🚀 go-pugleaf

gfraenke@magnus.acs.ohio-state.edu (Gideon A Fraenkel) writes:

>In article <3638.9209301031@rfsparc.essex.ac.uk> jenum@essex.ac.uk (Mohamed Jenu M Z) writes:
>>
>>Hi netters, 
>>
>>Is it ok to generate a 2D Gaussian distribution (mean 0 and variance 1)
>>pseudorandom numbers from a 1D pseudorandom numbers (Gaussian, mean 0,
>>variance 1) by arranging them into rows and columns? Any advice is
>>greatly appreciated. 
>>
>>Jenu
>>
>>jenum@ossparc.essex.ac.uk


>What about the 2D distribution is supposed to be Gaussian?  If it's
>the norm -- the distance from the origin -- , then the answer to your
>question is no.  In general, the distribution of a function f(x,y) is
>calculated as
>         /
>  P(z) = | delta(z-f(x,y)) P(x)P(y) dxdy   .
>         /
>You can see that if x and y are Gaussian, then P(z) will be
>proportional to z times a Gaussian, which among other things goes to
>zero at the origin.
A two-dimensional Gaussian has probability just like usual Gaussian:
1/(variance * sqrt(2pi)) exp(-(x-mean)^2/(2 variance^2)),
only that mean and x are vectors, and their sqauring is a simple dot
product. So you can use 1D Gaussians perfectly, since something
like exp(-x^2-y^2) can be split into exp(-x^2)exp(-y^2) without problem.
Apart from the post before being wrong (considering Gaussians definition),
I do not understand it. Too bad.