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I am facing a problem when trying to calculate the distribution of the sum of iid Beta-Negative-Binomial random variables or for that matter if only parameter $r$ is different. To get a hint to how they might be distributed I calculated the characteristic function of the BNB distribution:

$\varphi_X(t)=\int_0^1 \! \sum_{x=0}^{\infty} \binom{x+r-1}{x}p^r(1-p)^x\frac{p^{\alpha-1}(1-p)^{\beta-1}}{B(\alpha,\beta)}e^{itx} \, dp$ where $B$ is the Beta-function. One can now factor out the terms that do not contain $x$ and use the generating function $\sum_{n=0}^\infty \binom{n+k}{k}x^n=\frac{1}{(1-x)^{k+1}}$ to get the following expression for the characteristic function:

$\varphi_X(t)=\int_0^1 \! \frac{p^{\alpha-1}(1-p)^{\beta-1}}{B(\alpha,\beta)}\frac{p^r}{(1-(1-p)e^{it})^r} \, dp.$ The characteristic function of the sum of two independent variables $X$ and $Y$ is in this case the product of their corresponding characteristic functions.

$\varphi_{X+Y}(t)=\int_0^1 \! \frac{p^{\alpha-1}(1-p)^{\beta-1}}{B(\alpha,\beta)}\frac{p^{r_1}}{(1-(1-p)e^{it})^{r_1}} \, dp \,\,*\,\,\int_0^1 \! \frac{p^{\alpha-1}(1-p)^{\beta-1}}{B(\alpha,\beta)}\frac{p^{r_2}}{(1-(1-p)e^{it})^{r_2}} \, dp$

How can one now deduce the distribution of the sum?

Thank you in advance.

EDIT:

Probability mass function:

$\int_0^1 \! \binom{x+r-1}{x}p^r(1-p)^x\frac{p^{\alpha-1}(1-p)^{\beta-1}}{B(\alpha,\beta)}=\binom{x+r-1}{x}\frac{B(\alpha+r,\beta+x)}{B(\alpha,\beta)}=\frac{\Gamma(x+r)\Gamma(\alpha+r)\Gamma(\beta+x)\Gamma(\alpha+\beta)}{x!\Gamma(r)\Gamma(\alpha+\beta+r+x)\Gamma(\alpha)\Gamma(\beta)}$

Thus, the characteristic function can also be written as:

$\varphi_X(t) = \sum_{x=0}^{\infty}e^{itx} \binom{x+r-1}{x}\frac{\Gamma(\alpha+r)\Gamma(\beta+x)\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)\Gamma(\alpha+\beta+r+x)}$

EDIT2:

I made the non-trivial error to forget the integral in the pmf, which is also the reason why $\varphi_X(0)\ne0$.

$\varphi_X(0)=\frac{1}{B(\alpha,\beta)} \int_0^1 p^{\alpha-1}(1-p)^{\beta-1}*\frac{p^r}{(1-(1-p))^r} \, dp = 1$ with $B(x,y)=\int_0^1 p^{x-1}(1-p)^{y-1} \, dp$

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    Is there a clever way of solving this integral? I tried many ways but always failed. Your help would be greatly appreciated.2013-01-28

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