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exp-gamma distribution is defined as the density of the random variable log(X) when X is a gamma random variable.

I am trying to obtain its PDF. Unfortunaltely, the only formula I have found is from a link on wikipedia toward Wolfram page for that distribution and their formula is not the same as the one I obtain myself. They say that

"The exp-gamma distribution is mathematically defined to be the distribution that models $Y=log(X)$ whenever X $\sim$ GammaDistribution. "

So I suppose we are considering the same thing.

Here is their formula :

enter image description here

First of all there is a $\mu$ location parameter that is not in the Gamma distribution as presented in Wikipedia. But, Wolfram presents a more general gamma family with two additional parameters that they name GammaDistribution[α,β,γ,μ]. They explain :

GammaDistribution[α,β] (which is equivalent to GammaDistribution[α,β,1,0]) is often referred to as "the" gamma distribution.

So I will take 0 for $\mu$

Now, my calculus is as follows :

$f$ the PDF of $X$ is such that $f(x) = K.x^{k-1}e^{\frac{-x}{\theta}}$

with $Y=ln(X)$ we have $dx=e^ydy$

If $g$ is the PDF of $Y$ We must have $g(y)dy=f(x)dx = f(e^y)e^ydy$

So

$$g(y)=Ke^{-\frac{e^y}{\theta}+ky}$$

which is not the same formula as Wolfram's if you look carefully ...

Any idea ?

Arnaud Mégret
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  • I see no difference: simply substitute $(x-\mu)/\theta$ for your $y$ to obtain the Wolfram expression. Where do you think you have erred? – whuber Apr 07 '22 at 13:04
  • Why should I have the right to do such substitution ? Furthermore, doing so, I do not get exactly the Wolfram formula due to the $\theta$ already in $\frac{e^y}{\theta}$ – Arnaud Mégret Apr 07 '22 at 14:48
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    (1) That "right" is a basic rule of arithmetic: equal expressions can be substituted within free variables in other expressions. (2) That factor of $1/\theta$ is explained at https://stats.stackexchange.com/questions/4220. Maybe that thread addresses what is actually bothering you? – whuber Apr 07 '22 at 14:51
  • Sorry I do my best to understand. I suppose you are right and that I don't see something obvious. But I am perplex. X is a Gamma random variable and Y=log(X) is another one. Both Wolfram and I are calculating the PDF of Y. I name y the values of Y while Wolfram names them x. But there is no difference, just 2 names for the same thing. We should have found exactly the same PDF function. – Arnaud Mégret Apr 07 '22 at 15:11
  • Furthermore, your substitution provides $g(y)=Ke^{-\frac{e^\frac{x-\mu}{\theta}}{\theta}+k\frac{x-\mu}{\theta}}$ it is not the Worlfram formula. – Arnaud Mégret Apr 07 '22 at 15:18
  • The factor of $1/\Gamma(k)$ is the normalization factor that you have written "$K.$" Again, there is no apparent difference between what you obtained and the Wolfram value apart from the notation. – whuber Apr 07 '22 at 16:05
  • I have no problem with the normalization coefficient. My idea so far is that Wolfram formula is the PDF for $\theta Y + \mu$ not $Y$. But also, that there is maybe a mistake in their formula. I am going to reedit my question with that suggestion. – Arnaud Mégret Apr 07 '22 at 16:15
  • You will first want to learn about location-scale families of distributions. See https://stats.stackexchange.com/questions/265939, for instance. That will help you see there is no mistake. – whuber Apr 07 '22 at 16:41
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    I am very sorry for that misunderstanding :-) Let my try a last time to defend my point. You told me "substitute (x−μ)/θ for your y to obtain the Wolfram expression.". I did it and obtained the formula that you can see 4 comments above this one.
    1. Is my substitution correct ?
    2. If yes, is it the same as Wolfram's formula, even if you substitute $K$ by something ? (notice $\theta$ appears 3 times within the first level exponential while only 2 times in Worlfram's formula)

    By the way the link in your previous comment is very interesting.

     – Arnaud Mégret Apr 07 '22 at 17:23

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