A zero-inflated logarithmic series distribution of order k and its applications

Abstract

Here we develop an order k version of the zero-inflated logarithmic series distribution of Kumar and Riyaz [Staistica (accepted for publication), 2013b] through its probability generating function, and derive an expression for its probability mass function. Certain recurrence relation for its probabilities, raw moments and factorial moments are also obtained, and the maximum likelihood estimation of its parameters is discussed. We have tested the significance of the additional parameters of the distribution by generalized likelihood ratio test and illustrated all these procedures using certain real-life data sets.

Appendix

1.1 Proof of Proposition 1

From (2.5) we have

$$\begin{aligned} \displaystyle H(s)&= \sum \nolimits _{z=0}^\infty {p_z } s^z \end{aligned}$$

(5.1)

$$\begin{aligned} \ \displaystyle&= C\,\frac{-\ln \left( 1-\sum \nolimits _{j=1}^k {\theta _j } s^j \right) }{\sum \nolimits _{j=1}^k {\theta _j } s^j} \end{aligned}$$

(5.2)

On expanding the logarithmic function in (5.2), we get

$$\begin{aligned} \displaystyle H(s)&= C\,\sum \limits _{z=1}^\infty {\frac{\left( \sum \nolimits _{j=1}^k {\theta _j } s^j \right) ^{z-1}}{z}}\nonumber \\ \displaystyle&= C\sum \limits _{z=0}^\infty {\frac{\left( \sum \nolimits _{j=1}^k {\theta _j } s^j \right) ^z}{z+1}} . \end{aligned}$$

(5.3)

Now, by applying the multinomial expansion we have

$$\begin{aligned} H(s)=C\sum \limits _{z=0}^\infty {\sum \limits _{J_z } {\frac{z!}{z_1 !z_2 !\cdots z_k !}} } \frac{\theta _1^{z_1 } \theta _2^{z_2 } \cdots \theta _k^{z_k } }{(z+1)}s^\delta \end{aligned}$$

(5.4)

in which $\delta =\sum \nolimits _{j=1}^k {j\,z_j }$, $\sum \nolimits _{J_z }$ denote the summation over all tuples $(z_1 ,z_2, \dots ,z_k )$ of non-negative integers in the set $J_z =\{(z_1 ,z_2 ,\dots ,z_k ):\sum \nolimits _{j=1}^k {\,z_j } =z\}$. Now, on equating the coefficient of $s^z$ on the right hand side expressions of (5.1) and (5.4) we get (2.6).

1.2 Proof of Equation 2.10

In the light of (2.5), (2.7), (2.9), and (5.1) we have

$$\begin{aligned} \displaystyle H(s)&= \sum \limits _{z=0}^\infty {p_z } (\underline{v}+i)s^z \end{aligned}$$

(5.5)

$$\begin{aligned} \displaystyle&= R_i^{-1} { }_2F_1 \left( 1+i\;,\;1+i\;;\,\,2+i\;;\;\sum \limits _{j=1}^k {\theta _j }s^j\right) . \end{aligned}$$

(5.6)

On expanding the Gauss hypergeometric function in (5.6), we obtain

$$\begin{aligned} H(s)=R_i^{-1} \sum \limits _{z=0}^\infty {\frac{(1+i)_z (1+i)_z }{(2+i)_z }} \frac{\left( \sum \nolimits _{j=1}^k {\theta _j }s^j \right) ^z}{z!}, \end{aligned}$$

(5.7)

in which $(a)_k =a\,(a+1)\,\ldots \,(a+k-1)$ for $k\ge 1$ and $(a)_0 =1$. By multinomial expansions, we get the following from (5.7).

$$\begin{aligned} H(s)=R_i^{-1} \sum \limits _{z=0}^\infty {\frac{(1+i)_z (1+i)_z }{(2+i)_z }} \frac{\theta _1^{z_{_1 } } \theta _2^{z_{_2 } } \dots \theta _k^{z_{_k } } }{z_1 !z_2 !\,\dots \,z_k !}s^\delta \end{aligned}$$

(5.8)

where $\delta$ and $\sum \nolimits _{I_z }$ are as defined in (5.4). On equating the coefficients of $s^z$ on the right side expressions of (5.5) and (5.8) we get (2.10), since $(a-1)!(a)_k =\,(a+k-1)!$.

1.3 Proof of Proposition 2

From (2.5) we have

$$\begin{aligned} H(s)=R_0^{-1} \,_2 F_1 \left( 1\,\;,\;1\;;\;2\;;\;\sum \limits _{j=1}^k {\theta _j } s^j \right) . \end{aligned}$$

(5.9)

Differentiating (5.1) and (5.9) with respect to s, we get.

$$\begin{aligned} \sum \limits _{z=0}^\infty {z\,p_z } (\underline{v})s^{z-1}=D_0 R_0^{-1} \,_2 F_1 \left( 2\,\;,\;2\;;\;3\;;\;\sum \limits _{j=1}^k {\theta _j s^j} \right) \sum \limits _{j=1}^k {j\theta _j s^{j-1}} . \end{aligned}$$

(5.10)

From (5.1) and (5.9) we also have

$$\begin{aligned} R_1 \sum \limits _{z=0}^\infty {p_z (\underline{v}+1)\,s^z} =\;{ }_2F_1 \left( 2\,\;,\;2\,\,;\;\;3\,\;;\;\sum \limits _{j=1}^k {\theta _j s^j}\right) . \end{aligned}$$

(5.11)

By applying (5.11) in (5.10) we get

$$\begin{aligned} \sum \limits _{z=0}^\infty {z\,p_z (\underline{v})s^{z-1}} =D_0 R_0^{-1} R_1 \,\sum \limits _{z=0}^\infty {\sum \limits _{j=1}^k {j\theta _j p_z (\underline{v}+1)} s^{z+j-1}} . \end{aligned}$$

(5.12)

On equating the coefficient of $s^z$on both sides of (5.12) we obtain (2.13).

1.4 Proof of Proposition 3

On differentiating the right side expressions of (2.11) and (2.12) with respect to s, we obtain

$$\begin{aligned} \sum \limits _{r=1}^\infty {\mu _r (\underline{v})} \frac{(is)^{r-1}}{(r-1)!}=D_0 R_0^{-1} \sum \limits _{j=1}^k {j\theta _j } \mathrm{e}^{ijs}\;{ }_2F_1\left( 2\;,2\;;\;3\;;\sum \limits _{j=1}^k {\theta _j } \mathrm{e}^{ijs}\right) . \end{aligned}$$

(5.13)

From (2.11) and (2.12) we obtain the following.

$$\begin{aligned} R_1 \sum \limits _{r=0}^\infty {\mu _r (\underline{v}} +1)\frac{(is)^r}{r!}=\;{ }_2F_1 \left( 2\;,2\;;\;3\;;\;\sum \limits _{j=1}^k {\theta _j \mathrm{e}^{ijs}} \right) . \end{aligned}$$

(5.14)

Equations (5.13) and (5.14) together leads to

$$\begin{aligned} \sum \limits _{r=1}^\infty {\mu _r (\underline{v})} \frac{(is)^{r-1}}{(r-1)!}=D_0 R_0^{-1} R_1 \sum \limits _{r=0}^\infty {\sum \limits _{j=1}^k {j\theta _j \mathrm{e}^{ijs}\mu _z (\underline{v}+1)\frac{(is)^r}{r\,!}} } \; \end{aligned}$$

$$\begin{aligned} =D_0 R_0^{-1} R_1 \sum \limits _{r=0}^\infty {\sum \limits _{m=0}^\infty {\sum \limits _{j=1}^k {j^{m+1}\theta _j \mu _r (\underline{v}+1)} } } \frac{(is)^{r+m}}{r!m!}. \end{aligned}$$

(5.15)

Equating the coefficient of $\frac{(is)^r}{r!}$ on both sides of (5.15) we get (2.14).

1.5 Proof of Proposition 4

From (2.5) we have the following factorial moment generating function $F(s)$ of the ZILSD$_{k}$

$$\begin{aligned}&\displaystyle F(s)=\sum \limits _{r=0}^\infty {\mu _{\left[ r \right] } (\underline{v})\frac{s^{r}}{r!}} \end{aligned}$$

(5.16)

$$\begin{aligned}&\displaystyle =R_0^{-1}{ }_2 F_1 \left[ 1\,\;,\;1\;;\;2\;;\;\sum \limits _{j=1}^k {\theta _j (1+s)^j}\right] . \end{aligned}$$

(5.17)

On differentiating the right hand sides of (5.16) and (5.17) with respect to s, we obtain

$$\begin{aligned} \sum \limits _{r=1}^\infty {\mu _{\left[ r \right] } (\underline{v})\frac{s^{r-1}}{(r-1)!}} =D_0 R_0^{-1} \,\sum \limits _{j=1}^k {j\theta _j (1+s)^{j-1}} \,_2 F_1 [2\,\;,\;2\;;\;3\;;\;\sum \limits _{j=i}^k {\theta _j (1+s)^j}]. \end{aligned}$$

(5.18)

From (5.16) and (5.17) we have

$$\begin{aligned} R_1 \sum \limits _{r=0}^\infty {\mu _{[r]} (\underline{v}+1)\frac{s^r}{r!}} =_2 F_1 \left[ 2\,\;,\;2\;;\;3\;;\sum \limits _{j=1}^k {\theta _j (1+s)^j}\right] . \end{aligned}$$

(5.19)

Equations (5.18) and (5.19) together implies

$$\begin{aligned} \sum \limits _{r=1}^\infty {\mu \,_{\left[ r \right] } (\underline{v})\frac{s^{r-1}}{(r-1)!}} =D_0 R_0^{-1} R_1 \,\sum \limits _{r=0}^\infty {\sum \limits _{j=1}^k {j\theta _j (1+s)^{j-1}} } \mu _{\left[ r \right] } (\underline{v}+1)\frac{s^r}{r!} \end{aligned}$$

$$\begin{aligned} =D_0 R_0^{-1} R_1 \sum \limits _{r=0}^\infty {\sum \limits _{j=1}^k {\sum \limits _{m=0}^{j-1} {\left( \begin{array}{l} {j-1} \\ m \\ \end{array}\right) j\theta _j } } } \mu _{\left[ r \right] } (\underline{v}+1)\frac{s^{r+m}}{r!}. \end{aligned}$$

(5.20)

Now on equating the coefficient of $(r!)^{-1}s^r$ on both sides of (5.20) we get (2.15).