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From |
"Nick Cox" <[email protected]> |

To |
<[email protected]> |

Subject |
RE: st: exponential distribution |

Date |
Wed, 20 Feb 2008 13:14:08 -0000 |

Austin is naturally correct. But as (1 - uniform(x)) has the same distribution as uniform(x) his expression can be simplified without loss to g e1 = -ln(uniform()) Nick Austin Nichols Jon-- The general approach is to take the cumulative distribution function F(x) and solve for x in terms of F. Then replace F with uniform() and you are done: F=1-exp(-lambda*x) 1-F=exp(-lambda*x) -lambda*x=ln(1-F) x=-(1/lambda)*ln(1-F) so for lambda=1, you can g e1=-ln(1-uniform()) On Feb 19, 2008 9:02 PM, Jon Schwabish <[email protected]> wrote: > Does anyone know how to create an exponential > distribution with a mean of 1 to use as random > numbers? I believe it's a tweak on exp=-ln(uniform()), > but am not sure. * * For searches and help try: * http://www.stata.com/support/faqs/res/findit.html * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

**References**:**st: exponential distribution***From:*Jon Schwabish <[email protected]>

**Re: st: exponential distribution***From:*"Austin Nichols" <[email protected]>

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