| Date: | Tue, 4 Jun 2002 16:58:30 -0700 |
| Reply-To: | Cassell.David@EPAMAIL.EPA.GOV |
| Sender: | "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU> |
| From: | "David L. Cassell" <Cassell.David@EPAMAIL.EPA.GOV> |
| Subject: | Re: lapalcian [sic] distribution Vs Gaussian pdf |
| Content-type: | text/plain; charset=us-ascii |
|---|
Hristo Stevic <hristostev@YAHOO.COM> wrote [in part]:
> about the percentage values which show what percentage of data will be
> in a certain range of the pdf standard deviation, for example in
> gaussian distribution
> 68% of the observations fall within 1 standard deviation of the mean.
>
> want to know if there are similar reported percentage when using
> laplacian distribution
You can see for yourself. SAS now has the CDF() function, which will
let you get any 'percentage' that you want for a whole list of
distributions,
including the Laplace (or double exponential) function. Use it like
this:
P = cdf('LAPLACE',x,theta,lambda);
where x is the point for which you want P(X <= x),
theta is the optional location parameter, and
lambda is the optional scale parameter.
The mean of the Laplace distribution is equal to theta, and the
variance is equal to 2*lambda**2 . [So 1 standard deviation is
lambda * sqrt(2).]
You will find if you do the math that the probability of being within
1 s.d. of the mean of a Laplace distribution is about 0.7568833 ,
regardless
of the values of theta and lambda. Here's a trivial SAS check on that:
data temp;
do lambda = 1 to 5;
x = lambda*sqrt(2);
P = cdf('LAPLACE',x,0,lambda) - cdf('LAPLACE',-x,0,lambda);
put lambda= P= ;
end;
run;
lambda=1 P=0.7568832656
lambda=2 P=0.7568832656
lambda=3 P=0.7568832656
lambda=4 P=0.7568832656
lambda=5 P=0.7568832656
HTH,
David
--
David Cassell, CSC
Cassell.David@epa.gov
Senior computing specialist
mathematical statistician
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