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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.


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:

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


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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QUOTE: The characteristic function of the sum of two variables $X$ and $Y$ is defined as the product of their corresponding characteristic functions. END OF QUOTE. That's not true. If they're independent, then the charateristic function of their sum can be shown to be the product of their characteristic functions, but it is not defined to be that. Rather, that is a demonstrable fact, not a definition. The definition is still the same as for characteristic functions in general: it is $t\mapsto \mathbb E(e^{t(X+Y)})$. – Michael Hardy Dec 4 '12 at 17:34
Some of your characteristic functions $\varphi$ are such that $\varphi(0)\ne1$. This is odd. – Did Dec 4 '12 at 18:52
@MichaelHardy : That is obviously correct. I will correct that in my question. – Mark Dec 4 '12 at 19:26
@did: I will add the pmf of the BNB distribution too so it will be more obvious why i tried it this way. Trying to find the characteristic function when using the textbook definition of the pmf has proven to be too hard for me. – Mark Dec 4 '12 at 19:29
The trouble is not with what you tried or did not try but with the fact that characteristic functions are defined as $\varphi_X(t)=\mathbb E(\mathrm e^{itX})$. Hence $\varphi_X(0)=1$ for every random variable $X$. If $\psi(0)\ne1$, then $\psi\ne\varphi_X$ for every $X$. Thus, there is a problem with your formulas, which I suggest to get rid of before going any further. – Did Dec 4 '12 at 23:23

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