Testing for the bivariate Poisson distribution

Abstract

This paper studies goodness-of-fit tests for the bivariate Poisson distribution. Specifically, we propose and study several Cramér–von Mises type tests based on the empirical probability generating function. They are consistent against fixed alternatives for adequate choices of the weight function involved in their definition. They are also able to detect local alternatives converging to the null at a certain rate. The bootstrap can be used to consistently estimate the null distribution of the test statistics. A simulation study investigates the goodness of the bootstrap approximation and compares their powers for finite sample sizes. Extensions for testing goodness-of-fit for the multivariate Poisson distribution are also discussed.

Appendix

1.1 Proofs

Here we prove the results in Sects. 2–5, 7. Since the mathematical machinery employed in the proofs is quite standard, we just sketch the proofs. A detailed derivation of the results can be obtained from the authors upon request.

In order to prove Proposition 1, we first give a preliminary lemma. For an arbitrary set $S, \partial S$ and $int S$ denote the sets of boundary points and interior points of $S$, respectively.

Lemma 1

Let $\{f_n\}_{n\ge 1}$ and $f$ be a collection of nondecreasing real functions defined on $R=[b_1,c_1]\times [b_2,c_2]\times \cdots \times [b_d,c_d] \subseteq \mathbb {R}^d$, with $-\infty <b_j\le c_j<\infty , j=1,2,\ldots ,d$. Let $D=D_1\cup D_2 \cup D_3$, where $D_1$ is the vertex set of $R, D_2$ is a dense set in $\partial R$ and $D_3$ is a dense set in $int R$. If $f$ is continuous on $R$ and

$$\begin{aligned} \left| f_{n}(x)-f(x)\right| \rightarrow 0, \quad \forall x \in D, \end{aligned}$$

(14)

then

$$\begin{aligned} \displaystyle \sup _{x\in R}|f_n(x)-f(x)| \rightarrow 0. \end{aligned}$$

Proof

Let $\varepsilon >0$ be arbitrary but fixed. Because $f$ is a uniformly continuous function on $R, \exists \delta =\delta (\varepsilon )>0$, such that $\forall x,y\in R$ with $\Vert x-y\Vert <\delta$, we have that $|f(x)-f(y)|<\varepsilon$. Let

$$\begin{aligned} H_i=[u_{1i},v_{1i}]\times [u_{2i},v_{2i}]\times \cdots \times [u_{di},v_{di}],\quad i=1,2,\ldots ,M, \end{aligned}$$

with $\Vert u_i-v_i\Vert < \delta$, where $u_i=(u_{1i},u_{2i},\ldots ,u_{di}) \in D$ and $v_i=(v_{1i},v_{2i},\ldots ,v_{di}) \in D, 1\le i \le M$, such that $R=\bigcup _{i=1}^M H_i.$ Let $\Delta _n=\max _{x\in V} \left| f_n(x)-f(x)\right|$, where $V=\{u_i,v_i: i=1,2,\ldots ,M\}$. From (14), it follows that $\Delta _n \rightarrow 0$. Now, for each $x\in R$, there exists $i\in \{1,2,\ldots ,M\}$, such that $x\in H_i$, and thus $u_i\le x\le v_i$. Since $f_n$ and $f$ are nondecreasing, we get

$$\begin{aligned} f_n(x)-f(x)\le f_n(v_i)-f(u_i)=f_n(v_i)-f(v_i)+f(v_i)-f(u_i)\le \Delta _n+\varepsilon . \end{aligned}$$

Analogously, $f_n(x)-f(x)\ge -\Delta _n-\varepsilon$. Since $\varepsilon >0$ is arbitrary, these two inequalities imply the result. $\square$

Proof of Proposition 1

Let $D$ be a countable dense set in $R$ according to Lemma 1. By the SLLN, there exists a set $A_D \in \mathcal {A}$ such that $P(A_D)=1$ and for all $\omega \in A_D$: $g_n(u,\omega )\ \mathop {\longrightarrow } \ g(u), \forall u \in D$. Since $X_i\ge 0, 1\le i \le d$, we have that $g_n(u,\omega )$ and $g(u)$ are nondecreasing functions. Now (2) follows from Lemma 1. The proof of (3) follows similar steps. $\square$

Proof of Proposition 2

Let $(X_1,X_2)$ be a random vector and let $g(u_1,u_2)=E\left( u_1^{X_1} u_2^{X_2}\right)$ be its pgf. Then, from the first equation in (6)

$$\begin{aligned} \frac{\partial }{\partial u_1}\log g(u_1,u_2)=\theta _1+\theta _3(u_2-1). \end{aligned}$$

(15)

Integrating (15) over $u_1$, we obtain

$$\begin{aligned} g(u_1,u_2)&= \exp \{\phi _1(u_2)+\theta _1 u_1+\theta _3(u_2-1) u_1\},\\&= \exp \{\phi (u_2)+\theta _1(u_1-1)+\theta _3(u_1-1)(u_2-1)\}, \end{aligned}$$

where $\phi (u_2)=\phi _1(u_2)+\theta _1+\theta _3(u_2-1)$. Proceeding similarly, from the second equation in (6) we get

$$\begin{aligned} g(u_1,u_2)= \exp \{\varphi (u_1)+\theta _2(u_2-1)+\theta _3(u_1-1)(u_2-1)\}. \end{aligned}$$

Thus, necessarily $\varphi (u_1)=\theta _1(u_1-1)$ and $\phi (u_2)=\theta _2(u_2-1)$, in other words, the BDP is the only solution of (6) $\square$

Proof of Theorem 1

By definition, $R_{n,w}(\hat{{\theta }}_n)= ||G_n(\hat{{\theta }}_n)||_{\mathcal {H}}^2$. Note that

$$\begin{aligned} G_n(u;\hat{\theta }_n)=\frac{1}{\sqrt{n}}\sum _{i=1}^n G(\varvec{X}_i,\hat{\theta }_n; {u}), \end{aligned}$$

with

$$\begin{aligned} G(\varvec{X},\theta ; {u})= u_1^{X_{1}}u_2^{X_{2}}-g(u;{\theta }). \end{aligned}$$

(16)

To prove (7), we apply a Taylor expansion of $G(\varvec{X}_i,\hat{\theta }_n; {u})$ around $\hat{\theta }_n=\theta$. Now following the lines in the proof of Theorem 3.1 in Meintanis and Swanepoel (2007) we get (7). Observe that

$$\begin{aligned} \Vert R_{n}\Vert _{\mathcal {H}}^{2}=\frac{1}{n} \sum _{i=1}^n\sum _{j=1}^n h(\varvec{X}_i,\varvec{X}_j;\theta ), \end{aligned}$$

where, $h(\varvec{x},\varvec{y};\theta )$ is defined in (10) and satisfies $h(\varvec{x},\varvec{y};\theta )= h(\varvec{y},\varvec{x};\theta ), E_{\theta }\left\{ h^2(\varvec{X}_1,\right. \left. \varvec{X}_2;\theta )\right\} < \infty , E_{\theta } \left\{ |h(\varvec{X}_1,\varvec{X}_1;\theta )|\right\} < \infty$ and $E_{\theta }\{h(\varvec{X}_1,\varvec{X}_2;\theta )\}=0$. Thus, from Theorem 6.4.1.B in Serfling (1980),

$$\begin{aligned} \Vert R_{n}\Vert _{\mathcal {H}}^{2} \mathop {\longrightarrow } \limits ^{L}\ \sum _{j\ge 1}\lambda _j\chi ^2_{1j}, \end{aligned}$$

where $\chi ^2_{11},\chi ^2_{12},\ldots$ and the set $\{\lambda _j\}$ are as defined in the statement of the Theorem. In particular, $\Vert R_{n}\Vert _{_\mathcal {H}}^{2}=O_P(1)$, which implies (8). $\square$

Proof of Theorem 2

By definition $R_{n,w}^*(\hat{\theta }_n^*)= \Vert G_{n}^*(\hat{\theta }_n^*)\Vert _{\mathcal {H}}^{2}$, with

$$\begin{aligned} G_{n}^*(u;\hat{\theta }_n^*)=\frac{1}{\sqrt{n}}\sum _{i=1}^n G(\varvec{X}_i^*;\hat{\theta }_n^*; u) \end{aligned}$$

and $G(\varvec{X};\theta ; u)$ defined in (16). Following similar steps to those given in the proof of Theorem 1 it can be seen that $R_{n,w}^*(\hat{\theta }_n^*)= \Vert R_{n}^*\Vert _{\mathcal {H}}^{2}+r^*_{n}$ with $r_{n}^*=o_{P_*}(1)$ a.s., where $R_{n}^*(u)$ is defined as $R_{n}(u)$ with $\varvec{X}_i$ and $\theta$ replaced by $\varvec{X}^*_i$ and $\hat{\theta }_n$, respectively. Now, the result follows by applying Theorem 1.1 in Kundu et al. (2000) to $R_{n}^*$ and the continuous mapping theorem. $\square$

Proof of Theorem 3

We only present the proof of part (b), since the proof of part (a) is quite similar. By definition,

$$\begin{aligned} \frac{1}{n}S_{n,w}(\hat{\theta }_n)=\Vert E_{1n}\Vert ^2_{_\mathcal {H}}+ \Vert E_{2n}\Vert ^2_{_\mathcal {H}}. \end{aligned}$$

We have

$$\begin{aligned} \left| E_{1n}(u;\hat{\theta }_n )-E_1(u;\theta )\right|&\le r_1+|\theta _1-\hat{\theta }_{1n}| + |\theta _3-\hat{\theta }_{3n}|+ |\hat{\theta }_{1n}+\hat{\theta }_{3n}|r_0, \\ \left| E_{2n}(u;\hat{\theta }_n )-E_2(u;\theta )\right|&\le r_2+|\theta _2-\hat{\theta }_{2n}| + |\theta _3-\hat{\theta }_{3n}|+ |\hat{\theta }_{2n}+\hat{\theta }_{3n}|r_0, \end{aligned}$$

$\forall u \in [0,1]^2$, where $r_0=\sup _{u\in [0,1]^2}\left| g_n(u)- g(u)\right|$ and $r_i=\sup _{u\in [0,1]^2}\left| \frac{\partial }{\partial u_i}g_n(u)-\right. \left. \frac{\partial }{\partial u_i}g(u)\right| , i=1,2$. Since $r_i=o(1), i=0,1,2, |\theta _i-\hat{\theta }_{in}| =o(1), i=1,2,3$, and (5), we conclude that $\frac{1}{n}S_{n,w}(\hat{\theta }_n)=\Vert E_{1}\Vert ^2_{\mathcal {H}}+ \Vert E_{2}\Vert ^2_{\mathcal {H}}+o(1)$, which shows the result because $\Vert E_{1}\Vert ^2_{\mathcal {H}}+ \Vert E_{2}\Vert ^2_{\mathcal {H}}=\sigma (g;\theta )$. $\square$

Proof of Theorem 4

Let $A_n \subseteq \mathbb {N}_0^2$ so that $P_{\theta }(A_n)\rightarrow 0$, then it can be easily seen that $P_n(A_n)\rightarrow 0$, and thus $P_n$ is contiguous to $P_{\theta }$. Next we show that, under $P_n, \sqrt{n}(\hat{\theta }_n-\theta )$ converges in distribution to a normal law. With this aim, let $\varvec{X}_1, \varvec{X}_2, \ldots , \varvec{X}_n$ be iid with common pmf $P_{\theta }$ and let

$$\begin{aligned} l_n=\log \frac{P_n(\varvec{X}_1)P_n(\varvec{X}_2)\ldots P_n(\varvec{X}_n)}{P_{\theta }(\varvec{X}_1)P_{\theta }(\varvec{X}_2)\ldots P_{\theta }(\varvec{X}_n)}. \end{aligned}$$

Let $Z_{n,i}=b_n(\varvec{X}_i), 1\le i \le n$. Then, by a Taylor expansion,

$$\begin{aligned} l_n=\sum _{i=1}^n \log \left( 1+\frac{1}{\sqrt{n}}Z_{n,i} \right) =\frac{1}{\sqrt{n}}\sum _{i=1}^n Z_{n,i}-\frac{1}{2n}\sum _{i=1}^n Z_{n,i}^2+\varrho _n, \end{aligned}$$

(17)

where $\varrho _n=\varrho _n(Z_{n,1},Z_{n,2}, \ldots , Z_{n,n})$,

$$\begin{aligned} \varrho _n=\sum _{i=1}^n \frac{2}{3!\,n\sqrt{n}\,a^3_{in}} Z_{n,i}^3, \end{aligned}$$

with $a_{in}=a_{in}(Z_{n,i})$

$$\begin{aligned} a_{in}=1+\frac{\alpha _i}{\sqrt{n}}\,Z_{n,i}, \end{aligned}$$

for some $0<\alpha _i<1, 1\le i \le n$. Next, we study the limit of each term in the right side of (17).

Let $\varepsilon$ be a positive constant. Let $M_n=\sqrt{n}/2$ and $\tilde{\varrho }_n=\varrho _n(Z_{n,1}I\{|Z_{n,1}|\le M_n\}, \ldots , Z_{n,n}I\{|Z_{n,n}|\le M_n\})$. Note that $\tilde{a}_{in}=a_{in}(Z_{n,i}I\{|Z_{n,i}|\le M_n\})\ge 0.5, 1\le i \le n$, and therefore

$$\begin{aligned} E_{\theta }(|\tilde{\varrho }_n|)\le \frac{2^3}{3}\frac{1}{\sqrt{n}} \displaystyle \sup _m E_{\theta } \left[ |b_m(\varvec{X}_1)|^3 \right] , \end{aligned}$$

which implies

$$\begin{aligned} P_{\theta }(|\tilde{\varrho }_{n}|>\varepsilon )\rightarrow 0. \end{aligned}$$

(18)

We also have

$$\begin{aligned} P_{\theta }\left( |\tilde{\varrho }_{n} -\varrho _{n}|>\varepsilon \right)&\le P_{\theta }\left( \,\bigcup _{i=1}^n\left\{ |Z_{n,i}|> M_n \right\} \right) \nonumber \\&\le \sum _{i=1}^n P_{\theta }\left( |Z_{n,i}|> M_n \right) \le \frac{n}{M_n^4}\sup _m E_{\theta } \left[ b_m(\varvec{X}_1)^4 \right] \rightarrow 0. \end{aligned}$$

(19)

From (18) and (19), we conclude

$$\begin{aligned} P_{\theta }\left( |\varrho _{n}|>\varepsilon \right) \rightarrow 0. \end{aligned}$$

(20)

From Assumption 3, the central limit theorem for triangular arrays implies

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{i=1}^n Z_{n,i}\ \mathop {\longrightarrow } \limits ^{L}\ N(0,\sigma ^2), \end{aligned}$$

(21)

with $\sigma ^2=E_{\theta }\left\{ b(X_1,X_2)^2\right\} <\infty$.

From Assumption 3 we also have $E_{\theta }\left( \frac{1}{n}\sum _{i=1}^n Z_{n,i}^2\right) \rightarrow \sigma ^2$ and $Var_{\theta }\left( \frac{1}{n}\sum _{i=1}^n\right. \left. Z_{n,i}^2\right) = \frac{1}{n}Var_{\theta }(Z_{n,i}^2)\le \frac{1}{n}\sup _n E_{\theta }\left\{ b_n(X_1,X_2)^4\right\} \rightarrow 0$, which imply

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n Z_{n,i}^2 \mathop {\longrightarrow } \limits ^{P}\ \sigma ^2. \end{aligned}$$

(22)

Now, from Assumption 1, (17) and (20)–(22), the sequence $(\sqrt{n}(\hat{\theta }_n-\theta ), l_n)$ converges in law to a 4-dimensional normal variate when the data comes from $P_{\theta }$. Thus, from Le Cam’s Third Lemma (Corollary 12.3.2 in Lehmann and Romano 2005), we have that, when the data come from $P_n, \sqrt{n}(\hat{\theta }_n-\theta )$ converges in distribution to a normal law, which implies that it is bounded in probability. Now, following similar steps as those given in the proof of Theorem 1, it can be shown that (7) also holds when the data come from $P_n$, with $P_n(|r_n|>\varepsilon )\rightarrow 0, \forall \varepsilon >0$. Finally, applying Theorem 2.3 in Gregory (1977), we obtain that when the data have pmf $P_n$

$$\begin{aligned} \Vert R_{n}\Vert _{\mathcal {H}}^2 \mathop {\longrightarrow } \limits ^{L} \ \sum _{k=1}^{\infty } \lambda _j\left( Z_j+c_j\right) ^2, \end{aligned}$$

and thus the result. $\square$

Proof of Proposition 3

The proof follows is quite similar to that of Proposition 2, so we omit it. $\square$

1.2 Expressions for statistics $\varvec{T}, \varvec{I_B}$ and $\varvec{NI_B}$ and critical regions

We have

$$\begin{aligned} T&= \frac{n}{2}\frac{(\hat{\theta }_2)^{2} \left( S^2_{X_1}\!-\!\hat{\theta }_1\right) ^2\!-\! 2S^2_{X_1X_2}\left( S^2_{X_1}\!-\!\hat{\theta }_1\right) \left( S^2_{X_2}\!-\!\hat{\theta }_2\right) \!+\! (\hat{\theta }_1)^{2} \left( S^2_{X_2}\!-\!\hat{\theta }_2\right) ^2}{(\hat{\theta }_1)^{2} (\hat{\theta }_2)^{2}\!-\!S^4_{X_1X_2}}, \\&T \ge \chi ^2_{2,\alpha },\\ I_B&= n \frac{\hat{\theta }_2S^2_{X_1}-2S^2_{X_1X_2}+ \hat{\theta }_1S^2_{X_2}}{\hat{\theta }_1\hat{\theta }_2- S^2_{X_1X_2}},\quad I_B \ge \chi ^2_{2n-3,\alpha },\\ NI_B&= \frac{n}{1-r^2}\left( \frac{S^2_{X_1}}{\overline{X}_1} - 2r^2\sqrt{\frac{S^2_{X_1}S^2_{X_2}}{\overline{X}_1\overline{X}_2}}+ \frac{S^2_{X_2}}{\overline{X}_2}\right) , \,\, NI_B \ge \chi ^2_{2n-3,\alpha }, \end{aligned}$$

where $S^2_{X_i}$ is the sample variance, $i=1,2, S^2_{X_1X_2}$ is the sample covariance, $r$ is the sample correlation coefficient and $\chi ^2_{k, \alpha }$, for $0<\alpha <1$ and $k\in \mathbb {N}$, denotes the upper $\alpha$-percentile of the $\chi ^2$ distribution with $k$ degrees of freedom.

1.3 Computation of the test statistic $R_{n,w}(\hat{\theta }_n)$

For $w$ as in (12), for $a_1, a_2\in \mathbb {N}_0$, we get

$$\begin{aligned}&R_{n,w}(\hat{\theta }_n)=\frac{1}{n}\sum _{i=1}^n \sum _{j=1}^n\int \limits _0^1\int \limits _0^1 u_1^{X_{1i} +X_{1j}+a_1}u_2^{X_{2i}+X_{2j}+a_2}\ du\nonumber \\&\quad +\, ne^{2\left( \hat{\theta }_{3n}-\hat{\theta }_{1n}-\hat{\theta }_{2n}\right) } \int \limits _0^1 u_1^{a_1}e^{2\left[ \hat{\theta }_{1n} -\hat{\theta }_{3n}\right] u_1}\int \limits _0^1 u_2^{a_2}e^{2\left[ \hat{\theta }_{2n}-\hat{\theta }_{3n}+\hat{\theta }_{3n}u_1\right] u_2} du\nonumber \\&\quad -\, 2e^{\hat{\theta }_{3n}-\hat{\theta }_{1n}- \hat{\theta }_{2n}}\sum _{i=1}^n\int \limits _0^1 u_1^{X_{1i}+a_1}e^{\left[ \hat{\theta }_{1n} -\hat{\theta }_{3n}\right] u_1}\int \limits _0^1 u_2^{X_{2i}+a_2} e^{\left[ \hat{\theta }_{2n}-\hat{\theta }_{3n}+\hat{\theta }_{3n}u_1 \right] u_2}du.\qquad \end{aligned}$$

(23)

To compute $R_{n,w}(\hat{\theta }_n)$ we can calculate the integrals on the right hand side of (23). Nevertheless, for computational reasons, the following representation (which is obtained by interchanging the order of integration)

$$\begin{aligned} R_{n,w}(\hat{\theta }_n) = n \sum _{i,j,k,l=0}^{\infty } \frac{\left( p_n(i,j) -p(i,j;\hat{\theta }_n)\right) \left( p_n(k,l)- p(k,l;\hat{\theta }_n)\right) }{(i+k+a_1+1)(j+l+a_2+1)}\qquad \end{aligned}$$

(24)

turned out to be more appropriate, where $p(i, j;\theta )=P_{\theta }(X_1=i, X_2=j)$ and $p_n(i,j)$ is the relative frequency of the pair $(i,j)$. Besides, (24) is less restrictive, since it allows $a_1>-1$, and $a_2>-1$. A truncation of the four infinite series at $M+15$ yielded sufficiently precise values of the statistic and a good performance of the corresponding subroutine, where $\displaystyle M\!=\!\max \{ X_{1(n)},X_{2(n)}\}, X_{k(n)}\!=\!\max \nolimits _{1\le i\le n}X_{ki}, k=1,2$.

1.4 Computation of the test statistic $S_{n,w}(\hat{\theta }_n)$

When the weight function is chosen as in (12), for some $a=(a_1, a_2)$ with $a_i>-1, i=1,2$, we have

$$\begin{aligned} S_{n,w}(\hat{\theta }_n) =\frac{1}{n}\sum _{i=1}^n\sum _{j=1}^n (S_{1ij}+S_{2ij}), \end{aligned}$$

with

$$\begin{aligned} S_{kij}&= \frac{X_{ki}I_{B_{ki}}X_{1j}I_{B_{kj}}}{(X_{1i}+X_{1j}+a_1-1) (X_{2i}+X_{2j}+a_2+1)}\\&- \frac{(\hat{\theta }_{kn}- \hat{\theta }_{3n})\left( X_{ki}I_{B_{ki}}+ X_{kj}I_{B_{kj}}\right) }{(X_{1i}+X_{1j}+a_1)(X_{2i}+X_{2j}+a_2+1)} \\&-\frac{\hat{\theta }_{3n}(X_{ki} I_{B_{ki}}+X_{kj}I_{B_{kj}})}{(X_{1i}+X_{1j}+a_1)(X_{2i}+ X_{2j}+a_2+2)}\\&+\frac{2\hat{\theta }_{3n}(\hat{\theta }_{kn}- \hat{\theta }_{3n})}{(X_{1i}+X_{1j}+a_1+1)(X_{2i}+ X_{2j}+a_2+2)}\\&+ \frac{(\hat{\theta }_{kn}- \hat{\theta }_{3n})^2}{(X_{1i}+X_{1j}+a_1+1)(X_{2i}+X_{2j}+a_2+1)}\\&+\frac{\hat{\theta }_{3n}^{\ 2}}{(X_{1i}+X_{1j}+a_1+1)(X_{2i}+X_{2j}+a_2+3)}, \end{aligned}$$

$k=1,2, 1\le i,j \le n$, where $B_{rs}=\{X_{rs}\ge 1\}, r=1,2, 1\le s \le n$.