The beta prime distribution is infinitely divisible, as proved in Steutel and van Harn, 2003 (Appendix B). Sadly, in this book, there is no expression of the parameters of the distribution of $n$ variables iid following a beta prime distribution, as it is done for some other distributions (normal, gamma, etc).

I cannot find how to derive it by myself. As an example, I am trying to derive the probability of the sum of two variables, and I am stuck in solving the basic convolution integral:

$$ f_{X+Y}(z)=\int_{-\infty}^{+\infty}f_X(x)f_Y(z-x)dx=\int_0^{z}\frac{1}{B(a,b)^2}\frac{(x(z-x))^{a-1}}{((1+x)(1+z-x))^{a+b}}dx $$

This distribution must be beta prime and I want to compute the parameters of this joint distribution. Is there a mean to get it from the convolution of PDF? Is there another mean to get directly the parameters of the beta prime pdf of the sum of n random variables iid following a beta prime distribution?

  • 1
    $\begingroup$ cross posting is not allowed. please delete this question on mse $\endgroup$
    – Joey Doey
    May 28 '18 at 7:40
  • 1
    $\begingroup$ @JoeyDoey I don't intend to cross-post but I would like to transfer it from mse and I don't find how to. Not enough reputation maybe? I am deleting the one on mse right now. Thanks for the moderation. $\endgroup$
    – Bentoy13
    May 28 '18 at 8:13

Let $\{X_i\}_{i=1}^n$ with $X_i\overset{\text{i.i.d}}{\sim}\beta^\prime(\alpha,\beta)$ and $Z=X_1+\dots+X_n$. It follows from linearity of the expected value that $$ \mathsf EZ=\sum_{i=1}^n\mathsf EX_i=n\mathsf EX=\frac{n\alpha}{\beta-1},\quad\beta>1. $$ Furthermore, by mutual independence of the $X_i$'s we have $$ \mathsf{Var}Z=\sum_{i=1}^n\mathsf{Var}X_i=n\mathsf{Var}X=\frac{n\alpha(\alpha+\beta-1)}{(\beta-2)(\beta-1)^2},\quad\beta>2. $$ Since we know $Z\sim\beta^\prime(\gamma,\delta)$ we may write the system of equations $$ \begin{aligned} \frac{n\alpha}{\beta-1} &=\frac{\gamma}{\delta-1}\\ \frac{n\alpha(\alpha+\beta-1)}{(\beta-2)(\beta-1)^2} &=\frac{\gamma(\gamma+\delta-1)}{(\delta-2)(\delta-1)^2}. \end{aligned} $$ Solving this system for $\gamma$ and $\delta$ subsequently yields the following result: $$ \begin{aligned} \gamma &=\frac{\alpha n \left(\alpha +\beta ^2-2 \beta +\alpha \beta n-2 \alpha n+1\right)}{(\beta -1) (\alpha +\beta -1)}\\ \delta &=\frac{2 \alpha +\beta ^2-\beta +\alpha \beta n-2 \alpha n}{\alpha +\beta -1}. \end{aligned} $$ With some algebra you may be able to simplify these expressions.


Based on the discussion surrounding the exactness/correctness of the results I decided to perform an experiment in MATLAB. Here is the code used which performs the simulation for $\alpha=\beta=15$ and $n=5$:

a = 15; %alpha
b = 15; %beta
n = 5;
c = (a*n*(a-2*b-2*a*n+b^2+a*b*n+1))/((b-1)*(a+b-1)); %gamma
d = (2*a-b-2*a*n+b^2+a*b*n)/(a+b-1); %delta

Xdata = 1./betarnd(a,b,1e6,n)-1;
Xn = sum(Xdata,2); %Xn = X_1+X_2+...+X_n
ax = linspace(0,max(Xn),256);
f_Xn = @(x) x.^(c-1).*(1+x).^(-c-d)/beta(c,d);

hold on
xlabel(['X_' num2str(n)])
box on
hold off

Here we see that the histogram does indeed agree with the theoretical beta prime distribution with parameters $\gamma$ and $\delta$ as derived above. I tried other values of $\alpha$, $\beta$ and $n$ with similar results.

enter image description here

  • $\begingroup$ Thank you for the detailed computation. After performing a test with two and three such variables, it appears that this result is only approximate, so that I figure out that I was wrong assuming that the sum is beta prime distributed. Note that the approximation is worst as n increases. $\endgroup$
    – Bentoy13
    Oct 7 '20 at 8:36
  • 1
    $\begingroup$ I found two errors, one with my code (which is similar to yours) and one with yours, exactly at the same line (coincidence!): when we want to generate $X_n$ following the beta prime distribution. So the correct way to do is Xdata = 1./betarnd(a,b,1e6,n)-1; Xdata = 1./Xdata;. In my case, I put extra parentheses. In both cases, the resulting variables are following $\beta'(d,c)$ and not $\beta'(c,d)$. So I delete my little extra comment, and your answer will be perfect as soon as the little mistake will be corrected! Thnak you very much Aaron. $\endgroup$
    – Bentoy13
    Oct 8 '20 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.