An Absolute Continuous Bivariate Inverse Generalized Exponential Distribution: Properties, Inference and Extensions

Abstract

The aim of this paper is to introduce an absolutely continuous bivariate inverse generalized exponential (BIGE) distribution. The proposed distribution has been obtained by removing the singular component from the BIGE distribution similarly as the Block and Basu absolute continuous bivariate exponential distribution. This distribution has four parameters, and due to this, the joint probability density function can take variety of shapes. This distribution can be used quite effectively if there are no ties in the bivariate data set and particularly if the marginals are from a heavy tailed distribution. We have developed different properties of this distribution and provided classical inference of the unknown parameters. The maximum likelihood (ML) estimators cannot be obtained in closed form and one needs to solve a four-dimensional optimization problem to compute the ML estimators in this case. To avoid that, we propose to use the expectation maximization (EM) algorithm to compute the ML estimators of the unknown parameters. The analysis of one data set has been performed to see the effectiveness of the proposed algorithm and extended the results to the multivariate case also. Finally, we conclude the paper with several open problems for future research.

Appendices

Appendix 1: Proof of Theorem 1

Proof

(a) It is clear that $f_{U,V}(u,v)$ is continuous in $S_1 \cup S_2$. Since $f_{U,V}(x,x) = \lim _{u,v \rightarrow x} f_{U,V}(u,v)$, it follows that $f_{U,V}(u,v)$ is continuous in $S_0 \cup S_1 \cup S_2$. Since for all $0< u,v < \infty$,

$$ f_{U,V}(0,0) = f_{U,V}(\infty , \infty ) = f_{U,V}(u,0) = f_{U,V}(u, \infty ) = f_{U,V}(0, v) = f_{U,V}(\infty , v) = 0, $$

that $f_{U,V}(u,v)$ has a local maximum. It can be easily checked by taking derivatives of $\ln f_{U,V}(u,v)$ that $f_{U,V}(u,v)$ does not have any critical point in the region $S_1 \cup S_2$, hence $f_{U,V}(u,v)$ does not have any critical point in the region $S_1 \cup S_2$, hence it does not have any local maximum in $S_1 \cup S_2$. Therefore, in this case, the local maximum will be at $S_0$. By taking derivative with respect to x of $\ln f_{U,V}(x,x)$ and equating it to zero, we can get one needs to solve the Eq. (9). It can be easily seen that the left-hand side of (9) is a decreasing function of x, and it decreases from $\infty$ to -4. Hence, it has a unique solution.

(b) Note that since $\alpha _1 > 1$ and $\alpha _2 + \alpha _0 < 1$, it can be easily seen by taking partial derivatives of $\ln f_{U,V}(u,v)$ that $f_{U,V}(u,v)$ has a critical point at $(x_1, x_2)$, where $x_1$ and $x_2$ are solutions of the non-linear Eqs. (10) and (11), respectively. Clearly, $x_1 < 1/2$, since $\alpha _1 > 1$ and $x_2 < 1/2$, since $\alpha _2 + \alpha _0 < 1$. Hence, $(x_1, x_2) \in S_1$. Uniqueness follows using the same argument as in (a). It can be easily checked that $f_{U,V}(u,v)$ does not have a critical point in $S_2$.

(c) Follows similarly as in (b).     $\square$

Appendix 2: Observed Fisher Information Matrix

Using the same notation as Louis (1982), the observed Fisher information matrix can be written as

$$ {{\boldsymbol{F}}}_{obs} = {{\boldsymbol{B}}} - {{\boldsymbol{S}}}{{\boldsymbol{S}}}^{\top }, $$

here ${{\boldsymbol{B}}}$ is the negative of the second derivative of the log-likelihood function and ${{\boldsymbol{S}}}$ is the derivative vector. We provide the elements of the matrix ${{\boldsymbol{B}}}$ and the vector ${{\boldsymbol{S}}}$. We will use the following notation for brevity.

$$ a_{11} = \sum _{i \in I_1} \frac{1}{u_i^2 (1 - e^{-\frac{\widehat{\lambda }}{u_i}})^2}, \ \ \ a_{12} = \sum _{i \in I_2} \frac{1}{u_i^2 (1 - e^{-\frac{\widehat{\lambda }}{u_i}})^2}, $$

$$ a_{22} = \sum _{i \in I_2} \frac{1}{v_i^2 (1 - e^{-\frac{\widehat{\lambda }}{v_i}})^2}, \ \ \ a_{21} = \sum _{i \in I_1} \frac{1}{v_i^2 (1 - e^{-\frac{\widehat{\lambda }}{v_i}})^2} $$

$$ b_{11} = \sum _{i \in I_1} \frac{1}{u_i (1 - e^{-\frac{\widehat{\lambda }}{u_i}})}, \ \ \ b_{12} = \sum _{i \in I_2} \frac{1}{u_i (1 - e^{-\frac{\widehat{\lambda }}{u_i}})}, $$

$$ b_{22} = \sum _{i \in I_2} \frac{1}{v_i (1 - e^{-\frac{\widehat{\lambda }}{v_i}})}, \ \ \ b_{21} = \sum _{i \in I_1} \frac{1}{v_i (1 - e^{-\frac{\widehat{\lambda }}{v_i}})} $$

$$ c_{11} = \sum _{i \in I_1} \frac{1}{u_i}, \ \ \ c_{12} = \sum _{i \in I_2} \frac{1}{u_i}, c_{22} = \sum _{i \in I_2} \frac{1}{v_i}, \ \ \ c_{21} = \sum _{i \in I_1} \frac{1}{v_i}, $$

$$ d_{11} = \sum _{i \in I_1} \ln (1 - e^{-\frac{\lambda }{u_i}}), d_{12} = \sum _{i \in I_2} \ln (1 - e^{-\frac{\lambda }{u_i}}) $$

$$ d_{22} = \sum _{i \in I_2} \ln (1 - e^{-\frac{\lambda }{v_i}}), d_{21} = \sum _{i \in I_1} \ln (1 - e^{-\frac{\lambda }{v_i}}). $$

If the (i, j)-th element of the matrix ${{\boldsymbol{B}}}$ is B(i, j), then $B(i,j) = B(j,i)$, for $1 \le i,j \le 4$, and for $1 \le i \le j \le 4$,

$$ B(1,1) = \frac{n_1 + n_2 w_2}{\widehat{\alpha }_1^2}, \ \ \ B(2,2) = \frac{n_2 + n_1 w_1}{\widehat{\alpha }_2^2}, \ \ \ B(3,3) = \frac{n_1(1-w_1) + n_2 (1-w_2)}{\widehat{\alpha }_0^2}, $$

$$ B(4,4) = \frac{2}{\widehat{\lambda }^2} + a_{11}[\widehat{\alpha }_1-1] + a_{22}[\widehat{\alpha }_2-1] + a_{12}[\widehat{\alpha }_1+\widehat{\alpha }_0-1]+ a_{21}[\widehat{\alpha }_2+\widehat{\alpha }_0-1] $$

$$ B(1,4) = -\frac{1}{\widehat{\lambda }^2} - (c_{11}+b_{11}), B(2,4) = -\frac{1}{\widehat{\lambda }^2} - (c_{22}+b_{22}), B(3,4) = b_{12} + b_{21}, $$

$$ B(1,3) = B(2,3) = B(1,2) = 0. $$

If ${{\boldsymbol{S}}} = (S(1), S(2), S(3), S(4))^{\top }$, then

$$ S(1) = \frac{n_1 + w_2 n_2}{\widehat{\alpha }_1} + (d_{11} + d_{12}), S(2) = \frac{n_2 + w_1 n_1}{\widehat{\alpha }_2} + (d_{22} + d_{21}), S(3) = d_{12} + d_{21}, $$

$$\begin{aligned} S(4)= & {} \widehat{\lambda } (c_{11} + c_{12} + c_{21} + c_{22}) - \frac{2n}{\widehat{\lambda }} - \widehat{\alpha }_1 (b_{11} + b_{12}) + \widehat{\alpha }_2 (b_{22} + b_{21}) + \widehat{\alpha }_0 (b_{21} + b_{12}) \\&\ \ \ \ \ \ \ + (b_{11} + b_{12} + b_{21} + b_{22}). \end{aligned}$$

Appendix 3: Normalizing Constant c

In this section, we show that the normalizing constant c satisfies (18). We will show the result for $p = 3$, the general result easily follows from there. If $(Y_1,Y_2,Y_3)^{\top }$ follows a AMIGE with parameters $\alpha _0$, $\alpha _1$, $\alpha _2$, $\alpha _3$ and $\lambda$, then for ${{\boldsymbol{Y}}} = (Y_1,Y_2,Y_3)$ and ${{\boldsymbol{y}}} = (y_1,y_2,y_3)$

$$ f_{{\boldsymbol{Y}}}({{\boldsymbol{y}}}) = c \left\{ \begin{array}{ccc} f_{IGE}(y_1;\alpha _1,\lambda )f_{IGE}(y_2;\alpha _2,\lambda )f_{IGE}(y_3;\alpha _0+\alpha _3,\lambda ) &{} \hbox {if} &{} y_1< y_2< y_3 \\ f_{IGE}(y_1;\alpha _1,\lambda )f_{IGE}(y_3;\alpha _3,\lambda )f_{IGE}(y_2;\alpha _0+\alpha _2,\lambda ) &{} \hbox {if} &{} y_1< y_3< y_2 \\ f_{IGE}(y_2;\alpha _2,\lambda )f_{IGE}(y_1;\alpha _1,\lambda )f_{IGE}(y_3;\alpha _0+\alpha _3,\lambda ) &{} \hbox {if} &{} y_2< y_1< y_3 \\ f_{IGE}(y_2;\alpha _2,\lambda )f_{IGE}(y_3;\alpha _3,\lambda )f_{IGE}(y_1;\alpha _0+\alpha _1,\lambda ) &{} \hbox {if} &{} y_2< y_3< y_1 \\ f_{IGE}(y_3;\alpha _3,\lambda )f_{IGE}(y_1;\alpha _1,\lambda )f_{IGE}(y_2;\alpha _0+\alpha _2,\lambda ) &{} \hbox {if} &{} y_3< y_1< y_2 \\ f_{IGE}(y_3;\alpha _3,\lambda )f_{IGE}(y_2;\alpha _2,\lambda )f_{IGE}(y_1;\alpha _0+\alpha _1,\lambda ) &{} \hbox {if} &{} y_3< y_2 < y_1. \end{array} \right. $$

Now, note that

$$\begin{aligned}&\int _0^{\infty } \int _{y_1}^{\infty } \int _{y_2}^{\infty } f_{IGE}(y_1;\alpha _1,\lambda )f_{IGE}(y_2;\alpha _2,\lambda )f_{IGE}(y_3;\alpha _0+\alpha _3,\lambda ) dy_3 dy_2 dy_1 = \\&\int _0^{\infty } \int _{y_1}^{\infty } f_{IGE}(y_1;\alpha _1,\lambda )f_{IGE}(y_2;\alpha _2,\lambda )S_{IGE}(y_2;\alpha _0+\alpha _3,\lambda ) dy_3 dy_2 = \\&\frac{\alpha _2}{\alpha _2+\alpha _3+\alpha _0} \int _0^{\infty } \int _{y_1}^{\infty } f_{IGE}(y_1;\alpha _1,\lambda )S_{IGE}(y_1;\alpha _2+\alpha _3+\alpha _0,\lambda ) = \\&\frac{\alpha _1}{\alpha _1+\alpha _2+\alpha _3+\alpha _0} \times \frac{\alpha _2}{\alpha _2+\alpha _3+\alpha _0}. \end{aligned}$$

Similarly, the other integrations also can be obtained. Hence

$$\begin{aligned} c^{-1}= & {} \frac{\alpha _1}{\alpha _1+\alpha _2+\alpha _3+\alpha _0} \times \frac{\alpha _2}{\alpha _2+\alpha _3+\alpha _0} + \frac{\alpha _1}{\alpha _1+\alpha _2+\alpha _3+\alpha _0} \times \frac{\alpha _3}{\alpha _2+\alpha _3+\alpha _0} + \\&\frac{\alpha _2}{\alpha _1+\alpha _2+\alpha _3+\alpha _0} \times \frac{\alpha _1}{\alpha _1+\alpha _3+\alpha _0} + \frac{\alpha _2}{\alpha _1+\alpha _2+\alpha _3+\alpha _0} \times \frac{\alpha _3}{\alpha _1+\alpha _3+\alpha _0} + \\&\frac{\alpha _3}{\alpha _1+\alpha _2+\alpha _3+\alpha _0} \times \frac{\alpha _1}{\alpha _1+\alpha _2+\alpha _0} + \frac{\alpha _3}{\alpha _1+\alpha _2+\alpha _3+\alpha _0} \times \frac{\alpha _2}{\alpha _1+\alpha _2+\alpha _0}. \end{aligned}$$

Appendix 4: Proof of Theorem 2

Proof

(a) Follows from the definition.

(b) Proof follows along the same way as the proof of Part (a) of Theorem 2.1.

(c)

$$\begin{aligned} P(Z> z)= & {} P(U_1> z, \ldots , U_p> z, U_0 > z) \\= & {} S_{IGE}(z; \alpha _1, \lambda ) \times \ldots \times S_{IGE}(z; \alpha _p, \lambda ) S_{IGE}(z; \alpha _0, \lambda ) \\= & {} S_{IGE}(z; \alpha _1+\ldots +\alpha _p+\alpha _0, \lambda ). \end{aligned}$$

(d) Observe that $(Y_i,Y_j) \sim$ ABIGE$(\alpha _i,\alpha _j,\lambda )$. Hence,

$$\begin{aligned} P(Y_i < Y_j)= & {} \frac{\alpha _i+\alpha _j+\alpha _0}{\alpha _i+\alpha _j} \int _0^{\infty } \int _u^{\infty } f_{IGE}(u; \alpha _i,\lambda ) f_{IGE}(v; \alpha _j+\alpha _0,\lambda ) dv du \\= & {} \frac{\alpha _i+\alpha _j+\alpha _0}{\alpha _i+\alpha _j} \int _0^{\infty } f_{IGE}(u; \alpha _i,\lambda ) S_{IGE}(u; \alpha _j+\alpha _0,\lambda ) du \\= & {} \frac{\alpha _i}{\alpha _i+\alpha _j}. \end{aligned}$$

(e) Observe that $(Y_i,Y_j) \sim$ ABIGE$(\alpha _i,\alpha _j,\lambda )$. Hence,

$$\begin{aligned} P(Y_i > a| Y_i< Y_j)= & {} \frac{P(a< Y_i< Y_j)}{P(Y_i < Y_j)} \\= & {} \frac{\alpha _i+\alpha _j+\alpha _0}{\alpha _i} \int _a^{\infty } \int _u^{\infty } f_{IGE}(u; \alpha _i, \lambda ) f_{IGE}(v, \alpha _j +\alpha _0, \lambda ) dv du \\= & {} \frac{\alpha _i+\alpha _j+\alpha _0}{\alpha _i} \int _a^{\infty } f_{IGE}(u; \alpha _i, \lambda ) S_{IGE}(u, \alpha _j +\alpha _0, \lambda ) du \\= & {} S_{IGE}(a; \alpha _i+\alpha _j+\alpha _0, \lambda ). \end{aligned}$$

    $\square$