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.

Alexander: Is it then a log-Gamma distribution?

This is a healthy discussion.

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

P(Y