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On goodness-of-fit tests for the Bell distribution


The one-parameter Bell family of distributions, introduced by Castellares et al. (Appl Math Model 56:172–185, 2018), is useful for modeling count data. This paper proposes and studies a goodness-of-fit test for this distribution, which is consistent against fixed alternatives. The finite sample performance of the proposed test is investigated by means of several Monte Carlo simulation experiments, and it is also compared with other related ones. Real data applications are considered for illustrative purposes.

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The authors thank the Editor, the Associate Editor and two anonymous referees for their constructive comments and suggestions which helped to improve the presentation. M.D. Jiménez-Gamero has been partially supported by Grant MTM2017-89422-P of the Spanish Ministry of Economy, Industry and Competitiveness, the State Agency of Investigation, the European Regional Development Fund. Artur J. Lemonte acknowledges the financial support of the Brazilian agency CNPq (Grant 301808/2016–3).

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Correspondence to Apostolos Batsidis.

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Here we prove the results given in the previous sections.

Proof of Proposition 1

It can be checked that the PGF of \(X \sim \mathrm{Bell}(\theta )\) given in (1) satisfies the differential equation given in (2). Next, we proof that it is the only PGF in G satisfying such differential equation. It is well-known that the solution of the linear differential equation of order one of the form \(y^{\prime }+p(t) y=0\), where \(y=y(t)\), \(y'=\frac{\partial }{\partial t}y(t)\) and p(t) is a continuous function in t, is given by \(y=C \exp (-\int p(t) dt)\), where C is an arbitrary constant. Since the differential equation (2) is of this form, we have that

$$\begin{aligned} g(t)=C \exp \left( \int \theta e^{\theta \, t} dt \right) = C \exp \left( e^{\theta \, t}\right) . \end{aligned}$$

Taking into account that g is a PGF, it must satisfy \(g(1)=1\), implying that \(C=\exp \left( -e^{\theta }\right) \) and hence the desired result is obtained. \(\square \)

Let \(\phi (x;\theta )=(\phi (x; 0,\theta ),\phi (x; 1,\theta ), \ldots )\) and

$$\begin{aligned} f_r(x)=\sum _{u\ge 0}(u+r)\frac{x^u}{u!}=(x+r)e^x. \end{aligned}$$

We have the following lemmas.

Lemma 1

Let \(X_1, \ldots , X_n\) be independent and identically distributed from X, a random variable taking values in \(\mathbb {N}_{0}\) with probability mass function \(p(k)=\Pr (X=k)\), \(k\in \mathbb {N}_{0}\), so that \(E(X^2)<\infty \), then

$$\begin{aligned} E(\Vert \phi (X;\theta )\Vert _2^2) \le E(X^2) +\theta ^2 e^{\theta ^2} <\infty . \end{aligned}$$


By definition,

$$\begin{aligned} \Vert \phi (X;\theta )\Vert _2^2= & {} \sum _{k \ge 0}\phi (X; k,\theta )^2 \\= & {} \sum _{k \ge 0} (k+1)^2I(X=k+1)+ \sum _{k \ge 0} \sum _{u= 0}^{k}\frac{\theta ^{2u+2}}{(u!)^2}I(X=k-u), \end{aligned}$$

and thus

$$\begin{aligned} E(\Vert \phi (X;\theta )\Vert _2^2)= & {} E(X^2)+ \sum _{k \ge 0} \sum _{u= 0}^{k}\frac{\theta ^{2u+2}}{(u!)^2}p(k-u). \end{aligned}$$

Taking into account that

$$\begin{aligned} \sum _{k \ge 0} \sum _{u= 0}^{k}\frac{\theta ^{2u+2}}{(u!)^2}p(k-u)= \theta ^2\sum _{k \ge 0} p_k \sum _{u \ge 0}\frac{\theta ^{2u}}{(u!)^2} \le \theta ^2 \sum _{u \ge 0}\frac{\theta ^{2u}}{u!}=\theta ^2 e^{\theta ^2}, \end{aligned}$$

the result follows.\(\square \)

Lemma 2

Let \(X_1, \ldots , X_n\) be independent and identically distributed from X, a random variable taking values in \(\mathbb {N}_{0}\), then

$$\begin{aligned} \sum _{k \ge 0}\left[ \frac{\partial }{\partial \theta }\widehat{d}(k; \theta )\right] ^2 \le f_1^2(\theta ) <\infty , \quad \forall \, \theta >0. \end{aligned}$$


We have that

$$\begin{aligned} \sum _{k \ge 0}\left[ \frac{\partial }{\partial \theta }\widehat{d}(k; \theta )\right] ^2= \sum _{k \ge 0} \sum _{u,v=0}^k(v+1)(u+1)\frac{\theta ^v}{v!} \frac{\theta ^u}{u!}\widehat{p}(k-v) \widehat{p}(k-u). \end{aligned}$$

By interchanging the order of the sums, one gets

$$\begin{aligned} \sum _{k \ge 0}\left[ \frac{\partial }{\partial \theta }\widehat{d}(k; \theta )\right] ^2=\sum _{u,v \ge 0} (v+1)(u+1)\frac{\theta ^v}{v!}\frac{\theta ^u}{u!} \sum _{k\ge \max \{u,v\}}\widehat{p}(k-v) \widehat{p}(k-u). \end{aligned}$$

Taking into account that

$$\begin{aligned} \sum _{k\ge \max \{u,v\}}\widehat{p}(k-v) \widehat{p}(k-u) \le \sum _{k\ge 0}\widehat{p}(k)=1, \end{aligned}$$

the result follows.\(\square \)

Lemma 3

Let \(X_1, \ldots , X_n\) be independent and identically distributed from X, a random variable taking values in \(\mathbb {N}_{0}\). Assume that \(\widehat{\theta } {\mathop {\longrightarrow }\limits ^{a.s.(P)}} \theta \), for some \(\theta >0\). For each \(k\in \mathbb {N}_{0}\), let \(\theta _k=\alpha _{k}\theta +(1-\alpha _{k})\widehat{\theta }\), for some \(\alpha _{k} \in [0,1]\). Then,

$$\begin{aligned} \sum _{k \ge 0}\left[ \frac{\partial }{\partial \theta }\widehat{d}(k; \theta _k)\right] ^2 <\infty , \quad a.s.(P). \end{aligned}$$


Let \(\widetilde{\theta }=\max \{ \widehat{\theta }, \theta \}\). Proceeding as in the proof of Lemma 2, we get that

$$\begin{aligned} \sum _{k \ge 0}\left[ \frac{\partial }{\partial \theta }\widehat{d}(k; \theta _k)\right] ^2 \le f_1(\widetilde{\theta })^2. \end{aligned}$$

Since \(f_1(\widetilde{\theta })^2\) is a continuous function of \(\widehat{\theta }\), we have that

$$\begin{aligned} f_1(\widetilde{\theta })^2{\mathop {\longrightarrow }\limits ^{a.s.(P)}} f_1(\theta )^2 <\infty , \quad \forall \, \theta >0, \end{aligned}$$

and the result follows.\(\square \)

Lemma 4

Let \(X_1, \ldots , X_n\) be independent and identically distributed from X, a random variable taking values in \(\mathbb {N}_{0}\). Assume that \(\widehat{\theta } {\mathop {\longrightarrow }\limits ^{a.s.(P)}} \theta \), for some \(\theta >0\). Given the data, let \(X_1^*,\ldots , X_n^*\) be independent and identically distributed from \(X^* \sim \mathrm{Bell}(\widehat{\theta })\). Let \(\widehat{d}^*(k;\theta )\) be defined as \(\widehat{d}(k;\theta )\) with \(\widehat{p}(k)\) replaced by

$$\begin{aligned} \widehat{p}^*(k)=\frac{1}{n}\sum _{j=1}^nI(X^*_j=k), \quad k \ge 0. \end{aligned}$$


  1. (I)

    \(\displaystyle \sum _{k \ge 0} \left[ \frac{\partial }{\partial \theta }\widehat{d}^*(k;\widehat{\theta })-\mu (k; \widehat{\theta }) \right] ^2 {\mathop {\longrightarrow }\limits ^{P_*}} 0\),   a.s.(P),

  2. (II)

    \(\displaystyle \sum _{k \ge 0} \left[ \mu (k; {\theta })-\mu (k; \widehat{\theta }) \right] ^2 \rightarrow 0\),   a.s.(P).


(I) We have that

$$\begin{aligned}&\sum _{k \ge 0} \left[ \frac{\partial }{\partial \theta }\widehat{d}^*(k;\widehat{\theta })-\mu (k; \widehat{\theta }) \right] ^2\\&\quad = \sum _{k \ge 0}\left\{ \sum _{v=0}^k(v+1)\frac{\theta ^v}{v!} \left[ \widehat{p}^*(k-v)-p(k-v;\widehat{\theta })\right] \right\} ^2\\&\quad = \sum _{u,v \ge 0} (u+1)\frac{\theta ^u}{u!}(v+1)\frac{\theta ^v}{v!} \sum _{k\ge \max \{u,v\}}\left\{ \widehat{p}^*(k-v) - p(k-v;\widehat{\theta })\right\} \left\{ \widehat{p}^*(k-u)- p(k-u;\widehat{\theta })\right\} \\&\quad \le f_1(\widehat{\theta })^2 \sum _{k\ge 0}\left\{ \widehat{p}^*(k)- p(k;\widehat{\theta })\right\} ^2. \end{aligned}$$

Since \(f_1(\widehat{\theta })^2\) is a continuous function of \(\widehat{\theta }\), we have that

$$\begin{aligned} f_1(\widehat{\theta })^2{\mathop {\longrightarrow }\limits ^{a.s.(P)}} f_1(\theta )^2 <\infty , \quad \forall \, \theta >0. \end{aligned}$$

We also have that

$$\begin{aligned} E_*\left[ \sum _{k\ge 0}\left\{ \widehat{p}^*(k)- p(k;\widehat{\theta })\right\} ^2\right] = \frac{1}{n}\sum _{k\ge 0}p(k;\widehat{\theta })\left\{ 1- p(k;\widehat{\theta })\right\} \le \frac{1}{n} \rightarrow 0. \end{aligned}$$


$$\begin{aligned} \sum _{k\ge 0}\left\{ \widehat{p}^*(k)- p(k;\widehat{\theta })\right\} ^2 {\mathop {\longrightarrow }\limits ^{P_*}} 0, \end{aligned}$$

and the result in part (I) follows.

(II) We have that

$$\begin{aligned} \sum _{k \ge 0} \left[ \mu (k; {\theta })-\mu (k; \widehat{\theta }) \right] ^2=\varDelta _1+2\varDelta _2+\varDelta _3, \end{aligned}$$


$$\begin{aligned} \varDelta _1= & {} \sum _{k \ge 0} \sum _{u,v=0}^k (u+1)\frac{\widehat{\theta }^u}{u!}(v+1)\frac{\widehat{\theta }^v}{v!} \left\{ p(k-u; \widehat{\theta })- p(k-u; \theta )\right\} \left\{ p(k-v; \widehat{\theta })- p(k-v; \theta )\right\} ,\\ \varDelta _2= & {} \sum _{k \ge 0} \sum _{u,v=0}^k (u+1)\frac{\widehat{\theta }^u}{u!}\frac{v+1}{v!} \left\{ p(k-u; \widehat{\theta })- p(k-u; \theta )\right\} p(k-v; \theta ) \left\{ \widehat{\theta }^v-\theta ^v\right\} ,\\ \varDelta _3= & {} \sum _{k \ge 0} \sum _{u,v=0}^k \frac{u+1}{u!}\frac{v+1}{v!} p(k-u; \theta ) p(k-v; \theta ) \left\{ \widehat{\theta }^u-\theta ^u\right\} \left\{ \widehat{\theta }^v-\theta ^v\right\} . \end{aligned}$$

We first deal with \(\varDelta _1\). We have that

$$\begin{aligned} \varDelta _1 \le f_1(\widehat{\theta })^2 \sum _{k \ge 0} \left\{ p(k; \widehat{\theta })- p(k; \theta )\right\} ^2. \end{aligned}$$

To deal with the second term on the right-hand of (13), note that since

$$\begin{aligned} \frac{\partial }{\partial \theta } p(k; \theta )=\frac{p(k; \theta )}{\theta }\left\{ k-\theta e^\theta \right\} , \quad \forall \, k\ge 1, \end{aligned}$$

there exists \(k_1=k_1(\theta ) \ge 1\) such that \(\frac{\partial }{\partial \theta } p(k; \theta ) >0\), \(\forall \, k \ge k_1\). We are assuming that \(\widehat{\theta } {\mathop {\longrightarrow }\limits ^{a.s.(P)}} \theta \), for some \(\theta >0\), and therefore for any \(\varepsilon >0\) there exists \(n_0 \in \mathbb {N}\) (depending on \(\theta \), \(\varepsilon \) and the sequence \(X_1, X_2, \ldots \)) such that \(\widehat{\theta } \le \theta +\varepsilon \)a.s.(P) \(\forall \ n \ge n_0\). For any \(k_0 \ge k_1\), we have that

$$\begin{aligned} \sum _{k \ge 0} \left\{ p(k; \widehat{\theta })- p(k; \theta )\right\} ^2&= \sum _{k = 0}^{k_0} \left\{ p(k; \widehat{\theta })- p(k; \theta )\right\} ^2+ \sum _{k >k_0} \left\{ p(k; \widehat{\theta })- p(k; \theta )\right\} ^2\\&:= \delta _1(\widehat{\theta })+\delta _2(\widehat{\theta }). \end{aligned}$$

Since \(\delta _1(\widehat{\theta })\) is a continuous function of \(\widehat{\theta }\), it follows that

$$\begin{aligned} \delta _1(\widehat{\theta }){\mathop {\longrightarrow }\limits ^{a.s.(P)}} \delta _1(\theta )=0. \end{aligned}$$

As for \(\delta _2(\widehat{\theta })\), because \(k_0 \ge k_1\), we have that

$$\begin{aligned} \delta _2(\widehat{\theta })\le \sum _{k>k_0} p(k; \widehat{\theta })^2 + \sum _{k>k_0} p(k; {\theta })^2 \le 2\sum _{k >k_0} p(k;\theta +\varepsilon )^2, \quad a.s. \,(P), \quad \forall \, n \ge n_0, \end{aligned}$$

and the right-hand side of the above expression is as small as desired for large enough \(k_0\). Therefore, we have shown that

$$\begin{aligned} \sum _{k \ge 0} \left\{ p(k; \widehat{\theta })- p(k; \theta )\right\} ^2 {\mathop {\longrightarrow }\limits ^{a.s.(P)}} 0. \end{aligned}$$

This fact together with (12) shows that \(\varDelta _1 {\mathop {\longrightarrow }\limits ^{a.s.(P)}} 0\).

We have that

$$\begin{aligned} \varDelta _2 = \sum _{u \ge 0}(u+1)\frac{\widehat{\theta }^u}{u!} \sum _{v \ge 1}\frac{v+1}{v!}\left\{ \widehat{\theta }^v-\theta ^v \right\} M_2(u,v), \end{aligned}$$


$$\begin{aligned} M_2(u,v)=\sum _{k\ge \max \{u,v\}} \left\{ p(k-u; \widehat{\theta })- p(k-u; \theta )\right\} p(k-v; \theta ). \end{aligned}$$

Notice that \(0 \le |M_2(u,v)| \le 1\), \(\forall u,\, v \ge 0\). By applying the mean value theorem,

$$\begin{aligned} \widehat{\theta }^v-\theta ^v=v\widetilde{\theta }_v^{v-1}(\widehat{\theta }-\theta ), \quad \, \forall v \ge 1, \end{aligned}$$

where \(\widetilde{\theta }_v=\alpha _v\widehat{\theta }+(1-\alpha _v)\theta \), for some \(\alpha _v \in (0,1)\). As in the proof of Lemma 3, let \(\widetilde{\theta }=\max \{\theta , \widehat{\theta }\}\). Note that \(\widehat{\theta }_v \le \widetilde{\theta }\), \(\forall \, v \ge 1\). From the above considerations, we have that

$$\begin{aligned} |\varDelta _2| \le | \widehat{\theta }-\theta | \sum _{u \ge 0}(u+1)\frac{\widehat{\theta }^u}{u!} \sum _{v \ge 1}\frac{(v+1)v}{v!}\widetilde{\theta }^{v-1}= | \widehat{\theta }-\theta | f_1(\widehat{\theta })f_2(\widetilde{\theta }). \end{aligned}$$

Since the right-hand side of the above expression is a continuous function of \(\theta \), it follows that

$$\begin{aligned} | \widehat{\theta }-\theta | f_1(\widehat{\theta })f_2(\widetilde{\theta }) {\mathop {\longrightarrow }\limits ^{a.s.(P)}} | {\theta }-\theta |f_1({\theta })f_2({\theta })=0, \end{aligned}$$

and thus \(\varDelta _2 {\mathop {\longrightarrow }\limits ^{a.s.(P)}} 0\).


$$\begin{aligned} \varDelta _3 = \sum _{u,v \ge 0}\frac{u+1}{u!}\left\{ \widehat{\theta }^u-\theta ^u \right\} \frac{v+1}{v!}\left\{ \widehat{\theta }^v-\theta ^v \right\} M_3(u,v), \end{aligned}$$


$$\begin{aligned} 0 \le M_3(u,v)=\sum _{k\ge \max \{u,v\}} p(k-u; \theta ) p(k-v; \theta ) \le 1. \end{aligned}$$

By applying the mean value theorem (as done when studying \(\varDelta _2\)), we get

$$\begin{aligned} |\varDelta _3| \le (\widehat{\theta }-\theta )^2 f_2(\widetilde{\theta })^2. \end{aligned}$$

Since the right-hand side of the above expression is a continuous function of \(\theta \), it follows that

$$\begin{aligned} (\widehat{\theta }-\theta )^2 f_2(\widetilde{\theta })^2 {\mathop {\longrightarrow }\limits ^{a.s.(P)}} ({\theta }-\theta )^2 f_2({\theta })^2=0, \end{aligned}$$

and thus \(\varDelta _3 {\mathop {\longrightarrow }\limits ^{a.s.(P)}} 0\).\(\square \)

Lemma 5

Let \(X_1, \ldots , X_n\) be independent and identically distributed from X, a random variable taking values in \(\mathbb {N}_{0}\). Assume that \(\widehat{\theta } {\mathop {\longrightarrow }\limits ^{a.s.(P)}} \theta \), for some \(\theta >0\). For each \(k\in \mathbb {N}_{0}\), let \(\theta _k= \alpha _{k}\theta +(1-\alpha _{k})\widehat{\theta }\), for some \(\alpha _{k} \in [0,1]\). Then,

$$\begin{aligned} \sum _{k \ge 0}\frac{\partial }{\partial \theta }\left\{ \widehat{d}(k; \theta _k) -\widehat{d}(k; \theta )\right\} ^2 {\mathop {\longrightarrow }\limits ^{a.s.(P)}} 0. \end{aligned}$$


The proof is parallel to that of \(\varDelta _3 {\mathop {\longrightarrow }\limits ^{a.s.(P)}} 0\) in the proof of Lemma 4.\(\square \)

Proof of Theorem 1

By applying the mean value theorem, we get, for each \(k \in \mathbb {N}_{0}\), that

$$\begin{aligned} \widehat{d}(k;\widehat{\theta })=\widehat{d}(k;{\theta })+\frac{\partial }{\partial \theta } \widehat{d}(k;{\theta }_k)(\widehat{\theta }-{\theta }), \end{aligned}$$

with \(\theta _k=\alpha _{k}\theta +(1-\alpha _{k})\widehat{\theta }\), for some \(\alpha _{k} \in (0,1)\). From Lemma 1, \(E(\Vert \phi (X;\theta )\Vert _2^2)<\infty \) and thus by the SLLN in Hilbert spaces and the continuous mapping theorem, it follows that

$$\begin{aligned} \Vert \widehat{d}(k;{\theta })\Vert _2^2{\mathop {\longrightarrow }\limits ^{a.s.}}\Vert E\{\phi (X;\theta )\} \Vert _2^2=\eta <\infty . \end{aligned}$$

Finally, the result follows from (14), (15) and Lemma 3. \(\square \)

Proof of Theorem 2

From expansion (14),

$$\begin{aligned} \widehat{d}(k;\widehat{\theta })=\widehat{d}(k;{\theta })+\frac{\partial }{\partial \theta } \widehat{d}(k;{\theta })(\widehat{\theta }-{\theta })+\left\{ \frac{\partial }{\partial \theta } \widehat{d}(k;{\theta }_k)- \frac{\partial }{\partial \theta } \widehat{d}(k;{\theta })\right\} (\widehat{\theta }-{\theta }), \end{aligned}$$

with \(\theta _k=\alpha _{k}\theta +(1-\alpha _{k})\widehat{\theta }\), for some \(\alpha _{k} \in (0,1)\). Assumption 1 and Lemmas 2 and 4 imply that

$$\begin{aligned} \sqrt{n}\widehat{d}(\cdot ;\widehat{\theta }) = \sqrt{n}\widehat{d}(\cdot ;{\theta })+ \frac{\partial }{\partial \theta } \widehat{d}(\cdot ;{\theta })\sqrt{n}(\widehat{\theta }-{\theta })+r_1, \end{aligned}$$

with \(\Vert r_1\Vert _2=o_P(1)\). Now, by applying the SLLN in Hilbert spaces and Assumption 1, we get

$$\begin{aligned} \sqrt{n}\widehat{d}(\cdot ;{\theta })+\frac{\partial }{\partial \theta } \widehat{d}(\cdot ;{\theta })\sqrt{n}(\widehat{\theta }-{\theta })= \frac{1}{\sqrt{n}}\sum _{i=1}^n Y(X_i;\cdot , \theta )+r_2, \end{aligned}$$

with \(\Vert r_2\Vert _2=o_P(1)\). By the central limit theorem in Hilbert spaces,

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{i=1}^n Y(X_i;\cdot , \theta ) {\mathop {\longrightarrow }\limits ^{\mathcal {L}}} S(\theta ), \end{aligned}$$

where \(Y(X; \cdot , \theta ) =(Y(X;0, \theta ), Y(X;1, \theta ), \ldots )\). The result follows from (16)–(18) and the continuous mapping theorem. \(\square \)

Proof of Theorem 3

Proceeding as in the proof of Theorem 2, we have that

$$\begin{aligned} \sqrt{n}\widehat{d}^*(\cdot ;\widehat{\theta }^*) = \sqrt{n}\widehat{d}^*(\cdot ;{\theta })+ \frac{\partial }{\partial \theta } \widehat{d}^*(\cdot ;\widehat{\theta })\sqrt{n}(\widehat{\theta }^*-\widehat{\theta })+r^*_1, \end{aligned}$$

with \(\Vert r^*_1\Vert _2=o_{P_*}(1)\) a.s.(P). Let

$$\begin{aligned} Y_n^*=\frac{1}{\sqrt{n}}\sum _{i=1}^n Y(X_i^*;\cdot , \widehat{\theta }). \end{aligned}$$

By applying Lemma 4 and Assumption 2, we get

$$\begin{aligned} \sqrt{n}\widehat{d}^*(\cdot ;{\theta })+\frac{\partial }{\partial \theta } \widehat{d}^*(\cdot ;\widehat{\theta })\sqrt{n}(\widehat{\theta }^*-\widehat{\theta })= Y_n^*+r_2^*, \end{aligned}$$

with \(\Vert r_2^*\Vert _2=o_{P_*}(1)\)a.s.(P). To prove the result we derive the asymptotic distribution of \(Y_n^*\), showing that it coincides with the asymptotic distribution of \(S_{n}(\widehat{\theta } )\) when the data come from \(X\sim \mathrm{Bell}(\theta )\). With this aim, we apply Theorem 1.1 in Kundu et al. (2000). So, we will show that conditions (i)–(iii) in that theorem hold. For \(k \ge 0\), let \(e_k(j)= I(k=j)\). \(\{e_k\}_{k\ge 0}\) is an orthonormal basis of \(l^2\). We have that \(E_*\{\langle Y(X_i^*;\cdot , \widehat{\theta }),e_k \rangle _2\}= E_*\{Y(X^*;k, \widehat{\theta })\}=0\), \(\forall \, k \ge 0\), and by Lemma 1 and Assumption 2, \(E_*\{\Vert Y(X^*;\cdot , \widehat{\theta })\Vert _2^2\}<\infty \).

Let \(\mathcal {C}\) denote the operator defined in (11) and let \(\mathcal {C}_n\) be similarly defined by replacing \(\varrho (k,r)=Cov_{\theta }\{Y(X;k,\theta ),Y(X;r,\theta )\}\) with \(\varrho _n(k,r)=Cov_*\{\{Y(X^*; k,\widehat{\theta })Y(X^*; r,\widehat{\theta }) \}\), \(k\in \mathbb {N}_{0}\), \(r \in \mathbb {N}_{0}\). Assumption 2 and Lemma 4 imply that

$$\begin{aligned} \langle \mathcal {C}_n e_k,e_r\rangle _2= & {} E_*\{Y(X^*; k,\widehat{\theta })Y(X^*; r,\widehat{\theta }) \} \rightarrow E_{\theta }\{Y(X; k,{\theta })Y(X; r,{\theta }) \}\\= & {} \langle \mathcal {C} e_k,e_l\rangle _2\, \quad a.s.(P). \end{aligned}$$

Setting \(a_{k,r}=\langle \mathcal {C} e_k,e_r\rangle _2\) in the aforementioned Theorem 1.1, this proves that condition (i) holds. Similarly, condition (ii) holds since

$$\begin{aligned} \sum _{k\ge 0} \langle \mathcal {C}_n e_k,e_k\rangle _2 \rightarrow \sum _{k\ge 0}a_{kk}, \quad a.s.(P), \end{aligned}$$

and \(\sum _{k\ge 0}a_{kk}<\infty \). Finally, condition (iii) readily follows from Assumption 2. \(\square \)

Practical issues

Next, we describe some computational issues related to the calculation of the test statistics considered in the simulation study of Sect. 4. The test statistics \(S_{n}(\widehat{\theta })\), \(R_{n,w}(\widehat{\theta })\) and \(M_{n,w}(\widehat{\theta })\) are defined by means of infinite sums. However, these sums have to be truncated at some finite value, say M; that is,

$$\begin{aligned} S_n(\widehat{\theta })= & {} \sum _{k=0}^{M} \widehat{d}(k;\widehat{\theta })^2,\\ R_{n,w}(\widehat{\theta })= & {} \sum _{r,k\ge 0}^M \{p(r;\theta )-\widehat{p}(r)\} \{p(k;\theta )-\widehat{p}(k)\} \int _{0}^{1} t^{r+k} w(t) dt,\\ M_{n,w}(\widehat{\theta })= & {} \sum _{r,k\ge 0}^M\widehat{d}(r;\widehat{\theta }) \widehat{d}(k;\widehat{\theta }) \int _{0}^{1}t^{r+k} w(t) dt. \end{aligned}$$

From the numerical results, we have noted that taking \(M=20\) yields sufficiently precise values of these statistics. Finally, note that

$$\begin{aligned} \widehat{d}(k;\theta )=(k+1)\widehat{p}(k+1)-\sum _{u=0}^{k} {\textit{coef}}(u;\theta )\widehat{p}(k-u),\quad k\ge 0, \end{aligned}$$


$$\begin{aligned} \widehat{p}(k)=\frac{1}{n}\sum _{j=1}^{n}I(X_j=k),\quad k\ge 0, \end{aligned}$$

and, therefore,

$$\begin{aligned} {\textit{coef}}(u;\theta )=\frac{\theta ^{u+1}}{u!} \end{aligned}$$

can be recursively calculated as follows: \({\textit{coef}}(0;\theta )=\theta \), and \({\textit{coef}}(u;\theta )={\textit{coef}}(u-1;\theta )\theta /u\) for \(u\ge 1\).

Calculation of the bootstrap p-value

Let T denote any of the three test statistics and let \(T_{obs}\) stand for the observed value of such statistic. The bootstrap p-value, \(\hat{p}=P_*(T \ge T_{obs})\) cannot be exactly calculated. Nevertheless, it can be approximated as follows.

  1. 1.

    Calculate the observed value of the test statistics for the available dataset \(X_1,\ldots , X_n\), say \(S_{obs}(\widehat{\theta })\), \(M_{obs}(\widehat{\theta })\) and \(R_{obs}(\widehat{\theta })\).

  2. 2.

    Generate B bootstrap samples \(X_1^{*b},\ldots , X_n^{*b}\) from \(X^*\sim \mathrm{Bell}(\widehat{\theta })\), for \(b = 1,\ldots , B\).

  3. 3.

    Calculate the test statistics \(S_n(\widehat{\theta })\), \(M_{n,w}(\widehat{\theta })\) and \(R_{n,w}(\widehat{\theta })\) for each bootstrap sample and denote them, respectively, by \(S_b^*\), \(M_b^*\) and \(R_b^*\) for \(b = 1,\ldots , B\).

  4. 4.

    Compute the p-values of the tests based on the statistics \(S_n(\widehat{\theta })\), \(M_{n,w}(\widehat{\theta })\) and \(R_{n,w}(\widehat{\theta })\) by means, respectively, of the expressions

    $$\begin{aligned} \widehat{p}_S =\frac{\#\{S_b^*\ge S_{obs}(\widehat{\theta })\}}{B},\quad \widehat{p}_M =\frac{\#\{M_b^*\ge M_{obs}(\widehat{\theta })\}}{B},\quad \widehat{p}_R =\frac{\#\{R_b^*\ge R_{obs}(\widehat{\theta })\}}{B}. \end{aligned}$$

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Batsidis, A., Jiménez-Gamero, M.D. & Lemonte, A.J. On goodness-of-fit tests for the Bell distribution. Metrika 83, 297–319 (2020).

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  • Count data
  • Goodness-of-fit tests
  • Parametric bootstrap
  • Empirical probability generating function