inverse of F,

> and z = random out of [0,1] (or any appropriate interval if f is not

> normalized).

>

> I tried to do it the same way, just using complex numbers, but this

> does not seem to work, and could also not be generalized for n>2

> dimensions.

The problem with "doing it the same way" is that "the same way" only

applies in 1 dimension (as a consequence of the result that F(X) has

a uniform distribution, so that F*(X) has the distribution of X).

In more than 1 dimension, you still only have the result that F(X,Y)

has a uniform distribution, and this is only 1 equation. There is

no _straightforward_ way of basing it on two equations

(F1(X,Y), F2(X,Y)) = (Z1,Z1) where (Z1,Z2) is uniform on [0,1]x[0,1]

However, there are possible approaches. One is to use conditional

distributions: let F(X) be the marginal distribution for X, and let

G(Y;x) be the conditional distribution of Y given that X=x. Then

both U=F(X) and, for each x, V=G(Y;x) have uniform distributions,

so provided you can invert these as F*(U) and G*(V;x) you can then

sample U=u,V=v and obtain x=F*(u) and then y=G*(v;x).

The practical issues here depend a lot on the form of F(X,Y),

since both obtaining the conditional distribution and inverting

the functions F(X), G(Y;x) may be difficult in practice.

If it is straightforward to obtain the conditional distributions

for Y given X and X given Y, then Gibbs Sampling can enable you to

sample from (X,Y) without inverting the functions.

If you can't obtain the conditional distributions then it may be

impossible to follow such approaches, and then you may need to fall

back on simulating a random process with a known mechanism which has

the property of yielding (X,Y) distributed as f(x,y) -- if you know

of such a process!

If you could describe the density f(x,y) you are trying to sample

from to the list, then perhaps someone can suggest an approach.

> (PS. sorry if it is trivial.)

No, it isn't trivial!

Best wishes,

Ted.

--------------------------------------------------------------------

E-Mail: (Ted Harding) <Ted.Harding@...>

Fax-to-email: +44 (0)870 167 1972

Date: 10-Mar-03 Time: 08:29:19

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