Random variate generation and connected computational issues for the Poisson–Tweedie distribution

Abstract

After providing a systematic outline of the stochastic genesis of the Poisson–Tweedie distribution, some computational issues are considered. More specifically, we introduce a closed form for the probability function, as well as its corresponding integral representation which may be useful for large argument values. Several algorithms for generating Poisson–Tweedie random variates are also suggested. Finally, count data connected to the citation profiles of two statistical journals are modeled and analyzed by means of the Poisson–Tweedie distribution.

Appendix

Result 1

By expanding (1) in exponential and binomial series, it follows that

$$\begin{aligned} G_{X_{\text {PT}}}(s)= & {} e^{\frac{b}{a}(1-c)^a}\,\sum _{m=0}^\infty (-1)^m\, \frac{(b/a)^m}{m!}\,(1-cs)^{am}\\= & {} e^{\frac{b}{a}(1-c)^a}\,\sum _{m=0}^\infty (-1)^m\,\frac{(b/a)^m}{m!}\, \sum _{k=0}^\infty \left( {\begin{array}{c}am\\ k\end{array}}\right) (-cs)^k\\= & {} e^{\frac{b}{a}(1-c)^a}\,\sum _{k=0}^\infty (-cs)^k\,\sum _{m=0}^\infty (-1)^m \left( {\begin{array}{c}am\\ k\end{array}}\right) \,\frac{(b/a)^m}{m!} \end{aligned}$$

and hence

$$\begin{aligned} p_{X_{\text {PT}}}(k)=e^{\frac{b}{a}(1-c)^a}(-c)^k\,\sum _{m=0}^\infty (-1)^m \left( {\begin{array}{c}am\\ k\end{array}}\right) \,\frac{(b/a)^m}{m!}\,I_\mathbb {N}(k)\text {.} \end{aligned}$$

Moreover, by using the straightforward identity $\left( {\begin{array}{c}am\\ k\end{array}}\right) =\frac{1}{k!}\left. \frac{d^kx^{am}}{dx^k}\right| _{x=1}$, it also holds that

$$\begin{aligned} \sum _{m=0}^\infty (-1)^m\left( {\begin{array}{c}am\\ k\end{array}}\right) \,\frac{(b/a)^m}{m!}= & {} \frac{1}{k!}\left. \frac{d^ke^{-\frac{b}{a}x^a}}{dx^k}\right| _{x=1}=\frac{e^{-\frac{b}{a}}}{k!} \left. \frac{d^ke^{\frac{b}{a}(1-x^a)}}{dx^k}\right| _{x=1}\\= & {} \frac{e^{-\frac{b}{a}}}{k!}\,\sum _{m=0}^k\frac{(b/a)^m}{m!}\left. \frac{d^k(1-x^a)^m}{dx^k}\right| _{x=1}\\= & {} e^{-\frac{b}{a}}\,\sum _{m=0}^k\frac{(b/a)^m}{m!}\,\sum _{j=0}^m(-1)^j \left( {\begin{array}{c}m\\ j\end{array}}\right) \left( {\begin{array}{c}aj\\ k\end{array}}\right) \text {,} \end{aligned}$$

since $\left. \frac{d^k(1-x^a)^m}{dx^k}\right| _{x=1}=0$ when $k>m$. Thus, expression (9) promptly follows.

Result 2

On the basis of the expression of $p_{X_{\text {PT}}}$ given in Result 1, it follows

$$\begin{aligned} G_{X_{\text {PT}}}(s)=e^{\frac{b}{a}[(1-c)^a+1]}\,\sum _{k=0}^\infty (-cs)^k {{\mathrm{E}}}\left[ (-1)^M\left( {\begin{array}{c}aM\\ k\end{array}}\right) \right] \text {,} \end{aligned}$$

where M represents a $\mathcal {P}(b/a)$ r.v. Hence, it also holds that

$$\begin{aligned} p_{X_{\text {PT}}}(k)=e^{\frac{b}{a}[(1-c)^a+1]}(-c)^k{{\mathrm{E}}}\left[ (-1)^M \left( {\begin{array}{c}aM\\ k\end{array}}\right) \right] \text {.} \end{aligned}$$

Therefore, for $a\in ]0,1]$, from the previous expression we obtain

$$\begin{aligned} p_{X_{\text {PT}}}(k)\le & {} e^{\frac{b}{a}[(1-c)^a+1]}c^k{{\mathrm{E}}}\left[ \left| \left( {\begin{array}{c}aM\\ k\end{array}}\right) \right| \right] \\\le & {} \frac{b}{a}\left( 1+\frac{b}{a}\right) e^{ \frac{b}{a}[(1-c)^a+1]}c^kk^{-a-1}\text {.} \end{aligned}$$