On a General Class of Discrete Bivariate Distributions

Abstract

In this paper we develop a general class of bivariate discrete distributions. The basic idea is quite simple. The marginals are obtained by taking the random geometric sum of the baseline random variables. The proposed class of distributions is a flexible class of bivariate discrete distributions in the sense the marginals can take variety of shapes. The probability mass functions of the marginals can be heavy tailed, unimodal as well as multimodal. It can be both over dispersed as well as under dispersed. We discuss different properties of the proposed class of bivariate distributions. The proposed distribution has some interesting physical interpretations also. Further, we consider two specific base line distributions: Poisson and negative binomial distributions for illustrative purposes. Both of them are infinitely divisible. The maximum likelihood estimators of the unknown parameters cannot be obtained in closed form. They can be obtained by solving three and five dimensional non-linear optimizations problems, respectively. To avoid that we propose to use expectation maximization algorithm, and it is observed that the proposed algorithm can be implemented quite easily in practice. We have performed some simulation experiments to see how the proposed EM algorithm performs, and it works quite well in both the cases. The analysis of one real data set has been performed to show the effectiveness of the proposed class of models. Finally, we discuss some open problems and conclude the paper.

Appendices

Appendix A: Exact Expressions of C(a, j)

Note that for

$$ C(a,0) = \sum\limits_{k = 1}^{\infty} e^{-a k} = \frac{e^{-a}}{1 - e^{-a}} = \frac{1}{e^{a}-1}. $$

To compute C(a,1), first observe that

$$ C(a,1) = \sum\limits_{k = 1}^{\infty} e^{-a k} k = \frac{1}{e^{a}} + \frac{2}{e^{2a}} + \frac{3}{e^{3a}} + \cdots $$

and

$$ e^{a} C(a,1) = 1 + \frac{2}{e^{a}} + \frac{3}{e^{2a}} + \frac{4}{e^{3a}} + \cdots. $$

Hence

$$ (e^{a} - 1) C(a,1) = 1 + \frac{1}{e^{a}} + \frac{1}{e^{2a}} + \frac{1}{e^{3a}} + {\cdots} = \frac{e^{a}}{e^{a} - 1}. $$

Therefore

$$ C(a,1) = \frac{e^{a}}{(e^{a} - 1)^{2}}. $$

Now to compute C(a,2), note that

$$ C(a,2) = \sum\limits_{k = 0}^{\infty} e^{-a k} k^{2} = \sum\limits_{k = 0}^{\infty} e^{-a k} k(k-1) + \sum\limits_{k = 0}^{\infty} e^{-a k} k. $$

If we denote $\displaystyle S = \sum \limits _{k = 0}^{\infty } e^{-a k} k(k-1)$, then

$$ S = \frac{2 \cdot 1}{e^{2a}} + \frac{3 \cdot 2}{e^{3a}} + \frac{4 \cdot 3}{e^{4a}} + \cdots $$

and

$$ e^{a} S = \frac{2 \cdot 1}{e^{a}} + \frac{3 \cdot 2}{e^{2a}} + \frac{4 \cdot 3}{e^{3a}} + \cdots. $$

Hence

$$ S(e^{a}-1) = \frac{2 \cdot 1}{e^{a}} + \frac{2 \cdot 2}{e^{2a}} + \frac{2 \cdot 3}{e^{3a}} + {\cdots} = 2 \sum\limits_{k = 1}^{\infty} k e^{-ak} = \frac{2 e^{a}}{(e^{a}-1)^{2}}. $$

Therefore,

$$ C(a,2) = \frac{2 e^{a}}{(e^{a}-1)^{3}} + \frac{e^{a}}{(e^{a}-1)^{2}} = \frac{e^{2a} + e^{a}}{(e^{a}-1)^{3}}. $$

We will use the following notations:

$$ \begin{array}{@{}rcl@{}} S_{0}(a) &=& \sum\limits_{k = 0}^{\infty} k e^{-k a}, \ \ S_{1}(a) = \sum\limits_{k = 0}^{\infty} k(k-1) e^{-k a}, \ldots,\\ S_{m}(a) &=& \sum\limits_{k = 0}^{\infty} k(k-1) {\cdots} (k-m) e^{-k a}. \end{array} $$

Then using the fact

$$ e^{a} S_{m}(a) = \sum\limits_{k = 0}^{\infty} k(k-1) {\cdots} (k-m) e^{-(k-1) a}. $$

We can easily obtain the following relation

$$ S_{m}(a)(e^{a}-1) = (m + 1) S_{m-1}(a). $$

Further note that if we denote

$$ \begin{array}{@{}rcl@{}} k^{m} &=& C_{0m} k(k-1) {\cdots} (k-m + 1) + C_{1m} k(k-1) {\cdots} (k-m + 1) \\&&+ {\cdots} + C_{m-2,m} k (k-1) + C_{m-1,m} k, \end{array} $$

then C_0m, C_1m⋯ ,C_m− 1,m can be obtained recursively from the following set of linear equations. C_0m = 1 and

$$ \begin{array}{@{}rcl@{}} -C_{0m} \sum\limits_{1 \le i_1 \le m-1} i_1 + C_{1m} \!\!\!& = &\!\!\! 0 \\ C_{0m} \sum\limits_{1 \le i_1 < i_2 \le m-1} i_1 i_2 - C_{1m} \sum\limits_{1 \le i_1 \le m-2} i_1 + C_{2m} \!\!\!& = &\!\!\! 0 \\ - C_{0m} \sum\limits_{1 \le i_1 < i_2 < i_3 \le m-1} i_1 i_2 i_3 + C_{1m} \sum\limits_{1 \le i_1 < i_2 \le m-2} i_1 i_2 - C_{2m} \sum\limits_{1 \le i_1 \le m-3} i_1 + C_{3m} \!\!\!& = &\!\!\! 0 \\ \!\!\!& {\vdots} &\!\!\! \\ (-1)^{m-1} C_{0m} \prod\limits_{i = 1}^{m-1} i (-1)^{m-2} C_{1m} \prod\limits_{i = 1}^{m-2} i (-1)^{m-3} C_{2m} \prod\limits_{i = 1}^{m-3} i + {\cdots} - C_{m-2,m} + C_{m-1,m} \!\!\!& = &\!\!\! 0. \end{array} $$

If we use the following notations for n < m;

$$ a_{nm} = \sum\limits_{1 \le i_{1} < i_{2} < {\ldots} < i_{n} \le m} i_{1} i_{2} {\cdots} i_{n}, $$

$\displaystyle a_{mm} = \prod \limits _{i = 1}^{m} i$, then clearly

$$ a_{n,m + 1} = a_{n,m} + (m + 1)a_{n-1,m}, $$

and we obtain

$$ \begin{array}{@{}rcl@{}} C_{1m} & = & a_{1,m-1} \\ C_{2m} & = & C_{1m} a_{1,m-2} - a_{2,m-1} \\ C_{3m} & = & C_{2m} a_{1,m-3} - C_{1m} a_{2,m-2} + a_{3,m-1} \\ {\vdots} & = & {\vdots} \\ C_{m-1,m} & = & C_{m-2,m} a_{11} - C_{m-3,m} a_{2,2} + {\ldots} (-1)^{m-2} a_{m-1,m-1}. \end{array} $$

Since we have

$$ C(a,m) = C_{0m} S_{m-1}(a) + C_{1m}S_{m-2}(a) + {\ldots} + C_{m-1,m} S_{0}(a), $$

we can obtain recursively C(a, m + 1) from C(a, m).

Appendix B: Expressions of P(N = n|X = x, Y = y)

In this appendix we provide the expressions of P(N = n|X = x, Y = y) for both BPG and BNBG models. Suppose (X, Y ) ∼ BPG(λ₁, λ₂, p), then

$$ \begin{array}{@{}rcl@{}} P(N=n| X= x, Y= y) & = & \frac{P(X=x, Y=y|N=n) P(N = n)}{P(X = x, Y = y)} \\ & = & \frac{n^{x+y} e^{-n(\lambda_1+\lambda_2)} (1-p)^n}{C(\lambda_1+\lambda_2-\ln (1-p), x+y)}. \end{array} $$

Now to compute arg max_nP(N = n|X = x, Y = y), we consider

$$ g(n) = \frac{P(N=n + 1| X= x, Y= y)}{P(N=n| X= x, Y= y)} = \left( \frac{n + 1}{n} \right)^{x+y} e^{-(\lambda_{1}+\lambda_{2})} (1-p). $$

It is immediate that either g(n) is a decreasing function or it is an unimodal function and if n^∗ = arg max_nP(N = n|X = x, Y = y), then n^∗ is the smallest integer greater than

$$ \left( \left[ \frac{e^{\lambda_{1}+\lambda_{2}}}{1-p} \right]^{1/(x+y)} - 1 \right)^{-1}. $$

Now suppose (X, Y ) ∼BNBG(r₁, 𝜃₁, r₂, 𝜃₂, p), then

$$ \begin{array}{@{}rcl@{}} P(N=n| X= x, Y= y) & = & \frac{P(X=x, Y=y|N=n) P(N = n)}{P(X = x, Y = y)} \\ & = & \frac{(1-p)^n (1-\theta_1)^{nr_1} (1-\theta_2)^{nr_2}}{D(r_1, r_2, \theta_1, \theta_2, x, y, p)} \\&&\times \frac{{\Gamma}(x+nr_1)}{x! {\Gamma}(nr_1)} \frac{{\Gamma}(y+nr_2)}{y! {\Gamma}(nr_2)}. \end{array} $$

Hence,

$$ \begin{array}{@{}rcl@{}} g(n) & = & \frac{P(N=n + 1| X= x, Y= y)}{P(N=n| X= x, Y= y)} \\ & = & (1-p) (1-\theta_1)^{r_1} (1-\theta_2)^{r_2} \times \\ & & \frac{{\Gamma}(x+(n + 1)r_1) {\Gamma}(y+(n + 1)r_2)}{{\Gamma}((n + 1)r_2) {\Gamma}((n + 1)r_2)} \times \frac{{\Gamma}(nr_1) {\Gamma}(nr_2)}{{\Gamma}(x+nr_1) {\Gamma}(y+nr_2)}. \end{array} $$

In this case because of the complicated nature of g(n), it is not possible to show that g(n) has a unique maximum. But in all our numerical experiments it has been observed that g(n) has a unique maximum. We have chosen n^∗ to be the minimum n, such that g(n) < 1.

Appendix C: EM Algorithm for Non-Identical NB Distribution

In this Appendix we will show that if X_i ∼ NB(n_ir, 𝜃), X_i’s are independent, n_i’s are known for i = 1,…,m, then how to obtain MLEs of r and 𝜃, based on a sample {x₁,…,x_m}. In this case we will be using an EM algorithm very similar to Adamidis (1999). We use the following notation: α = −(ln(1 − 𝜃))^− 1, r = αλ and provide the algorithm to compute the MLEs of λ and 𝜃. It is observed that in this case at each E-step, the corresponding M-step can be obtained in explicit forms. Using the same notation as in Adamidis (1999), it can be easily seen that for i = 1,…,m,

$$ X_{i} \overset{d}{=} \sum\limits_{j = 1}^{M_{i}} Y_{ij}, $$

here Y_ij’s are i.i.d. logarithmic series distribution (LSD) with PDF

$$ f_{LSD}(y; \theta) = \frac{\alpha \theta^{y}}{y}; \ \ \ y \in \mathbb{N} = \{1, 2, 3, \ldots\}, $$

M_i ∼PO(n_iλ) and all the random variables are independently distributed. Further, if Z_ij’s are i.i.d. random variables with PDF

$$ f(z; \theta) = \alpha^{-1} \frac{(1-\theta)^{z}}{\theta}; \ \ \ z \in (0,1), $$

and 0, otherwise, then the log-likelihood function of the ‘complete data’ (Y_ij, Z_ij, M_i;i = 1,…,m, j = 1,…,M_i), without the additive constant can be written as

$$ \begin{array}{@{}rcl@{}} l^{*}(\alpha, \lambda) &=& - \lambda \widetilde{n} + \ln \lambda \sum\limits_{i = 1}^{m} m_{i} + \ln \theta \left[ \sum\limits_{i = 1}^{m} \sum\limits_{j = 1}^{m_{i}} y_{ij} - \sum\limits_{i = 1}^{m} m_{i} \right] \\&&+ \ln (1-\theta) \left[ \sum\limits_{i = 1}^{n} \sum\limits_{j = 1}^{m_{i}} z_{ij} \right], \end{array} $$

here $\displaystyle \widetilde {n} = \sum \limits _{i = 1}^{m} n_{i}$. Hence, the MLEs of λ and 𝜃 based on the complete observations can be easily obtained as

$$ \widehat{\lambda} = \frac{{\sum}_{i = 1}^{m} m_{i}}{\widetilde{n}} \ \ \ \text{and} \ \ \ \widehat{\theta} = \frac{{\sum}_{i = 1}^{m} {\sum}_{j = 1}^{m_{i}} y_{ij} - {\sum}_{i = 1}^{m} m_{i}}{{\sum}_{i = 1}^{m} {\sum}_{j = 1}^{m_{i}} y_{ij} + {\sum}_{i = 1}^{m} {\sum}_{j = 1}^{m_{i}} z_{ij} - {\sum}_{i = 1}^{m} m_{i}}. $$

Hence, following the same way as in Adamidis (1999), it can be easily seen that if at the k-th stage the estimates of λ and 𝜃 are λ^(k) and 𝜃^(k), respectively, and if we denote

$$ a_{i}(\alpha, \lambda) = n_{i} \alpha \lambda \sum\limits_{l = 1}^{x_{i}} (\alpha n_{i} \lambda + l-1)^{-1}, $$

then

$$ \begin{array}{@{}rcl@{}} \lambda^{(k + 1)} &=& \frac{{\sum}_{i = 1}^{m} a_{i}(\alpha^{(k)}, \lambda^{(k)})}{\widetilde{n}} \ \ \ \text{and} \\ \theta^{(k + 1)} &=& \frac{\widetilde{x} - {\sum}_{i = 1}^{m}a_{i}(\alpha^{(k)}, \lambda^{(k)})}{\widetilde{x} + {\sum}_{i = 1}^{m} a_{i}(\alpha^{(k)}, \lambda^{(k)}) \left( \frac{\alpha^{(k)}(1-\theta^{(k)})}{\theta^{(k)}} -1 \right)}. \end{array} $$

Here $\displaystyle \widetilde {x} = \sum \limits _{i = 1}^{m} x_{i}$.