Modelling bivariate lifetime data using copula

Abstract

Generally modelling lifetime data is carried out using probability distributions with the aid of reliability functions such as hazard rate, mean residual life, etc. In the present work an alternative approach is proposed by considering bivariate copulas instead of bivariate distributions. We define the analogues of reliability functions that are expressed in terms of copulas and study their properties. The results of the study are applied to case of the copulas of a bivariate exponential family of distributions.

Appendices

Appendix A: Proof of Proposition 3.2

Since $G_1(u,v) =\dfrac{{A_1 (u,v)}}{{A_1 (u,1)}}$, we have $G_1(p,1)=1$ and

$$\begin{aligned} C(u,1)=\exp \left[ - \int \limits _u^1 {\frac{{dp}}{p}}\right] =u. \end{aligned}$$

The properties $C(1,v)=v$ and $C(u,0)=0$ are obvious. Further,

$$\begin{aligned} C(0,v)=v \exp \left[ - \int \limits _0^1 {\frac{{{G_1}(p,v)}}{p}} dp\right] =0, \quad \text {by (ii)}. \end{aligned}$$

Condition (iii) implies that for $v_2 \ge v_1$,

$$\begin{aligned}&v_2\left[ \exp \left( - \int \limits _{u_2}^1 {\frac{{{G_1}(p,v_2)}}{p}} dp\,\right) - \exp \left( - \int \limits _{u_1}^1 {\frac{{{G_1}(p,v_2)}}{p}} dp\,\right) \right] \\&\quad \ge v_1\left[ \exp \left( - \int \limits _{u_2}^1 {\frac{{{G_1}(p,v_1)}}{p}} dp\,\right) - \exp \left( - \int \limits _{u_1}^1 {\frac{{{G_1}(p,v_1)}}{p}} dp\,\right) \right] \end{aligned}$$

which is the same as

$$\begin{aligned} C(u_2,v_2)- C(u_1,v_2)-C(u_2,v_1)+C(u_1,v_1) \ge 0 \end{aligned}$$

and thus C(u, v) is a copula.

Appendix B: Proof of Proposition 3.3

Assume (3.9). Then by direct calculation,

$$\begin{aligned} (G_1,G_2)= (1-\beta \log v,1-\beta \log u ) \end{aligned}$$

which is of the form (3.10). On the other hand, if we assume that (3.10) holds then formulas (3.6) and (3.7) lead to the functional equation

$$\begin{aligned} v\,\, u^ {B_1(v)}= u\,\, v^{B_2(u)}, \end{aligned}$$

(B1)

which is equivalent to

$$\begin{aligned} u^{\frac{1}{B_2(u)-1}}= v^{\frac{1}{B_1(v)-1}}. \end{aligned}$$

The solution is

$$\begin{aligned} u^{\frac{1}{B_2(u)-1}}= k = v^{\frac{1}{B_1(v)-1}} \end{aligned}$$

giving

$$\begin{aligned} B_1(v) = 1- \beta \log v\quad \text{ and }\quad B_2(u) = 1-\beta \log u\quad \text {where}\quad \beta = (-\log k)^{-1}. \end{aligned}$$

Substituting into (3.6) and (3.7), we recover the copula (3.9). This completes the proof.

Appendix C: Proof of Proposition 3.4

To prove the necessary part, we note that for the Clayton survival copula,

$$\begin{aligned} G_1(u,v)= \dfrac{u^{-\frac{1}{\theta }}}{u^{-\frac{1}{\theta }}+v^{-\frac{1}{\theta }}-1} \end{aligned}$$

and

$$\begin{aligned} G_2(u,v)= \dfrac{v^{-\frac{1}{\theta }}}{u^{-\frac{1}{\theta }}+v^{-\frac{1}{\theta }}-1} \end{aligned}$$

so that (3.11) holds. To prove the converse part, we observe that (3.12) is an Archimedean copula with generator $\phi (u)=\theta (u^{\frac{-1}{\theta }}-1)$. From Theorem 4.3.8 in Nelsen [18], for Archimedean copula with generator $\phi$,

$$\begin{aligned} \phi ^{'}(u)\dfrac{\partial C}{\partial v}= \phi ^{'}(v)\dfrac{\partial C}{\partial u} \end{aligned}$$

(C1)

for almost all u,v in I.

Using

$$\begin{aligned} \dfrac{G_1}{G_2}= \dfrac{u \frac{1}{c}\frac{\partial C}{\partial u}}{v \frac{1}{c}\frac{\partial C}{\partial v}}=\dfrac{u^{-\frac{1}{\theta }}}{v^{-\frac{1}{\theta }}} \end{aligned}$$

(C1) becomes

$$\begin{aligned} \phi ^{'}(u)u ^{\frac{1}{\theta }+1}=\phi ^{'}(v)v ^{\frac{1}{\theta }+1}, \end{aligned}$$

for all u, v.

The solution of the above functional equation is

$$\begin{aligned} \phi ^{'}(u)u ^{\frac{1}{\theta }+1}=\phi ^{'}(v)v ^{\frac{1}{\theta }+1}=k \end{aligned}$$

(C2)

where k is a constant that may depend on $\theta$. Also k is less than zero since $\phi$ is decreasing. Solving (C2) with the boundary condition $\phi (1)=0$ and setting $k=-1,$

$$\begin{aligned} \phi (u)= \theta \left( u^{\frac{-1}{\theta }}-1\right) \end{aligned}$$

and accordingly C is a Clayton survival copula.

Appendix D: Proof of Proposition 4.1

Differentiating (4.1) with respect to u,

$$\begin{aligned} C\,\frac{{\partial M_1 }}{{\partial u}} + \,M_1 \frac{{\partial C}}{{\partial u}} = - C(u,v)\frac{{dS_1^{ - 1} }}{{du}}. \end{aligned}$$

(D1)

Also

$$\begin{aligned} M_1(u,1)&= m_1(S_1^{-1}(u),0)\\ =&- \frac{1}{u}\int \limits _0^u {p_1\,} \frac{{dS_1^{ - 1} (p_1)}}{{dp_1}}dp_1, \end{aligned}$$

or

$$\begin{aligned} \frac{{dS_1^{ - 1} (u)}}{{du}} = - \frac{1}{u}\frac{d (u M_1(u,1))}{du}. \end{aligned}$$

Substituting in (D1),

$$\begin{aligned} \frac{{\partial \log C}}{{\partial u}} = \frac{{\frac{d }{{d u}}(u\,M_1 (u,1))}}{{u\,M_1 (u,v)}} - \frac{{\partial \log M_1 (u,v)}}{{\partial u}}. \end{aligned}$$

(D2)

Similarly

$$\begin{aligned} \frac{{\partial \log C}}{{\partial v}} = \frac{{\frac{d }{{d v}}(v\,M_2 (1,v))}}{{v\,M_2 (u,v)}} - \frac{{\partial \log M_2 (u,v)}}{{\partial v}}. \end{aligned}$$

(D3)

Integrating (D2) from u to 1 and (D3) from v to 1 we have (4.3) and (4.4).

Appendix E: Proof of Proposition 4.2

Condition (i) follows from the identity

$$\begin{aligned} G_1(u,v)=\frac{\frac{d}{du}u M_1(u,1)}{M_1(u,v)} -u \frac{\partial \log M_1(u,v)}{\partial u} \ge 0. \end{aligned}$$

Now

$$\begin{aligned} C(u,1)=\frac{M_1(1,1)}{M_1(u,1)}\exp \left[ - \int \limits _u^1 {\frac{\partial }{{\partial p_1}}\log p_1 {M_1}(p_1,1)} dp_1\right] =u. \end{aligned}$$

It is easy to see that $C(1,v)=v$ and $C(u,0)=0$. Also

$$\begin{aligned} C(0,v) =v \exp \left[ - \int \limits _0^1 {\frac{{\tfrac{\partial }{{\partial p_1}}p_1{M_1}(p_1,1)}}{{p_1{M_1}(p_1,v)}}} dp_1\right] =0 \end{aligned}$$

by (ii).

Condition (iii) implies that for $v_2 \ge v_1$

$$\begin{aligned} v_2[B(u_2,v_2)-B(u_1,v_2)]\ge v_1[B(u_2,v_1)-B(u_1,v_1)] \end{aligned}$$

which is equivalent to the 2-increasing property of C(u, v). Thus C(u, v) is a copula.

Appendix F: Proof of Proposition 4.3

The necessary part of the theorem follows by direct calculations using (3.9) in (4.5) and (4.6). This gives

$$\begin{aligned} (L_1, L_2)=\left( \frac{1}{1-\beta \log v} ,\frac{1}{1-\beta \log u}\right) , \end{aligned}$$

which is of the form (4.9). We prove the converse part by assuming (4.9). Equations (4.7) and (4.8) yield after some algebra

$$\begin{aligned} C(u,v)= v \exp \left( \frac{\log u}{K_1(v)}\right) =u \exp \left( \frac{\log v}{K_2(u)}\right) . \end{aligned}$$

(F1)

This leads to the functional equation

$$\begin{aligned} \frac{\log v}{1-\frac{1}{K_1(v)}}=\frac{\log u}{1-\frac{1}{K_2(u)}} \end{aligned}$$

(F2)

which is true for all u, v. This happens if and only if (F2) is a constant, say $\dfrac{1}{\beta }$, $\beta >0$. Thus from (F2),we obtain

$$\begin{aligned} K_1(v)=\frac{1}{1-\beta \log v} \end{aligned}$$

and

$$\begin{aligned} K_2(u)=\frac{1}{1-\beta \log u}. \end{aligned}$$

Substituting $K_1(v)$ and $K_2(u)$ in (F1), we obtain the Gumbel–Barnett survival copula.

There exist some identities connecting $(G_1,G_2)$, $(M_1,M_2)$ and $(L_1,L_2)$. From (D2) and (D3)

$$\begin{aligned} G_1(u,v)= [M_1(u,v)]^{-1}\left[ \frac{d}{du}u M_1(u,1)-u\frac{\partial M_1(u,v)}{\partial u}\right] \end{aligned}$$

and

$$\begin{aligned} G_2(u,v)= [M_2(u,v)]^{-1}\left[ \frac{d}{dv}v M_2(1,v)-{v}\frac{\partial M_2(u,v)}{\partial v}\right] \end{aligned}$$

Differentiating (4.5)

$$\begin{aligned} L_1 \frac{\partial C}{\partial u}+C \frac{\partial L_1}{\partial u}= \dfrac{C}{u} \end{aligned}$$

or

$$\begin{aligned} L_1 \frac{\partial \log C}{\partial u}+ \frac{\partial L_1}{\partial u}= \dfrac{1}{u}. \end{aligned}$$

Using the definition of $G_1$,

$$\begin{aligned} L_1(u,v) G_1(u,v)+u \frac{\partial L_1}{\partial u}=1. \end{aligned}$$

(F3)

Similarly

$$\begin{aligned} L_2(u,v) G_2(u,v)+v\frac{\partial L_2}{\partial v}=1. \end{aligned}$$

(F4)

Also $(L_1,L_2)$ is related to $(M_1,M_2)$ by the identities

$$\begin{aligned} \dfrac{1}{L_1}\left( \dfrac{\partial L_1}{\partial u}-\dfrac{1}{u}\right) = \dfrac{1}{M_1}\left( \dfrac{\partial M_1}{\partial u}-\dfrac{1}{u}\dfrac{d(u M_1(u,1))}{d u}\right) \end{aligned}$$

and

$$\begin{aligned} \dfrac{1}{L_2}\left( \dfrac{\partial L_2}{\partial v}-\dfrac{1}{v}\right) = \dfrac{1}{M_2}\left( \dfrac{\partial M_2}{\partial v}-\dfrac{1}{v}\dfrac{d(v M_2(1,v))}{d v}\right) . \end{aligned}$$