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how to produce a 2-D point distribution with normal density distribution

Asked by adameye adameye on 30 Apr 2013

Hi: I am trying to simulate a galaxy distribution, in which all units are identical, but the local densities are different. Simply speaking, I need to produce a 2-D point distribution, say, 1000 identical points in a 10*10 square, then if I measure the local densities (for example, within a circle with radius=1), I could get a normal distribution of densities.

   Maybe I did not explain it clearly, but just do not know how to do it, I can easily generate a data set with normal distribution value, but how to apply normal distribution on the position of 2-D points (move some together and take some apart) to get normal density distribution? thanks! 

2 Comments

Iman Ansari on 30 Apr 2013

See this: (mean zero and variance 1)

A=randn(1000,2);
x=A(:,1);
y=A(:,2);
plot(x,y,'Marker','.','LineStyle','none')
var(x)+1i*var(y)
mean(x)+1i*mean(y)
adameye adameye on 1 May 2013

thanks, this is very interesting, but this data looks like one "big ball", right? If I want to have a full pic of data, with points cluster somewhere, voids somewhere, how can I do it? Maybe I can combine some of those results to get a pic with a bunch of "big balls"? thanks again!

adameye adameye

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3 Answers

Answer by Iman Ansari on 1 May 2013

With adding a number in x and y directions you can change their position, and the number multiplied changes their radius:

A=randn(3000,2);
x=A(:,1)';
y=A(:,2)';
x=[x(1:800)+3 1/2.*x(801:1800)-3 1.5.*x(1801:3000)+2];
y=[y(1:800)+4 1/2.*y(801:1800)   1.5.*y(1801:3000)-3];
plot(x,y,'Marker','.','LineStyle','none')
axis('equal')
var(x)+1i*var(y)
mean(x)+1i*mean(y)

0 Comments

Iman Ansari
Answer by adameye adameye on 1 May 2013

Thanks a lot, This is a different way to solve the problem.

btw, do you know how to produce the needed distribution from power spectrum? because in cosmos people normaly make gaussian simulation a kind of stationary distributing (Cov(x,y)==Cov(x+u,y+u), only related with the distance between x and y, not relation with the shift u), in other words, the distribution will look like no obvious center....

thanks again

1 Comment

Iman Ansari on 1 May 2013

What do you mean from power spectrum?

adameye adameye
Answer by adameye adameye on 14 May 2013

Hi: Sorry I have to read some articles to investigate this problem and it took long time, I got http://www.astro.rug.nl/~weygaert/tim1publication/lss2007/computerIII.pdf I am still trying to understand the right procedure.... thank you very much! "

4 Generating Gaussian eld in Fourier Space Generating a Random Gaussian eld is easiliest done in Fourier space. Then the complex Fourier amplitudes are Y~ = Y~ exp(i*phi). Where  is a random phase and the modules are Rayleigh distributed

f(x) =2x/(sigma^2)*exp(-x^2/sigma^2)

The dispersion is of course related to the Power Spectrum as

sigma^2=(delat_k^3)*P(k) "

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adameye adameye

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