A Negative-Binomial Index Considering Dispersion and Zero Probability

Abstract

We develop the so-called NB index,  which can distinguish the negative binomial from other distributions for overdispersion and zero inflation like Consuls’s generalized Poisson distribution. Although we concentrate on the i.i.d. case, it is also pointed out how to extend the results to time series models. After deriving the asymptotic distribution of the NB index, we investigate the finite-sample performance with simulations. We conclude with a real-data application.

Appendices

7 Proof of Theorem 1

We state here the proof of Theorem 1.We use the same notation and the method for deriving the asymptotics as in the paper [6]. We start with a central limit theorem (CLT) for $i.\,i.\,d.$ random variables $Y_{1,r},\ldots ,Y_{T,r}$ with $r=0,1,2$, which are defined by

$$\begin{aligned} \varvec{Y}_t\ :=\ \left( \begin{array}{c} Y_{t,0}\\ Y_{t,1}\\ Y_{t,2}\\ \end{array}\right) \ :=\ \left( \begin{array}{c} \mathbbm {1}_{\{X_t=0\}} - p_0\\ X_t - \mu \\ X_t^2 - \sigma ^2-\mu ^2\\ \end{array}\right) \end{aligned}$$

(6)

with mean $\mu$, variance $\sigma ^2$ and zero probability $p_0$ . Since $X_1,\ldots , X_T$ denotes an $i.\,i.\,d.$ sample of random variables with the NB distribution, it holds that

$$ \begin{array}{l} \frac{1}{\sqrt{T}} \sum _{t=1}^T \varvec{Y}_t {\mathop {\longrightarrow }\limits ^{\mathcal {D}}} {N }\left( \mathbf {0}, \varvec{\Sigma }\right) \qquad \text {with } \varvec{\Sigma }= \left( \sigma _{ij} \right) \text { given by } \sigma _{ij}\ =\ E\big [ Y_{0,i} Y_{0,j} \big ]. \end{array} $$

We can derive the asymptotic distribution of the statistic $\hat{I}_{NB }$ from (3) with the help of the delta method in the next step after we calculated the matrix $\varvec{\Sigma }$. So we need to calculate the entries $\sigma _{ij}$ with $0\le i\le j\le 2$ from the covariance matrix $\varvec{\Sigma }$.

$$\begin{aligned}&\sigma _{0,0} =\textstyle E\left[ (\mathbbm {1}_{\{X_t=0\}}-p_0)^2\right] = E[\mathbbm {1}_{\{X_{t}=0\}}]-p_0^2=p_0(1-p_0)=p^n(1-p^n),\\&\sigma _{1,1} =\textstyle E[(X_t-\mu )^2]=\frac{n(1-p)}{p^2},\\&\sigma _{2,2}=E\big [\big (X_t^2-\sigma ^2-\mu ^2\big )^2\big ]=E[X_t^4]-(\sigma ^2+\mu ^2)^2\\&=\frac{n (1-p) \{n^3 (1-p)^3+6 n^2 (1-p)^2+n \left( 4 p^2-15 p+11\right) +p^2-6 p+6\}}{p^4}\\&-\left[ \frac{n (1-p) \{n (1-p)+1\}}{p^2}\right] ^2\\&=\frac{n (1-p) \{4 n^2 (1-p)^2+2 n (5-2p)(1-p)+p^2-6 p+6\}}{p^4}, \end{aligned}$$

where we used the moment generating function.

$$\begin{aligned}&\sigma _{1,2}= E\big [(X_t-\mu )\big (X_t^2-\sigma ^2-\mu ^2\big )\big ]=E[X_t^3]-(\sigma ^2+\mu ^2)\mu \\&=\frac{n (1-p) \{n^2 (1-p)^2+3 n (1-p)-p+2\}}{p^3}-\frac{n (1-p) \{n (1-p)+1\}}{p^2}\cdot \frac{n(1-p)}{p}\\&=\frac{n (1-p) \{2 n (1-p)-p+2\}}{p^3}, \end{aligned}$$

where we used the moment generating function.

$$\begin{aligned} \sigma _{0,1}= & {} E\left[ (\mathbbm {1}_{\{X_t=0\}}-p_0)(X_t-\mu )\right] =\underbrace{E[\mathbbm {1}_{\{X_t=0\}}X_t]}_{=0}-p_0\mu =-\frac{p^nn(1-p)}{p}. \end{aligned}$$

Analogously,

$$\begin{aligned}&\sigma _{0,2}=E\left[ (\mathbbm {1}_{\{X_t=0\}}-p_0)(X_t^2-\sigma ^2-\mu ^2)\right] \\&=E[\mathbbm {1}_{\{X_t=0\}}X_t^2]-p_0(\sigma ^2+\mu ^2) =-\frac{p^n \cdot n (1-p) \{n (1-p)+1\}}{p^2}. \end{aligned}$$

After having computed the required $\sigma _{ij}$, we compute the asymptotic distribution of $\hat{I}_{NB }$. For this purpose, let us define the function $g: \mathbb {R}^3\rightarrow \mathbb {R}$ by

$$\begin{aligned} g(y_0,y_1,y_2)\ :=\ \frac{\left( \frac{y_1}{y_2-y_1^2}\right) ^\frac{y_1^2}{y_2-y_1^2-y_1}}{y_0},\qquad \end{aligned}$$

(7)

since $g\big (p_0,\mu ,\sigma ^2+\mu ^2\big )=1$. First, we need to compute the gradient $\nabla \varvec{g} (y_0,y_1,y_2)$,

Next, we evaluate the gradient in $(p^n,\ \frac{n(1-p)}{p}, \frac{n (1-p) \{-n (1-p)-1\}}{p^2})$ and obtain as

$$ \mathbf D =\left( \begin{array}{ccc} -p^{-n},\,\frac{p [(1-p) \{2 n (1-p)+1\}+\{2 n (1-p)+2-p\} \log (p)]}{(1-p)^2},\,\frac{p^2 (p-\log (p)-1)}{(1-p)^2} \end{array} \right) ^\top . $$

The application of the Delta method implies that, asymptotically, $\hat{I}_{NB }$ is normally distributed with variance given as

$$ p^{-n}+\frac{2 n (n+1) \log (p) \left\{ (1-p)(3-p)+\log (p)\right\} }{(p-1)^2}+n \{2 n (2-p)+3-p\}-1. $$

This concludes the proof of Theorem 1.

Remark 1

Note that we can extend our results to count time series models like INAR(1), INARCH(1) or INMA(1), see [7] for further description of these models. In these cases, in analogy to [6], we can apply the CLT by [2] to $\frac{1}{\sqrt{T}} \sum _{t=1}^T \varvec{Y}_t$, provided that moments of order $>4$ exist, i.e.,

$$ \begin{array}{l} \frac{1}{\sqrt{T}} \sum _{t=1}^T \varvec{Y}_t {\mathop {\longrightarrow }\limits ^{\mathcal {D}}} {N }\left( \mathbf {0}, \varvec{\Sigma }\right) \qquad \text {with } \varvec{\Sigma }= \left( \sigma _{ij} \right) \text { given by}\\[3mm] \sigma _{ij}\ =\ E\big [ Y_{0,i} Y_{0,j} \big ] + \sum _{r=1}^\infty \Big (E\big [Y_{0,i} Y_{r,j}\big ] + E\big [ Y_{r,i} Y_{0,j}\big ] \Big ) \end{array} $$

and proceed analogously as we did in Sect. 7.

8 Proof for Theorem 2 (GPD)

We use the same procedure as in Sect. 7.

$X_1,\ldots , X_T$ denotes an $i.\,i.\,d.$ sample of random variables with GPD. We can derive the asymptotic distribution of the statistic $\hat{I}_{NB }$ with the help of the delta method in the next step after we calculated the matrix $\varvec{\Sigma }$.

$$\begin{aligned}&\sigma _{0,0} = \textstyle E\left[ (\mathbbm {1}_{\{X_t=0\}}-p_0)^2\right] =E[\mathbbm {1}_{\{X_{t}=0\}}]-p_0^2=p_0(1-p_0)=e^{-\theta }(1-e^{-\theta }),\\&\sigma _{1,1} = E[(X_t-\mu )^2]=\frac{\theta }{(1-\lambda )^3},\\&\sigma _{2,2}=E\big [\big (X_t^2-\sigma ^2-\mu ^2\big )^2\big ]=E[X_t^4]-(\sigma ^2+\mu ^2)^2\\&=\frac{\theta ^4}{(1-\lambda )^4}+\frac{6 \theta ^3}{(1-\lambda )^5}+\frac{\theta \left\{ \theta \left( -8 \lambda ^2+\lambda +7\right) +6 \lambda ^2+8 \lambda +1\right\} }{(1-\lambda )^7}\\&-\frac{\theta ^2 (\theta \lambda -\theta -1)^2}{(1-\lambda )^6}=\frac{\theta \left\{ 4 \theta ^2 (\lambda -1)^2+\theta \left( -8 \lambda ^2+2 \lambda +6\right) +6 \lambda ^2+8 \lambda +1\right\} }{(1-\lambda )^7}, \end{aligned}$$

where we used the moment generating function.

$$\begin{aligned}&\sigma _{1,2}= E\big [(X_t-\mu )\big (X_t^2-\sigma ^2-\mu ^2\big )\big ]=E[X_t^3]-(\sigma ^2+\mu ^2)\mu \\&=\frac{\theta \left\{ \theta ^2 (\lambda -1)^2-3 \theta (\lambda -1)+2 \lambda +1\right\} }{(1-\lambda )^5}-\frac{\theta ^2 (1+\theta -\theta \lambda )}{(1-\lambda )^4}\\&=\frac{\theta \{1+2 \theta (1-\lambda )+2 \lambda \}}{(1-\lambda )^5}, \end{aligned}$$

where we used the moment generating function.

$$\begin{aligned} \sigma _{0,1}= & {} E\left[ (\mathbbm {1}_{\{X_t=0\}}-p_0)(X_t-\mu )\right] = \underbrace{E[\mathbbm {1}_{\{X_t=0\}}X_t]}_{=0}-p_0\mu =-\frac{e^{-\theta }\theta }{1-\lambda }. \end{aligned}$$

Analogously,

$$\begin{aligned}&\sigma _{0,2}=E\left[ (\mathbbm {1}_{\{X_t=0\}}-p_0)(X_t^2-\sigma ^2-\mu ^2)\right] \\&=E[\mathbbm {1}_{\{X_t=0\}}X_t^2]-p_0(\sigma ^2+\mu ^2) =\frac{e^{-\theta } \theta (\theta \lambda -\theta -1)}{(1-\lambda )^3}. \end{aligned}$$

After having computed the required $\sigma _{ij}$, we compute the asymptotic distribution of $\hat{I}_{NB }$ under the alternative hypothesis. We use the same function g as in Eq. (7), so the gradient remains the same. Next, we obtain as

$$ \varvec{D}=\left( \begin{array}{ccc} -e^{2 \theta } (1-\lambda )^{\frac{2\theta (1-\lambda )}{\lambda (2-\lambda )}}\\ \frac{e^{\theta } (1-\lambda )^{\frac{2\theta (1-\lambda )}{\lambda (2-\lambda )}+2} \big [ \{2 \theta (1-\lambda )-\lambda ^2+2 \lambda +1\} \log \left( (1-\lambda )^2\right) +(\lambda -2) \lambda \{-2 \theta (1-\lambda )-1\}\big ]}{(\lambda -2)^2 \lambda ^2}\\ \frac{e^{\theta } (1-\lambda )^4 (1-\lambda )^{\frac{2\theta (1-\lambda )}{\lambda (2-\lambda )}} \{(\lambda -2) \lambda -\log \left( (1-\lambda )^2\right) \}}{(\lambda -2)^2 \lambda ^2} \end{array} \right) . $$

The application of the Delta method implies that, asymptotically, $\hat{I}_{NB }$ under the alternative hypothesis is normally distributed with variance given as

This concludes the proof of Theorem 2.