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From: rabbits77 on 28 Mar 2010 21:05
The cauchy ditribution has pdf

f(x)=1/(pi(1+x^2))

In order to generate random observations
for this distribution we need to find
F^-1(u) which in this case is:

F(x)=arctan(x)/pi

F^-1(u)=tan(pi*u) -.5<u<.5

So, if u has uniform (-.5,.5) distribution
than X=tan(pi*U) has a cauchy distribution.
Is this correct?
From: Robert Israel on 28 Mar 2010 22:50
rabbits77 <rabbits77(a)my-deja.com> writes:

> The cauchy ditribution has pdf
>
> f(x)=1/(pi(1+x^2))
>
> In order to generate random observations
> for this distribution we need to find
> F^-1(u) which in this case is:
>
> F(x)=arctan(x)/pi

Not quite. That would go to -1/2 as
x -> -infinity and +1/2 as x -> +infinity.

> F^-1(u)=tan(pi*u) -.5<u<.5
>
> So, if u has uniform (-.5,.5) distribution
> than X=tan(pi*U) has a cauchy distribution.
> Is this correct?

This result is correct, though the intermediate
steps are wrong.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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