On the Convolution of Scaled Sibuya Distributions

Abstract

We introduce a new heavy tailed distribution on $\mathbb {Z}_+$ that arises as the infinite convolution of scaled Sibuya distributions. We provide closed form expressions for its probability mass function, its cumulative distribution function, and its probability generating function. We interpret our main results in terms of the weak convergence of partial sums of a binomially thinned sequence of i.i.d. random variables with a common scaled Sibuya distribution. Properties of infinite divisibility and discrete self-decomposability of the new distribution are also discussed. As an application, we briefly describe an integer-valued autoregressive process of order one with a scaled Sibuya innovation sequence. Finally, we discuss some partial extensions of our results to the case of the generalized Sibuya distribution introduced by Kozubowski and Podgórski., Ann. of the Inst. Statist. Math., 70(4), 855-887., 2018.

Appendix

Lemma 5

Assume $n\ge 2$, $a_{i}\in [0,1]$ for $i=0,1,2,\cdots ,n-1$ and $a\in (0,1)$. Then,

• (i)
  $$\begin{aligned} \prod _{i=0}^{n-1}(1-a_{i})=1+\sum _{k=1}^{n}(-1)^{k}\Bigl [\sum \limits _{0\le j_{1}<j_{2}<\cdots <j_{k}\le n-1}\prod _{l=1}^{k}a_{j_{l}}\Big ]. \end{aligned}$$

  (42)
• (ii)
  $$\begin{aligned} \sum _{0\le j_{1}<j_{2}<\cdots <j_{k}\le n-1} a^{j_{1}} a^{j_{2}}\cdots a^{j_{k}}=a^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) }\prod _{l=0}^{k-1}{\frac{1-a^{n-l}}{1-a^{l+1}}} \end{aligned}$$

  (43)

  for every $k\in \{1.\cdots ,n\}$.

Proof

See Appendix (Lemma 2) in Aly and Bouzar (2023).$\square$

Lemma 6

Let $n\ge 1$, and for $0\le i \le n-1$, $\lambda _i\in (0,1]$ and $\gamma _i\in (0,1]$. Let $p_r^{(n)}=\bigl (f(\lambda _0,\gamma _0)*f(\lambda _1,\gamma _1) *\cdots *f(\lambda _{n-1},\gamma _{n-1})\bigr )_r$, $r\ge 0$, be the pmf of the convolution of the scaled Sibuya distributions $(\{f_r(\lambda _i,\gamma _i\}$, $0\le i\le n-1)$. The the following assertions are true.

• (i) If $\gamma _i\in (0,1)$ for all $0\le i\le n-1$, or if $\gamma _i=1$ for some but not all i’s, then

  $$\begin{aligned} p_r^{(n)}= {\left\{ \begin{array}{ll} \prod _{i=0}^{n-1}(1-\lambda _i) &{}\hbox {if\ } r=0\\ \sum _{k=1}^n (-1)^{r+k} \beta _k^{(n)}(r) &{} \hbox {if } r\ge 1,\\ \end{array}\right. } \end{aligned}$$

  (44)

  where

  $$\begin{aligned} \beta _k^{(n)}(r)=\left( {\begin{array}{c}\sum _{l=1}^k\gamma _{j_l}\\ r\end{array}}\right) \sum _{0\le j_1<j_2<\cdots <j_k\le n-1}\Bigl (\prod _{l=1}^k \lambda _{j_l}\Bigr ). \end{aligned}$$

  (45)

• (ii) If $\gamma _0=\gamma _1=\cdots =\gamma _{n-1}=1$, i.e., $\{ p_r^{(n)}\}$ is the convolution of Bernoulli distributions, then its support reduces to $0\le r\le n$ and formula Eq. 44 holds with

  $$\begin{aligned} \beta _k^{(n)}(r)= {\left\{ \begin{array}{ll} 0 &{} \hbox {if\ } r\ge k+1\\ \displaystyle \left( {\begin{array}{c}k\\ r\end{array}}\right) \sum _{0\le j_1<j_2<\cdots <j_k\le n-1}\Bigl (\prod _{l=1}^k \lambda _{j_l}\Bigr ) &{} \hbox {if\ } 1\le r\le k\le n.\\ \end{array}\right. } \end{aligned}$$

  (46)

• (iii) If $\lambda _0=\lambda _1=\cdots =\lambda _{n-1}=1$, i.e, $\{p_r^{(n)}\}$ is the convolution of standard Sibuya distributions, then formula Eq. 44 holds with

  $$\begin{aligned} \beta _k^{(n)}(r)=\sum _{0\le j_1<j_2<\cdots <j_k\le n-1}\left( {\begin{array}{c}\sum _{l=1}^k\gamma _{j_l}\\ r\end{array}}\right) , \qquad 1\le k\le n. \end{aligned}$$

  (47)

Proof

(i) The pgf of $\{p_r^{(n)}\}$ is $P(z)=\prod _{i=0}^{n-1} (1-\lambda _i(1-z)^{\gamma _i})$. Since $p_0^{(n)}=P(0)$, Eq. 44 holds for $r=0$. Let $a_i=\lambda _i(1-z)^{\gamma _i}$, $0\le i\le n-1$. Clearly, $a_i\in (0,1)$ for any $z\in [0,1)$. Therefore, by Eq. 42 in Lemma 5,

$$\begin{aligned} P(z)=1+\sum _{k=1}^{n}(-1)^{k}\Bigl [\sum \limits _{0\le j_{1}<j_{2}<\cdots <j_{k}\le n-1}\Bigl (\prod _{l=1}^k \lambda _{j_l}\Bigr ) (1-z)^{\sum _{l=1}^k\gamma _{j_l}}\Big ] \end{aligned}$$

(48)

for any $z\in [0,1)$. By the binomial series expansion, we have

$$ (1-z)^{\sum _{l=1}^k\gamma _{j_l}}=\sum _{r=0}^\infty (-1)^r\left( {\begin{array}{c}\sum _{l=1}^k\gamma _{j_l}\\ r\end{array}}\right) z^r \quad 0\le z<1. $$

This implies that for any $z\in [0,1)$,

$$ P(z)=1+\sum _{r=0}^\infty (-1)^r\biggl (\sum _{k=1}^{n}(-1)^{k}\Bigl [\sum \limits _{0\le j_{1}<j_{2}<\cdots <j_{k}\le n-1}\Bigl (\prod _{l=1}^k \lambda _{j_l}\Bigr )\left( {\begin{array}{c}\sum _{l=1}^k\gamma _{j_l}\\ r\end{array}}\right) \Big ]\biggr )z^r, $$

thus establishing (44) for any $r\ge 1$. Part (ii) follows from Eqs. 44–45 and the fact that $\sum _{l=1}^k\gamma _{j_l}=k$, and part (iii) from the fact that $\prod _{l=1}^k \lambda _{j_l}=1$. $\square$

Proof of Theorem 1

To prove (i), we note by Proposition 1 that $\{p_r^{(n}\}$ of Eq. 4 has pgf

$$\begin{aligned} \varphi _n(z)=\prod _{k=0}^{n-1} (1-\lambda \alpha ^{k\gamma }(1-z)^\gamma )\quad (0\le z\le 1). \end{aligned}$$

(49)

Therefore, the convergence in distribution of the sequence $(\{p_r^{(n)}\}, n\ge 1)$ of Eq. 4 will follow by establishing that $\varphi (z)$ of Eq. 3 is indeed a pgf. To prove this fact we call on Theorem (part (iii)) in Foster and Williamson (1971). Translating their notation in our context, we have $h(z)=1-\lambda (1-z)^\gamma$ and $f(z)=1-\alpha +\alpha z$. Then

$$ \frac{1-h(z)}{f(z)-z}=\frac{1-\Psi (z)}{(1-\alpha )(1-z)}=\lambda (1-z)^{\gamma -1}/(1-\alpha )\qquad 0\le z<1. $$

The sequence of pgf’s $\{\varphi _n(z)\}$ converges to a pgf if and only if $\int _0^1\frac{1-\Psi (z)}{(1-\alpha )(1-z)}\,$ $dz <\infty$. The latter condition clearly holds. Therefore, $\varphi (z)$ is a pgf. Next, we prove Eq. 6. First, we note that $\lim _{n\rightarrow \infty }p_0^{(n)}=\prod _{k=0}^\infty (1-\lambda \alpha ^{k\gamma }) <\infty$, since $\sum _{k=0}^\infty \lambda \alpha ^{k\gamma }<\infty$. Let $r\ge 1$. We define a purely atomic measure, we denote meas, on $\mathbb {Z}_+$ and its power set $\mathcal {P}(\mathbb {Z_+})$ as follows:

$$\begin{aligned} meas(\{k\})={\frac{\lambda ^{k}\alpha ^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) \gamma }}{\prod _{l=1}^{k}(1-\alpha ^{l\gamma })}}\quad (k\ge 1), \end{aligned}$$

(50)

with $meas(\{0\})=1$. Note that meas is a finite measure, i.e., $\sum _{k=0}^{\infty } meas$ $(\{k\})<\infty$ by the ratio test, as $meas(\{k+1\})/meas(\{k\})=\lambda \alpha ^{k\gamma }/(1-\alpha ^{(k+1)\gamma })$. For a fixed integer $r\ge 0$, we define a the sequence of functions $\{f_{n}^{(r)}(\cdot )\}$ on $\mathbb {Z}_+$ as follows:

$$ f_n^{(r)}(k)=(-1)^{r+k}\left( {\begin{array}{c}k\gamma \\ r\end{array}}\right) \prod _{l=0}^{k-1}(1-\alpha ^{(n-l)\gamma }) I_{\{1,2,\cdots ,n\}}(k) \qquad (k\ge 0). $$

Then, $p_r^{(n)}$ of Eq. 4 admits the representation $p_r^{(n)}\!=\!I_{\{0\}}(r)+ \int _{\mathbb {Z}_+}f_n^{(r)}(k)\,meas$ (dk) on the measure space $(\mathbb {Z}_+,\mathcal {P}(\mathbb {Z}_+),meas)$. Next, we define a function h on $\mathbb {Z}_+$: $h(k)=\left( {\begin{array}{c}k\gamma \\ r\end{array}}\right)$. It is clear that $|f_{n}^{(r)}(k)|\le |h(k)|$ (recall $\alpha \in (0,1)$) and that $\sum _{k=0}^{\infty }|h(k)|meas(\{k\})<\infty$ by the ratio test (as $|h(k+1)/h(k)| \rightarrow 1$ as $k\rightarrow \infty$), implying the integrability of |h(k)|. Moreover, for every $k\in \mathbb {Z}_+$, $\lim _{n\rightarrow \infty }f_{n}^{(r)}(k)=(-1)^{r+k}\left( {\begin{array}{c}k\gamma \\ r\end{array}}\right)$. It follows by the dominated convergence theorem that

$$ p_{r}=I_{\{0\}}(r)+\lim _{n\rightarrow \infty }\int _{\mathbb {Z}_+}f_{n}^{(r)}(k)\,meas(dk)=I_{\{0\}}(r)+\int _{\mathbb {Z}_+}(-1)^{r+k}\left( {\begin{array}{c}k\gamma \\ r\end{array}}\right) \,meas(dk), $$

which is precisely Eq. 6, thus completing the proof of (i). For part (ii), we note that Eq. 7 is trivially true for $r=0$. We have by Eq. 6 that for any $r\ge 1$,

$$ \sum _{j=0}^r p_j=1+ \sum _{j=0}^r \sum _{k=1}^\infty (-1)^{k+j} \frac{\lambda ^k\alpha ^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) \gamma }}{\prod \limits _{l=1}^k (1-\alpha ^{l\gamma })}\left( {\begin{array}{c}k\gamma \\ j\end{array}}\right) , $$

or

$$ \sum _{j=0}^r p_j=1+ \sum _{k=1}^\infty (-1)^{k} \frac{\lambda ^k\alpha ^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) \gamma }}{\prod \limits _{l=1}^k (1-\alpha ^{l\gamma })} \Bigl [\sum _{j=0}^r (-1)^j \left( {\begin{array}{c}k\gamma \\ j\end{array}}\right) \Bigr ]. $$

Using the identity $\left( {\begin{array}{c}k\gamma \\ j\end{array}}\right) =\left( {\begin{array}{c}k\gamma -1\\ j-1\end{array}}\right) +\left( {\begin{array}{c}k\gamma -1\\ j\end{array}}\right)$ ($j\ge 1$), it is easily shown that $\sum _{j=0}^r (-1)^j \left( {\begin{array}{c}k\gamma \\ j\end{array}}\right) =(-1)^r \left( {\begin{array}{c}k\gamma -1\\ r\end{array}}\right)$, completing the proof of Eq. 7. To prove Eq. 8 in part (iii), we note $\varphi (z)$ of Eq. 3 can be rewritten as

$$\begin{aligned} \varphi (z)=\exp \left\{ \sum _{k=0}^{\infty }\ln (1-\lambda \alpha ^{k\gamma }(1-z)^\gamma )\right\} . \end{aligned}$$

(51)

The representation Eq. 8 of $\varphi (z)$ follows by way of the power series expansion $-\ln (1-x)=\sum _{n=1}^{\infty }x^{n}/n$, $0\le x<1$ applied at $x=\lambda \alpha ^{k\gamma }(1-z)^\gamma$ to the summands in Eq. 51 combined with the fact that the uniform convergence over [0, 1] of the resulting double series (summands dominated by $\lambda ^n\alpha ^{k\gamma }$) allows for the interchanging of the order of summation. To prove Eq. 9, let $a_{i}=\lambda \alpha ^{i\gamma }(1-z)^\gamma$ in Eq. 42. Using Eq. 43 yields the following expression for $\varphi _{n}(z)$ of Eq. 49:

$$\begin{aligned} \varphi _n(z)=1+\sum _{k=1}^n (-1)^k \lambda ^k (1-z)^{k\gamma }\alpha ^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) \gamma }\prod _{l=0}^{k-1}{\frac{1-\alpha ^{(n-l)\gamma }}{1-\alpha ^{(l+1)\gamma }}}. \end{aligned}$$

(52)

We now proceed as in the proof of Eq. 6. We define a sequence of functions $\{g_n(\cdot )\}$ on the finite measure space $(\mathbb {Z}_+,\mathcal {P}(\mathbb {Z}_+),meas)$, with meas of Eq. 50, as follows: $g_n(0)=1$ and

$$ g_n(k)=(-1)^k(1-z)^{k\gamma }\Bigl (\prod _{l=0}^{k-1}(1-\alpha ^{(n-l)\gamma })\Bigr )I_{\{1,2,\cdots ,n\}}(k) \qquad (k\ge 1). $$

It is easily seen that $|g_{n}(k)|\le 1$ (recall $\alpha \in (0,1)$ and $z\in [0,1]$) and that $g(k)=\lim _{n\rightarrow \infty }g_{n}(k)$ takes the following values: $g(0)=1$ and $g(k)=(-1)^k(1-z)^{k\gamma }$ if $k\ge 1$. Rewriting Eq. 52 in terms of the discrete integral on $(\mathbb {Z}_+,\mathcal {P} (\mathbb {Z}_+),meas)$ and calling on the dominated convergence theorem, we have

$$ \varphi (z)=\lim _{n\rightarrow \infty }\varphi _n(z)=\lim _{n\rightarrow \infty }\int _{\mathbb {Z_+}}g_{n}(k)\,meas(dk)=\int _{\mathbb {Z}_+}g(k)\,meas(dk), $$

which establishes Eq. 9, and thus completes the Proof of Theorem 1.