If X is a random variable has a Gamma distribution with (N,K), what is the distribution of Y, where Y=ln (X)?
Many thanks.
Transformation formulae are given in most theoretical statistics books. A simple case is a continuous one-to-one transformation (like ln) of a continuous distribution. We just divide by the absolute derivative, |dy/dx|.
In your case, dy/dx = 1/x = 1/exp(y)
fX(x) = xN-1exp(-x/K)/(KNGamma(N))
fY(y) = exp(Ny - exp(y)/K)/(KNGamma(N))
I think -Y has a Gumbel distribution which is one of the limiting extreme value distributions.
I do not have an answer for you, Huda. I do know that the Gumbel distribution is used to approximate p-values for the likelihood ratio test in the satscan cluster analysis software.
Why are you considering ln(X)?
I have found this, though: http://www.math.wm.edu/~leemis/chart/UDR/PDFs/GammaLoggamma.pdf
Is it the log-gamma distribution?
Thanks for all, really it is a valuable comments.
Dear Raid, I need the distribution of ln(x), where X has Gamma p.d.f. to prove some estimators are minimax.
I think the density of Y would be of the form:
Let f(x) and g(y) be the densities,respectively, of X and Y. Set x=Exp(y) and use the following transformation.
g(y)=f(x) |dx/dy| . Note that x=Exp(y) and f(x) is the density of Gamma Distribution.
After little simplification you will arrive at the density of Y. At this point, I leave u to do the identification of the prob. density function.
The correct form of required distribution slightly differs from Gumbel...
I've tested the solution by Wolfram Mathematica. A good excercise.
Many thanks for all. According to one-one transformation, Y is distributed as Log Gamma, the chart in this URL clarifies Univariate Distribution Relationships
http://www.math.wm.edu/~leemis/chart/UDR/UDR.html
Putting N = K = 1 and Z = -Y in
fY(y) = exp(Ny - exp(y)/K)/(KNGamma(N))
we get
fZ(y) = exp(-z - exp(-z))
which is the standardised form of the Gumbel distribution. So Alexander and Eliardo are correct. The log of a gamma is a slight generalisation of the Gumbel. As it is the log of a gamma, the term log gamma seems appropriate.
(Terminology can be confusing--a lognormal isn't the log of a normal, instead its log is normal--in fact you can't take the log of a normal because the log of a negative value is a complex number).
A nice related excercise is the following: for independent random variables X and Y uniformly distributed in [0,1], prove that P( X Y < z ) = - ln z.
As there are two variables, this is a little more advanced. Depending on how you solve it, it can be a related problem. You can transform X and Y to U = -ln X, V = -ln Y so that XY < z becomes U + V > -ln z. Having obtained the distributions of U and V we have a two variable problem.
We can also solve it directly. We want 1 - the area between the lines y = 1, x = 1 and the hyperbola, xy = z. Alternatively it is the complemetary area which can be broken up as a rectangle with area z and the area between z and 1 under the hyperbola.
Huda, I'll leave you to do this both ways. It will help contribute to your expertise in transfromations.
Sorry for a misprint in the left hand side of the equality. The derivative operation was missed and correct statement is P'( X Y < z ) = - ln z. Demo and the Monte-Carlo test are in the attached file.
This is a simple case that follows from a change of variables: The following refer to pdf's (prob. density functions)
p_x = gamma pdf wıth parameters N & K
y = ln(x)
p_y = p_x*dx/dy = x*p_x where x = exp(y)>0 & -inf< y
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