Suppose $X_i\sim GAM(\alpha,\beta)$ with PDF $$f_{X_i}(x)=\frac{x^{\alpha-1}e^{-x/\beta}}{\Gamma(\alpha)\beta^\alpha}$$ Then, from univariate transformation, the random variable $Y_i:=ln(X_i)\sim LOGGAM(\alpha,\beta)$ with PDF $$f_{Y_i}(y)=\frac{e^{\alpha y}\cdot e^{-e^{y}/\beta}}{\Gamma(\alpha)\beta^\alpha}$$

Suppose I have a random sample of $Y_i$ that are independent and identically distributed from this log-gamma distribution, $i\in\{1,2,...,n\}$.

What is the distribution of $\sum_{i=1}^n Y_i$? That is, what is the distribution of the sum of independent and identically distributed log-gamma random variables?

From the MGF technique, we can find that $$M_{\sum Y_i}(t)=\left(\frac{\beta^t\Gamma(\alpha+t)}{\Gamma(\alpha)}\right)^n$$

But I have no idea how to derive the PDF of a distribution that has this particular MGF. We know that $\alpha>0, \beta>0,$ and $Y_i\in(-\infty, \infty)$, but even with those assumptions I'm still lost. How can we get this PDF?

  • $\begingroup$ Presumably this is related to the logarithm of the product of iid gamma random variables. That distribution can apparently be expressed with Bessel functions $\endgroup$ – Henry Apr 22 '18 at 0:13

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