You write: "The difference between two independent identically distributed exponential random variables is governed by a Laplace distribution." How exactly? Can you give the pdf-function of the Laplace distribution in tems of lambda, where lambda is the parameter of the exponential distribution (f(t) = lambda*exp(-lambda*t)?
The Laplace distribution is the distribution of the difference of two random variables (rv) each with an exponential distribution.
Let X be one of the rvs and let Y be the other. X is distributed with a pdf of (f(t) = lambda*exp(-lambda*t) and Y is distributed with a pdf of (f(t) = gamma*exp(-gamma*t). Then the rv (X - Y) is distributed as a Laplace distribution.
If gamma == lambda then the Laplace distribution is symmetric (a classical Laplace distribution); if gamma != lambda then X - Y is distributed as skew symmetric Laplace distribution.
Hope that helps.DrMicro (talk) 19:06, 10 January 2012 (UTC)
Isn't the median of the data also the maximum likelihood estimator for the location parameter (mean)? Shouldn't this be stated or did I miss it? Fjhickernell (talk) 01:26, 18 February 2010 (UTC)
I think the median already notes that the sample median is the MLE for the location parameter, and gives the formula for the MLE of b. But if it isn't clear, then perhaps it should be edited to clarify. Rlendog (talk) 01:59, 18 February 2010 (UTC)
Generation of a sample of Laplace random variables
The equation provided for generating a sample of Laplace-distributed random variables does not seem to provide the desired sample. The method described here ... http://www.math.uah.edu/stat/special/Laplace.html ... seems to work better. ... In "matlab" ...
U = rand(r,c);
in = find(U<=0.5);
ip = find(U>0.5);
x(in) = muX + sigmaX/sr2 * log(2*U(in));
x(ip) = muX - sigmaX/sr2 * log(2*(1-U(ip)));