Re: distance between points from 2d normal distribution

Hi,
Actually the way I did it was:

D^2 = X^2 + Y^2

For my case, the pdf's for X^2 and Y^2 are identical, both:

p(x)= 1/(2*sigma*sqrt(pi)) * exp(-x^2/(4*sigma^2))

In other words std = sqrt(2)*sigma (variance is 2*sigma^2) and then I
used the formula for chi-square distribution of order 2 to get

f(x) = 1/(4*sigma^2) * exp(-x / (4*sigma^2))

But this is the distribution for D^2 isn't it? Do I need to take
positive square root? The thing is, when I plot f(x) I get a
negative exponential curve (obviously) but a simulation for distances
is clearly not that shape. You can compare the simulations in
matlab:

For D^2:

hist((randn(1,5000000) - randn(1,5000000)).^2 + (randn(1,5000000) -
randn(1,5000000)).^2, 250);

For D:

hist(sqrt((randn(1,5000000) - randn(1,5000000)).^2 +
(randn(1,5000000) - randn(1,5000000)).^2) , 250)

My desire is to analytically get the distribution for the latter, but
I don't know how to manipulate pdfs! Sorry for more troubling, but
any more direction would be of great help!

Many thanks,
Jeff
.

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