In article <[EMAIL PROTECTED]>,
Andrew Morse <[EMAIL PROTECTED]> wrote:
> I tried to derive an expression for the log-gamma distribution using
>the same procedure that can be used to derive an expression for the
>log-normal distribution.
> 1. I started with the expression for a gamma [normal] distribution,
>Gamma(x; alpha, beta) [Normal(x; mu, sigma)].
> 2. I replaced x with log y, yielding Gamma(log y; alpha, beta)
>[Normal(log y; mu, sigma)].
> 3. I wrote out the expression for the integral of Gamma(log y;
>alpha, beta) [Normal(log y; mu, sigma)] over (d[log y] = dy/y).
> 4. The integrand of the resulting integral over y should be the
>log-gamma [log-normal] distribution.
> This algorithm yields the right answer for the log-normal
>distribution, but my result for the log-gamma distribution is different
>from any form of the log-gamma that I have ever seen. Given a gamma
>distribution of the form...
> Gamma(x; a, b) = x^{a-1} * exp{-x/b} / {b^a * G(a)}...
> ...I obtain a log-gamma distribution of...
> LogGamma(x; a, b) = [log x]^{a-1} * x^{-(b+1)/b} / {b^a * G(a)},
>defined for 1 <= x <= infinity.
On the other hand, the logarithm of a gamma random variable
has a rather nice useful distribution, with density
exp(ax - exp(x)/b)/(b^a * G(a).
The reason for the difference is that the gamma distribution
goes from 0 to infinity, and so it make more sense to take
the logarithm. Also, the sufficient statistics for a sample
from a gamma distribution are the sum and the sum of the
logarithms.
...............
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
[EMAIL PROTECTED] Phone: (765)494-6054 FAX: (765)494-0558
.
.
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