In this section, we introduce two new families of bivariate distributions with marginals having distributions with forms (1) or (2). We apply a similar technique of that proposed by Marshall and Olkin [15], for obtaining these families.
The construction of the families (models)
Lifetime model
Suppose that a system consists of two subsystems, say A and B. Subsystem A contains two components, say A1, and C, connected in series (parallel) with lifetimes U1 and U0 , respectively. Subsystem B contains the two components, say B1 and C, connected in series (parallel), where the lifetime of component B1 is U2.
Suppose that Ui, i = 0, 1, 2. , are independent random variables following EF(θi, c), i = 0, 1, 2 for the series case and IEF(βi; c), i = 0, 1, 2. , for the parallel case, i.e.,
$$ {\overline{F}}_{U_i}(u)=\left\{\begin{array}{c}{\overline{F}}_{\mathrm{EF}}\left(u;\theta, c\right)={e}^{-{\theta}_i{g}_1\left(u;c\right)},i=0,1,2,\mathrm{for}\ \mathrm{the}\ \mathrm{series}\ \mathrm{case},\\ {}{\overline{F}}_{\mathrm{IEF}}\left(u;\beta, c\right)=1-{e}^{-{\beta}_i{g}_2\left(u;c\right)},i=0,1,2,\mathrm{for}\ \mathrm{the}\ \mathrm{parallel}\ \mathrm{case}.\end{array}\right. $$
(3)
If X and Y are the lifetimes of the two subsystems A and B, respectively, then we have X = min {U0, U1} and Y = min {U0, U2}., for the series case, while X = max {U0, U1} and Y = max {U0, U2}, for the parallel case.
Stress model
Consider a two-component system and consider three independent stresses say U0, U1, and U2. Each component is subject to an individual stress say U1 and U2, respectively, while U0 is an overall stress transmitted to both the components equally. Then,
- 1.
The observed stress on the two components is X = max {U0, U1} and Y = max {U0, U2}., respectively.
- 2.
If the stresses are always fatal, then the lifetime of the two components are X = min {U0, U1} and Y = min {U0, U2}.
We can observe that in the two models there is the possibility of having X = Y; thus, the two models have both an absolute continuous part and a singular part, similar to M-O bivariate exponential model.
Theorems 1–3 present the survival functions and the probability density functions of the proposed bivariate families.
Theorem 1 Suppose Ui, i = 0, 1, 2., are independent random variables following EF(θi; c), i = 0, 1, 2., and let X = min {U0, U1} and Y = min {U0, U2}; then, the bivariate vector (X, Y) will have the survival function
$$ {\overline{F}}_{BEF}\left(X,Y\right)=\exp \left\{-{\theta}_1{g}_1\left(x;c\right)-{\theta}_2{g}_1\left(y;c\right)-{\theta}_0{g}_1\left(\max \left\{x,y\right\};c\right)\right\}. $$
(4)
Proof Obviously, from \( {\overline{F}}_{X,Y}\left(x,y\right)=P\left(X>x,Y>y\right) \), we can write \( {\overline{F}}_{\mathrm{BEF}}\left(X,Y\right) \) as
$$ P\left(\min \left\{{U}_0,{U}_1\right\}>x,\min \left\{{U}_0,{U}_2\right\}>y\right)=P\left({U}_1>x,{U}_2>y,{U}_0>\min \left(x,y\right)\right). $$
Since Ui are independent random variables following EF(θi; c), i = 0, 1, 2. Hence, (4) holds. ∎
We will denote the bivariate distribution with survival function having the form (4) by BEF(θ0, θ1, θ2; c). Clearly, X and Y are independent if and only if (iff) θ0 = 0. The joint survival function can also written as
$$ {\overline{F}}_{\mathrm{BEF}}\left(X,Y\right)=\left\{\begin{array}{c}\exp \left\{-\left({\theta}_0+{\theta}_1\right){g}_1\left(x;c\right)-{\theta}_2{g}_1\left(y;c\right)\right\},\kern0.75em \mathrm{if}\ x\ge y\\ {}\exp \left\{-{\theta}_1{g}_1\left(x;c\right)-\left({\theta}_0+{\theta}_2\right){g}_1\left(y;c\right)\right\},\kern0.5em \mathrm{if}\ y>x\ \end{array}\right. $$
Theorem 2 Suppose Ui, i = 0, 1, 2. , are independent random variables following EIF(βi; c), i = 0, 1, 2. , and let X = max {U0, U1} and Y = max {U0, U2}; then, the bivariate vector (X, Y) has the cumulative function
$$ {F}_{\mathrm{BIEF}}\left(X,Y\right)=\exp \left\{-{\beta}_1{g}_2\left(x;c\right)-{\beta}_2{g}_2\left(y;c\right)-{\beta}_0{g}_2\left(\min \left\{x,y\right\};c\right)\right\}. $$
(5)
Proof Similarly as in Theorem 1, using FX, Y(x, y) = P(X < x, Y < y), we can show that (5) holds. ∎
We will denote the bivariate distribution with cumulative function with the form (5) by BIEF(β0, β1, β2; c). Clearly, X and Y are independent iff β0 = 0. The joint cumulative function can also be written as
$$ {F}_{\mathrm{BIEF}}\left(X,Y\right)=\left\{\begin{array}{c}\exp \left\{-{\beta}_1{g}_2\left(x;c\right)-\left({\beta}_0+{\beta}_2\right){g}_2\left(y;c\right)\right\},\mathrm{if}\ x\ge y\\ {}\exp \left\{-\left({\beta}_0+{\beta}_1\right){g}_2\left(x;c\right)-{\beta}_2{g}_2\left(y;c\right)\right\},\mathrm{if}\ y>x\ \end{array}\right. $$
Theorem 3 If the vector (X, Y) has either BEF(θ0, θ1, θ2; c) or BIEF(β0, β1, β2; c), then their joint pdf is given by
$$ {f}_{X,Y}\left(x,y\right)=\left\{\begin{array}{c}{f}_1\left(x,y\right),\mathrm{if}\ x>\mathrm{y}\\ {}{f}_2\left(x,y\right),\mathrm{if}\ x<y\\ {}{f}_0(x),\mathrm{if}\ x=y\end{array}\right. $$
(6)
where \( {\displaystyle \begin{array}{c}{f}_1\left(x,y\right)=\left\{\begin{array}{c}{\theta}_2\left({\theta}_0+{\theta}_1\right){g}_1^{\prime}\left(x;c\right){g}_1^{\prime}\left(y;c\right)\ {\mathrm{e}}^{-\left({\theta}_0+{\theta}_1\right){g}_1\left(x;c\right)-{\theta}_2{g}_1\left(y;c\right)},\mathrm{for}\ \mathrm{BEF}\left({\theta}_0,{\theta}_1,{\theta}_2;c\right)\\ {}\ {\beta}_1\left({\beta}_0+{\beta}_2\right){g}_2^{\prime}\left(x;c\right){g}_2^{\prime}\left(y;c\right){e}^{-{\beta}_1{g}_2\left(x;c\right)-\left({\beta}_0+{\beta}_2\right){g}_2\left(y;c\right)},\mathrm{for}\ \mathrm{BIEF}\left({\upbeta}_0,{\upbeta}_1,{\upbeta}_2;\mathrm{c}\right)\end{array}\right.\\ {}{f}_2\left(x,y\right)=\left\{\begin{array}{c}{\theta}_1\left({\theta}_0+{\theta}_2\right){g}_1^{\prime}\left(x;c\right){g}_1^{\prime}\left(y;c\right)\ {\mathrm{e}}^{-{\theta}_1{g}_1\left(x;c\right)-\left({\theta}_0+{\theta}_2\right){g}_1\left(y;c\right)},\mathrm{for}\ \mathrm{BEF}\left({\theta}_0,{\theta}_1,{\theta}_2;c\right)\\ {}\ {\beta}_2\left({\beta}_0+{\beta}_1\right){g}_2^{\prime}\left(x;c\right){g}_2^{\prime}\left(y;c\right){e}^{-\left({\beta}_0+{\beta}_1\right){g}_2\left(x;c\right)-{\beta}_2{g}_2\left(y;c\right)},\mathrm{for}\ \mathrm{BIEF}\left({\upbeta}_0,{\upbeta}_1,{\upbeta}_2;\mathrm{c}\right)\end{array}\right.\end{array}} \)
and
$$ {f}_0(x)=\left\{\begin{array}{c}{\theta}_0{g}_1^{\prime}\left(x;c\right){\mathrm{e}}^{-\theta {g}_1\left(x;c\right)},\mathrm{for}\ \mathrm{BEF}\left({\theta}_0,{\theta}_1,{\theta}_2;c\right)\\ {}-{\beta}_0{g}_2^{\prime}\left(x;c\right){e}^{-\beta {g}_2\left(x;c\right)\Big)},\mathrm{for}\ \mathrm{BIEF}\left({\upbeta}_0,{\upbeta}_1,{\upbeta}_2;\mathrm{c}\right)\end{array}\right. $$
With θ = θ0 + θ1 + θ2, β = β0 + β1 + β2 and \( {g}_i^{\prime}\left(t;c\right),i=1,2, \) is the first derivative of gi(t; c) with respect to t.
Proof Clearly, for the two models, f1(x, y) and f2(x, y) can be easily obtained by using \( \frac{\partial^2{\overline{F}}_{X,Y}\left(x,y\right)}{\partial x\partial y} \) or \( \frac{\partial^2{F}_{X,Y}\left(x,y\right)}{\partial x\partial y} \) for x > y and y > x respectively. For f0(x), we use the relation
\( {\int}_0^{\infty }{\int}_0^x{f}_1\left(x,y\right) dydx+{\int}_0^{\infty }{\int}_0^y{f}_2\left(x,y\right) dx dy+{\int}_0^{\infty }{f}_0(x) dx=1 \). So, for the BEF, we have
$$ {\int}_0^{\infty }{\int}_0^x{f}_1\left(x,y\right) dydx=1-\left({\theta}_0+{\theta}_1\right){\int}_0^{\infty }{g}_1^{\prime}\left(t;c\right){\mathrm{e}}^{-\theta {g}_1\left(t;c\right)} dt $$
and
$$ {\int}_0^{\infty }{\int}_0^y{f}_2\left(x,y\right) dxdy=1-\left({\theta}_0+{\theta}_2\right){\int}_0^{\infty }{g}_1^{\prime}\left(t;c\right){\mathrm{e}}^{-\theta {g}_1\left(t;c\right)} dt, $$
Thus,
$$ {\int}_0^{\infty }{f}_0(x) dx=1-\left[2-\left({\theta}_0+\theta \right){\int}_0^{\infty }{g}_1^{\prime}\left(t;c\right){\mathrm{e}}^{-\theta {g}_1\left(t;c\right)} dt\right]={\theta}_0{\int}_0^{\infty }{g}_1^{\prime}\left(t;c\right){\mathrm{e}}^{-\theta {g}_1\left(t;c\right)} dt. $$
Similarly for the BIEF, we have \( {\int}_0^{\infty }{f}_0(x) dx=1+\left({\beta}_1+{\beta}_2\right){\int}_0^{\infty }{g}_2^{\prime}\left(t;c\right){\mathrm{e}}^{-\beta {g}_2\left(t;c\right)} dt=-{\beta}_0{\int}_0^{\infty }{g}_2^{\prime}\left(t;c\right){\mathrm{e}}^{-\beta {g}_2\left(t;c\right)} dt \).
Hence, the proof is complete. ∎
Notice that both distribution BEF(θ0, θ1, θ2; c) and BIEF (β0, β1, β2; c) are singular on the line X = Y, since P(X = Y) ≠ 0. Thus the two models have a singular part and an absolute continuous part, similar to Marshall and Olkin’s model. The following theorem provides explicitly the absolute continuous part and the singular part of BEF and BIEF.
Theorem 4 If the vector (X, Y) has BEF(θ0, θ1, θ2; c) or BIEF(β0, β1, β2; c), then
- (i)
The survival function for the BEF is
$$ {\overline{F}}_{\mathrm{BEF}}\left(x,y\right)=\frac{\theta_1+{\theta}_2}{\theta }{\overline{F}}_{\mathrm{BEF}(a)}\left(x,y\right)+\frac{\theta_0}{\theta }{\overline{F}}_{\mathrm{BEF}(s)}\left(x,y\right), $$
(7)
Where, θ = θ0 + θ1 + θ2, \( {\overline{F}}_{\mathrm{BEF}(s)}\left(x,y\right)={\mathrm{e}}^{-\theta {g}_1\left(\max \left\{x,y\right\};c\right)} \) is the singular part, and \( {\overline{F}}_{\mathrm{BEF}(a)}\left(x,y\right)=\frac{\theta }{\theta_1+{\theta}_2}{\mathrm{e}}^{-{\theta}_1{g}_1\left(x;c\right)-{\theta}_2{g}_1\left(y;c\right)-{\theta}_0{g}_1\left(\max \left\{x,y\right\};c\right)}-\frac{\theta_0}{\theta_1+{\theta}_2}{\mathrm{e}}^{-\theta {g}_1\left(\max \left\{x,y\right\};c\right)} \) is the absolute continuous part.
- (ii)
The cumulative function for the BIEF is
$$ {F}_{\mathrm{BIEF}}\left(x,y\right)=\frac{\beta_1+{\beta}_2}{\beta }{F}_{\mathrm{BIEF}(a)}\left(x,y\right)+\frac{\beta_0}{\beta }{F}_{\mathrm{BIEF}(s)}\left(x,y\right), $$
(8)
where, β = β0 + β1 + β2, \( {F}_{\mathrm{BIEF}(s)}\left(x,y\right)={\mathrm{e}}^{-\beta {g}_2\left(\min \left\{x,y\right\};c\right)} \) is the singular part and \( {F}_{\mathrm{BIEF}(a)}\left(x,y\right)=\frac{\beta }{\beta_1+{\beta}_2}{\mathrm{e}}^{-{\beta}_1{g}_2\left(x;c\right)-{\beta}_2{g}_2\left(y;c\right)-{\beta}_0{g}_2\left(\min \left\{x,y\right\};c\right)}-\frac{\beta_0}{\beta_1+{\beta}_2}{\mathrm{e}}^{-\beta {g}_2\left(\min \left\{x,y\right\};c\right)} \) is the absolute continuous part.
Proof (i) For the BEF, using the fact that \( {\overline{F}}_{\mathrm{BEF}}\left(x,y\right)=\alpha {\overline{F}}_{\mathrm{BEF}(a)}\left(x,y\right)+\left(1-\alpha \right){\overline{F}}_{\mathrm{BEF}(s)}\left(x,y\right) \)
$$ \frac{\partial^2{\overline{F}}_{\mathrm{BEF}}\left(x,y\right)}{\partial x\partial y}=\alpha {f}_{\mathrm{BEF}(a)}\left(x,y\right)=\left\{\begin{array}{c}{f}_{\mathrm{EF}}\left(x;{\theta}_0+{\theta}_1,c\right){f}_{\mathrm{EF}}\left(y;{\theta}_2,c\right),\mathrm{if}\ x>y\\ {}{f}_{\mathrm{EF}}\left(x;{\theta}_1,c\right){f}_{\mathrm{EF}}\left(y;{\theta}_0+{\theta}_2,c\right),\mathrm{if}\ x<y\end{array}\right. $$
Hence α may be obtained as
$$ \alpha =\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{x}{\int }}{f}_{\mathrm{EF}}\left(x;{\theta}_0+{\theta}_1,c\right){f}_{\mathrm{EF}}\left(y;{\theta}_2,c\right) dydx+\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{y}{\int }}{f}_{\mathrm{EF}}\left(x;{\theta}_1,c\right){f}_{\mathrm{EF}}\left(y;{\theta}_0+{\theta}_2,c\right) dxdy=\frac{\theta_1+{\theta}_2}{\theta }, $$
and \( {\overline{F}}_{\mathrm{BEF}(a)}\left(x,y\right)=\underset{y}{\overset{\infty }{\int }}\underset{x}{\overset{\infty }{\int }}{f}_{\mathrm{BEF}(a)}\left(u,v\right) dudv \); hence, with α and \( {\overline{F}}_{\mathrm{BEF}(a)}\left(x,y\right) \) known, the singular part \( {\overline{F}}_{\mathrm{BEF}(s)}\left(x,y\right) \) can be obtained by subtraction.
(ii) Similarly for the BIEF, FBIEF(a)(x, y) is computed by using FBIEF(x, y) = γFBIEF(a)(x, y) +(1 − γ)FBIEF(s)(x, y), 0 ≤ γ ≤ 1. Using a similar manner as in part (i), we can show that (8) holds. ∎
The marginal distributions of X and Y and the conditional distributions are given by Theorems 5 and 6, while the distributions of min{X, Y}, for the BEF, and max{X, Y}, for the BIEF, are given by Theorem 7.
Theorem 5 If the vector (X, Y) has either BEF(θ0, θ1, θ2; c) or BIEF(β0, β1, β2; c), then the marginal distributions of X and Y are either EF(θ0, θi; c) or IEF (β0, βi; c), i=1,2, respectively.
Proof If (X, Y) has BEF(θ0, θ1, θ2; c), then from (6) we have
$$ {f}_X(x)=\underset{0}{\overset{x}{\int }}{f}_1\left(x,y\right) dy+\underset{x}{\overset{\infty }{\int }}{f}_2\left(x,y\right) dy+{f}_0(x)=\left({\theta}_0+{\theta}_1\right){g}_1^{\prime}\left(x;c\right)\ {\mathrm{e}}^{-\left({\theta}_0+{\theta}_1\right){g}_1\left(x;c\right)}. $$
Similarly we can derive fY(y). In a similar manner, fX(x) and fY(y) can be shown to have IEF (β0, βi; c), i=1,2, respectively for the BIEF. ∎
Notice that the marginal distributions of X and Y can also be obtained using the next lemma.
Lemma 1
(i) Let X = min {U0, U1}, then X ∼ EF(θ0 + θ1; c) iff U0 and U1 are independent and U0 ∼ EF(θ0; c), U1 ∼ EF(θ1; c).
(ii) Let X = max {U0, U1}, then X ∼ IEF(β0 + β1; c) iff U0 and U1 are independent and U0 ∼ IEF(β0; c), U1 ∼ IEF(β1; c).
Here “∼” means follows or has the distribution.
Proof (i) for X = min {U0, U1}, we have
$$ P\left(X>x\right)=P\left(\min \left\{{U}_0,{U}_1\right\}>x\right)=P\left({U}_0>x,{U}_1>x\right). $$
If U0 and U1 are independent and U0 ~ EF(θ0; c) and U1 ~ EF(θ1; c) U1 ∼ EF(θ1; c), then
$$ P\left(X>x\right)=P\left({U}_0>x\right)P\left({U}_1>x\right)={e}^{-\left({\theta}_0+{\theta}_1\right){g}_1\left(x;c\right)}. $$
Conversely, if X ∼ EF(θ0 + θ1; c), then
$$ P\left(X>x\right)={e}^{-\left({\theta}_0+{\theta}_1\right){g}_1\left(x;c\right)}={e}^{-{\theta}_0{g}_1\left(x;c\right)}{e}^{-{\theta}_1{g}_1\left(x;c\right)}. $$
Then, U0 and U1 are independent and \( {\overline{F}}_{U_0}(x)={e}^{-{\theta}_0{g}_1\left(x;c\right)} \) and \( {\overline{F}}_{U_1}(x)={e}^{-{\theta}_1{g}_1\left(x;c\right)}, \) i.e. U0 ∼ EF(θ0; c) and U1 ∼ EF(θ1; c).
- (ii)
Similarly for the BIEF. ∎
Consequently, from Theorems 1 and 2 and Lemma 1, we have the following lemma, Lemma 2.
Lemma 2
(i) (X, Y) ∼ BEF(θ0, θ1, θ2; c) iff there exist independent EF random variables Ui, i = 0, 1, 2, such that X = min {U0, U1} and Y = min {U0, U2}.
(ii) (X, Y) ∼ BIEF(β0, β1, β2; c) if and only if there exist independent IEF random variables Ui, i = 0, 1, 2, such that X = max {U0, U1} and Y = max {U0, U2}.
Theorem 6 The conditional distribution of X given Y = y, is given by
$$ {f}_{X\mid Y}\left(x|y\right)=\left\{\begin{array}{c}\frac{\theta_2\left({\theta}_0+{\theta}_1\right)}{\theta_0+{\theta}_2}\ {g}_1^{\prime}\left(x;c\right)\ {\mathrm{e}}^{-\left({\theta}_0+{\theta}_1\right){g}_1\left(x;c\right)+{\theta}_0{g}_1\left(y;c\right)},\mathrm{if}\ x>y\\ {}{\theta}_1{g}_1^{\prime}\left(x;c\right)\ {\mathrm{e}}^{-{\theta}_1{g}_1\left(x;c\right)},\mathrm{if}\ x<y\\ {}\frac{\theta_0}{\theta_0+{\theta}_2}\ {\mathrm{e}}^{-{\theta}_1{g}_1\left(x;c\right)},\mathrm{if}\ x=y\end{array}\right. $$
(9)
for the BEF, while for the BIEF is given by
$$ {f}_{X\mid Y}\left(x|y\right)=\left\{\begin{array}{c}-{\beta}_1{g}_2^{\prime}\left(x;c\right){e}^{-{\beta}_1{g}_2\left(x;c\right)},\mathrm{if}\ x>y\\ {}\frac{-{\beta}_2\left({\beta}_0+{\beta}_1\right)}{\ \left({\beta}_0+{\beta}_2\right)}{g}_2^{\prime}\left(x;c\right){e}^{-\left({\beta}_0+{\beta}_1\right){g}_2\left(x;c\right)+{\beta}_0{g}_2\left(y;c\right)},\mathrm{if}\ x<y\\ {}\frac{\beta_0}{\beta_0+{\beta}_2}\ {\mathrm{e}}^{-{\beta}_1{g}_2\left(x;c\right)},\mathrm{if}\ x=y\end{array}\right. $$
(10)
Proof The proof is trivial so it is omitted.
Theorem 7 If (X, Y) is a bivariate vector of continuous random variables, then
- (i)
min{X, Y} ∼ EF(θ; c), if (X, Y) ∼ BEF(θ0, θ1, θ2; c),
- (ii)
$$ \max \left\{X,Y\right\}\sim \mathrm{IEF}\left(\beta; c\right),\mathrm{if}\ \left(X,Y\right)\sim \mathrm{BIEF}\left({\beta}_0,{\beta}_1,{\beta}_2;c\right). $$
Proof (i) if (X, Y) ∼ BEF(θ0, θ1, θ2; c), then using (4), we have
$$ P\left(\min \left\{X,Y\right\}>t\right)=P\left(X>t,Y>t\right)={e}^{-{\theta}_1{g}_1\left(t;c\right)-{\theta}_2{g}_1\left(t;c\right)-{\theta}_0{g}_1\left(t;c\right)}={e}^{-\theta {g}_1\left(t;c\right)}. $$
Similarly by using (5) for the BIEF, we can show that max{X, Y} ∼ IEF(β; c). ∎