Where can I find the explicit expression of the distribution of the sum of n i.i.d. logistic random variables, for n=2,3,4…
The expressions given in "On the convolution of logistic random variables, "Metrika
Volume 30, Issue 1, December 1983, Pages 1-13. are rather cumbersome and not explicit.
Best Answer
A Remark on the Convolution of the Generalised Logistic Random Variables, Matthew Oladejo Ojo (2003).
There are typos in the article linked above. In Equation (1.1), the denominator of $f$ is $(1+e^x)^{p+q}$. In Section 2, in the unnumbered second equation expressing the caracteristic function of $Y$, the argument of the Gamma function in the denominator is $p$ and not $i p$.
Probability density function $$ f(x) = \frac{\Gamma(p+q)}{\Gamma(p)\Gamma(q)} \frac{e^{px}}{(1+e^x)^{p+q}}, \quad -\infty<x<\infty,\ p>0,\ q>0$$ for each of $n$ independent random variables gives the probability distribution $f_n(y)$ for the sum:
$$ f_n(x) = e^{px} (\Gamma(p)\Gamma(q))^{-n} \sum_{k=0}^\infty \frac{(-1)^{n(k+1)+1}}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}}\left[ e^{xz}\frac{\Gamma(z+p+q)^n}{\Gamma(z+1)^n} \right]_{\vert z=k} $$
For example, when $p=q=1$, we get
$$ f_n(x) = e^{x} \sum_{k=0}^\infty \frac{(-1)^{n(k+1)+1}}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}}\left[ e^{xz} (z+1)^n \right]_{\vert z=k} $$
Some explicit expressions of $f_n(y)$ for this case $p=q=1$:
$$f_{1}(y)=\frac{e^y}{\left(e^y+1\right)^2}$$
$$f_{2}(y)=\frac{e^y \left(e^y (y-2)+y+2\right)}{\left(e^y-1\right)^3}$$
$$f_{3}(y)=\frac{e^{2 y} \left[-2 y^2+\left(y^2+6\right) \cosh y-6 y \sinh y+6\right]}{\left(e^y+1\right)^4}$$
$$f_{4}(y)=\tfrac{1}{3}e^{5y/2}(e^y-1)^{-5}[y (11 y^2-36) \cosh(y/2) + y (36 + y^2) \cos(3 y/2) - 24 (2 y^2-2 + (2 + y^2) \cosh y) \sinh(y/2)]$$