Re: approximate a pdf function with exponential distribution
From: Julian V. Noble (jvn_at_virginia.edu)
Date: 01/11/05
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Date: Tue, 11 Jan 2005 12:01:21 -0500
Robert Israel wrote:
>
[ snip ]
>
> >If the distribution is exponential,
>
> > dp(x) = A exp(-Ax) dx ,
>
> >then the mean is 1/A and the variance is 1/A^2 .
>
> >Since this is a 1-parameter distribution you can't fit both the
> >mean and variance of your pdf. However, if you use the distribution
>
> > dp(x) = A^{\alpha + 1} exp(-Ax) dx / \Gamma( \alpha + 1)
>
> >you have two parameters to play with and can thus fit these
> >aspects of your pdf.
>
> Huh? That's not a probability measure.
>
You're right, that's a typo! (Or a sin of omission, if you prefer.)
I meant to type
dp(x) = A^{\alpha + 1} x^ {\alpha} exp(-Ax) dx / \Gamma( \alpha + 1)
^^^^^^^^^^^
which is a probability measure (positive, integrates to 1). Thanks for
noticing.
--
Julian V. Noble
Professor Emeritus of Physics
jvn@lessspamformother.virginia.edu
^^^^^^^^^^^^^^^^^^
http://galileo.phys.virginia.edu/~jvn/
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