Journal of Statistical Planning and Inference 139 (2009) 2084 -- 2090

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Journal of Statistical Planning and Inference

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Stochastic comparisons of multivariate frailty models

Neeraj Misraa, Nitin Guptaa,∗, Rameshwar D. Guptab

aDepartment of Mathematics and Statistics, Indian Institute of Technology Kanpur, Kanpur 208 016, IndiabDepartment of Computer Science and Applied Statistics, University of New Brunswick, St. John, New Brunswick, Canada E2L 4L5

A R T I C L E I N F O A B S T R A C T

Article history:Received 29 April 2008Received in revised form16 September 2008Accepted 20 September 2008Available online 2 October 2008

MSC:60E1562N05

Keywords:Hazard rate orderIncreasing concave orderLikelihood ratio orderUpper orthant orderUsual stochastic orderReversed hazard rate order

A multivariate frailty model in which survival function depends on baseline distributions ofcomponents and the frailty random variable is considered. Since misspecification in choiceof frailty distribution and/or baseline distribution may affect the distribution of multivariatefrailtymodel, using theory of stochastic orders, we comparemultivariate frailtymodels arisingfrom different choices of frailty distribution.

© 2008 Elsevier B.V. All rights reserved.

1. Introduction to frailty models

The proportional hazard model is widely used in reliability theory and survival analysis. It may happen in some situationsthat some factors (also called covariates) are typically unknown and hence cannot be explicitly included in the analysis. Theproportional hazard rate model may be inappropriate in such situations. Frailty models provide an alternative to proportionalhazard rate model where misspecified or omitted covariates are described by an unobservable random variable, called the frailtyrandom variable. Although frailty models are mainly studied in the context of survival analysis, they seem to be appropriate invarious reliability studies also.

Frailty models have two versions, namely, univariate frailty models and multivariate frailty models. In the sequel, we providea brief discussion on multivariate frailty models in the context of reliability theory. In the following discussion, for a positiveinteger n, Rn

+ will denote the product space [0,∞)n and Rn will denote the n-dimensional Euclidean space.Consider a system consisting of n components, which have random lifetimes X∗

1,X∗2, . . . ,X

∗n in an ideal environment. Let Fi, fi

and Fi =1−Fi be the survival function, the Lebesgue probability density function and the distribution function, respectively, of X∗i ,

i = 1, 2, . . . ,n. Suppose that these components share the same random environment described by non-negative random variableV (say, the shocks of random intensity V). Suppose that the random variable V has the Lebesgue probability density functionh, distribution function H, and survival function H = 1 − H. Under the random environment, let the lifetimes of componentsbe denoted by X1,X2, . . . ,Xn. In the shared frailty model (see Clayton, 1978; Hougaard, 2000), conditional on V = v, the joint

∗ Corresponding author. Tel.: +915122597636; fax: +915122597500.E-mail addresses: neeraj@iitk.ac.in (N. Misra), guptan@iitk.ac.in (N. Gupta), gupta@unbsj.ca (R.D. Gupta).

0378-3758/$ - see front matter © 2008 Elsevier B.V. All rights reserved.doi:10.1016/j.jspi.2008.09.006

N. Misra et al. / Journal of Statistical Planning and Inference 139 (2009) 2084 -- 2090 2085

conditional survival function of random vector X = (X1, . . . ,Xn) is given by

FX|V (t|v) = P(X1 > t1, . . . ,Xn > tn|V = v) =n∏

i=1

(Fi(ti))v, t ∈ Rn, v�0,

where t=(t1, t2, . . . , tn). Thismeans that, given the random frailty, the lifetimes of components are statistically independent.We callV (or its distribution) the frailty randomvariable (or frailty distribution) and randomvariables X∗

1,X∗2, . . . ,X

∗n (or their distributions)

baseline random variables (or baseline distributions). Clearly, the survival function of random vector X = (X1, . . . ,Xn) is given by

F(t) = P(X1 > t1, . . . ,Xn > tn) = E

⎛⎜⎝

⎛⎝ n∏

i=1

Fi(ti)

⎞⎠

V⎞⎟⎠ , t ∈ Rn. (1.1)

The frailty model (1.1) is a multivariate extension of univariate frailty model, introduced by Vaupel et al. (1979) in the contextof survival analysis. For a survey of the literature on frailty models and their applications in survival analysis, one may refer toHougaard (2000) and references cited therein.

Gupta and Kirmani (2006) and Xu and Li (2008) made stochastic comparisons between univariate frailty models, arising fromdifferent choices of frailty/baseline distributions, and also studied their aging properties.

In this paper, using theory of stochastic orders, we comparemultivariate frailty models arising from different choices of frailtydistribution. Now, we introduce some notations and definitions which will be used throughout this paper.

For vectors x= (x1, . . . , xn), y= (y1, . . . , yn) ∈ Rn, the notation x�ymeans that xi�yi, i=1, 2, . . . ,n, and the notation y >xmeansthat xi < yi, i = 1, 2, . . . ,n. For real numbers x and y, x ∧ y (x ∨ y) denote the minimum (maximum) of x and y. Also, for x, y ∈ Rn,x ∧ y = (x1 ∧ y1, x2 ∧ y2, . . . , xn ∧ yn) and x ∨ y = (x1 ∨ y1, x2 ∨ y2, . . . , xn ∨ yn).

All n-dimensional (where n is a positive integer) random vectors, considered in this paper, will be assumed to have supportRn

+ and will be denoted by bold letters (e.g., Z). Moreover, any one-dimensional random vector will be referred to as a randomvariable and will be denoted by non-bold letter (e.g., Z). Similar notation of bold and non-bold letters will be used for scalar(non-random) elements of Rn and R1, respectively. A function � : A → R1, A ⊆ Rn, is said to be increasing (decreasing) on A if�(x)� (� )�(y), whenever x, y ∈ A and x�y.

For completeness in the presentation, we provide standard definitions of those related stochastic orders. Details of thesenotions can be found in Shaked and Shanthikumar (2007) and Mullar and Stoyan (2002).

Let S=(S1, S2, . . . , Sn) (T=(T1, T2, . . . , Tn)) be an n-dimensional randomvectorwith distribution functionG1(x)=P(S�x) (G2(x)=P(T�x)), x ∈ Rn, survival function G1(x) = P(S >x) (G2(x) = P(T >x)), x ∈ Rn, and the Lebesgue probability density functiong1(x) (g2(x)), x ∈ Rn.

Random vector S is said to be smaller than T in the

(a) usual stochastic (st) order (written as S� stT) if

E(�(S))�E(�(T)),

for all increasing functions � : Rn → R1 for which the indicated expectations exist;(b) upper orthant (uo) order (written as S�uoT) if

G1(x)�G2(x) for every x ∈ Rn;

(c) likelihood ratio (lr) order (written as S� lrT) if

g1(x)g2(y)�g1(x ∧ y)g2(x ∨ y) for all x, y ∈ Rn;

(d) hazard rate (hr) order (written as S�hrT) if

G1(x)G2(y)�G1(x ∧ y)G2(x ∨ y) for all x, y ∈ Rn;

(e) reversed hazard rate (rh) order (written as S� rhT) if

G1(x)G2(y)�G1(x ∧ y)G2(x ∨ y) for all x, y ∈ Rn;

(f) increasing concave (icv) order (written as S� icvT) if

E(�(S))�E(�(T)),

for all increasing concave functions � : Rn → R1 for which the indicated expectations exist.

2086 N. Misra et al. / Journal of Statistical Planning and Inference 139 (2009) 2084 -- 2090

For random vectors S and T, it is known that

(i) S� lrT ⇒ S� stT ⇒ S�uoT ⇒ E(S)�E(T);(ii) S� lrT ⇒ S�hrT ⇒ S�uoT; and(iii) S� lrT ⇒ S� rhT.

2. Stochastic orders on multivariate frailty models

We will compare frailty models

F(x) = E

⎛⎜⎝

⎛⎝ n∏

i=1

Fi(xi)

⎞⎠

V1⎞⎟⎠ , x ∈ Rn (2.1)

and

G(x) = E

⎛⎜⎝

⎛⎝ n∏

i=1

Fi(xi)

⎞⎠

V2⎞⎟⎠ , x ∈ Rn, (2.2)

arising from different choices of frailty random variables in model (1.1). Suppose that frailty random variable V1 (V2) has theLebesgue probability density function h1 (h2), distribution function H1 (H2), and survival function H1 (H2). For frailty models(2.1) and (2.2), let corresponding random vectors be X = (X1,X2, . . . ,Xn) and Y = (Y1,Y2, . . . ,Yn), respectively. Since the resultsderived in this paper depend on the joint distributions of (V1,V2) only through marginal distributions of V1 and V2, without lossof generality we assume that V1 and V2 are statistically independent.

Let us consider a situation in which two companies, say A and B, are manufacturing machines consisting of similar com-ponents. Each of these machines from company A (company B) are made up of n-components having random lifetimesX∗1,X

∗2, . . . ,X

∗n (Y∗

1,Y∗2, . . . ,Y

∗n), where random variables X∗

i and Y∗i , i = 1, . . . ,n, have the same distribution. Suppose that the ma-

chines manufactured by two companies are installed in different random environments (i.e., different frailty random variablesV1 and V2). In such situations it will be interesting to investigate how a stochastic ordering between random variables V1 and V2translates into a stochastic ordering between random vectors X and Y.

Random vectors X and Y have probability density functions

f (x) =∫ ∞

0vn

⎧⎨⎩

n∏i=1

(fi(xi)Fv−1i (xi))

⎫⎬⎭h1(v) dv, x ∈ Rn (2.3)

and

g(x) =∫ ∞

0vn

⎧⎨⎩

n∏i=1

(fi(xi)Fv−1i (xi))

⎫⎬⎭h2(v) dv, x ∈ Rn, (2.4)

respectively. The following theorem deals with the situationwhere random variables V1 and V2 are ordered in terms of likelihoodratio order.

Theorem 2.1. For frailty models (2.1) and (2.2), if V1� lrV2, then Y� lrX.

Proof. For fixed x, y ∈ Rn, define �1 = f (x ∨ y)g(x ∧ y) − f (y)g(x). Using (2.3) and (2.4), we have

�1 = E(�2(V1,V2)) − E(�1(V1,V2)),

where, for (x, y) ∈ R2+,

�1(x, y) = xnynn∏

i=1

{fi(xi)fi(yi)F

x−1i (yi)F

y−1i (xi)

}

and

�2(x, y) = xnynn∏

i=1

{fi(xi)fi(yi)F

x−1i (xi ∨ yi)F

y−1i (xi ∧ yi)

}.

N. Misra et al. / Journal of Statistical Planning and Inference 139 (2009) 2084 -- 2090 2087

Define ��21(x, y) = �2(x, y) − �1(x, y), (x, y) ∈ R2+. If we can show that ��21(x, y)�0 and ��21(x, y)� − ��21(y, x), whenever

0�x < y <∞, then using Theorem 1.C.22 of Shaked and Shanthikumar (2007, p. 51), it would follow that�1�0 and hence Y� lrX.Let x and y be real numbers such that 0�x < y <∞. Then,

��21(x, y) = xnyn

⎡⎣ n∏i=1

{fi(xi)fi(yi)Fx−1i (xi)F

x−1i (yi)}

⎤⎦

⎡⎣ n∏i=1

Fy−xi (xi ∧ yi) −

n∏i=1

Fy−xi (xi)

⎤⎦

and

��21(x, y) + ��21(y, x) = xnyn

⎡⎣ n∏i=1

{fi(xi)fi(yi)Fx−1i (xi)F

x−1i (yi)}

⎤⎦

×⎡⎣ n∏i=1

Fy−xi (xi ∧ yi) +

n∏i=1

Fy−xi (xi ∨ yi) −

n∏i=1

Fy−xi (xi) −

n∏i=1

Fy−xi (yi)

⎤⎦ .

Clearly ��21(x, y)�0. To show that ��21(x, y) + ��21(y, x)�0, consider the following three cases:Case I: ��21(x, y) + ��21(y, x) = 0 for xi�yi, i = 1, . . . ,n;Case II: ��21(x, y) + ��21(y, x) = 0 for yi�xi, i = 1, . . . ,n;Case III: for some l ∈ {1, 2, . . . ,n−1} and a permutation (i1, . . . , in) of (1, . . . ,n),−∞ < xij �yij <∞, j=1, . . . , l, and−∞ < yij �xij <∞,

j = l + 1, . . . ,n.In this case,

��21(x, y) + ��21(y, x) = xnyn

⎡⎣ n∏i=1

{fi(xi)fi(yi)Fx−1i (xi)F

x−1i (yi)}

⎤⎦

⎡⎣ l∏j=1

Fy−xij (xij ) −

l∏j=1

Fy−xij (yij )

⎤⎦

×⎡⎣ n∏j=l+1

Fy−xij (yij ) −

n∏j=l+1

Fy−xij (xij )

⎤⎦

� 0

since 0 < Fij (yij )�Fij (xij ), j = 1, . . . , l, 0 < Fij (xij )�Fij (yij ), j = l + 1, . . . ,n and y − x >0. �

The following theorem deals with the situation where random variables V1 and V2 are ordered in terms of reversed hazardrate ordering.

Theorem 2.2. For frailty models (2.1) and (2.2), if V1� rhV2, then Y�hrX.

Proof. For fixed x, y ∈ Rn, define �2 = F(x ∨ y)G(x ∧ y) − F(y)G(x). Then

�2 = E(�2(V1,V2)) − E(�1(V1,V2)),

where, for (x, y) ∈ R2+,

�1(x, y) =n∏

i=1

(Fxi (yi)F

yi (xi)) and �2(x, y) =

n∏i=1

(Fxi (xi ∨ yi)F

yi (xi ∧ yi)).

Define, ��21(x, y) = �2(x, y) − �1(x, y), (x, y) ∈ R2+. If we can show that, for each fixed y ∈ R1

+, ��21(x, y) decreases in x on [0, y)and ��21(x, y)� − ��21(y, x), whenever 0�x < y <∞, then using Theorem 1.B.48 of Shaked and Shanthikumar (2007, p. 38) itwould follow that �2�0 and hence Y�hrX. For 0�x < y <∞,

��x

��21(x, y) =⎧⎨⎩

n∏i=1

(Fxi (xi)F

xi (yi))

⎫⎬⎭

⎡⎣

⎧⎨⎩

n∏i=1

Fy−xi (xi ∧ yi)

⎫⎬⎭ ln

⎛⎝ n∏

i=1

Fi(xi ∨ yi)

⎞⎠ −

⎧⎨⎩

n∏i=1

Fy−xi (xi)

⎫⎬⎭ ln

⎛⎝ n∏

i=1

Fi(yi)

⎞⎠

⎤⎦

and

��21(x, y) + ��21(y, x) =⎧⎨⎩

n∏i=1

(Fxi (xi)F

xi (yi))

⎫⎬⎭

⎡⎣ n∏i=1

Fy−xi (xi ∧ yi) +

n∏i=1

Fy−xi (xi ∨ yi) −

n∏i=1

Fy−xi (xi) −

n∏i=1

Fy−xi (yi)

⎤⎦ .

2088 N. Misra et al. / Journal of Statistical Planning and Inference 139 (2009) 2084 -- 2090

In a similar manner to the proof of Theorem 2.1, it holds that

��x

��21(x, y)�0 and ��21(x, y) + ��21(y, x)�0 whenever 0�x�y <∞.

Hence the result follows. �

The following lemma will be useful in proving the next result of this section.

Lemma 2.1. (i) For 0 <� <�, �1(x) = (1 − x�)/(1 − x�) is increasing in x ∈ (0, 1).(ii) For � >0, �2(x) = x�(− ln x)/(1 − x�) is increasing in x ∈ (0, 1).

Proof. (i) For 0 < x <1,

�′1(x) = d

dx�1(x) = x�−1�3(x)

(1 − x�)2,

where �3(x) = (� − �)x� − �x�−� + �, 0 < x <1. Clearly �′3(x) = �(� − �)x�−�−1(x� − 1) <0, ∀0 < x <1. It follows that

�3(x)� limx→1−�3(x) = 0, ∀0 < x <1. Therefore �′1(x)�0, ∀0 < x <1.

(ii) Let 0 < z1�z2 <1. Consider the functions k1(x) = 1 − zx1, x ∈ R1+, and k2(x) = 1 − zx2, x ∈ R1

+. Using the generalized meanvalue theorem, it follows that

1 − z�11 − z�2

= k1(�) − k1(0)k2(�) − k2(0)

= k′1(�)

k′2(�)

= z�1(ln z1)

z�2(ln z2)

for some � ∈ (0,�). Since 0 < z1/z2�1, � ∈ (0,�) and ln z1/ln z2 >0,

1 − z�11 − z�2

= z�1(ln z1)

z�2(ln z2)�

z�1(ln z1)z�2(ln z2)

i.e.,z�2(− ln z2)1 − z�2

�z�1(− ln z1)1 − z�1

. �

Using (2.3), (2.4) and the Fubini theorem, it follows that random vectors X and Y have distribution functions

F(x) =∫ ∞

0

⎧⎨⎩

n∏i=1

(1 − Fvi (xi))

⎫⎬⎭h1(v) dv, x ∈ Rn (2.5)

and

G(x) =∫ ∞

0

⎧⎨⎩

n∏i=1

(1 − Fvi (xi))

⎫⎬⎭h2(v) dv, x ∈ Rn, (2.6)

respectively. In the following theorem we deal with the situation where random variables V1 and V2 are ordered in terms ofhazard rate ordering.

Theorem 2.3. For frailty models (2.1) and (2.2), if V1�hrV2, then Y� rhX.

Proof. For fixed x, y ∈ Rn, define�3=F(x∨y)G(x∧y)−F(y)G(x). If Fi(xi)=1 for some i ∈ {1, . . . ,n} or Fj(yj)=1 for some j ∈ {1, . . . ,n}then, using (2.5) and (2.6), F(x) = F(y) = G(x) = G(y) = 0 and so �3 = 0. Now suppose that, for every i, j ∈ {1, . . . ,n}, Fi(xi) <1 andFj(yj) <1. Then, using (2.5) and (2.6), we have

�3 = E(�2(V1,V2)) − E(�1(V1,V2)),

where, for (x, y) ∈ R2+, �1(x, y) = ∏n

i=1{(1 − Fxi (yi))(1 − F

yi (xi))} and �2(x, y) = ∏n

i=1{(1 − Fxi (xi ∨ yi))(1 − F

yi (xi ∧ yi))}. Define,

��21(x, y) = �2(x, y) − �1(x, y), (x, y) ∈ R2+. If we show that, for each fixed x ∈ R1

+, ��21(x, y) is an increasing function of y on(x,∞) and ��21(x, y)� − ��21(y, x), whenever 0�x < y <∞, then using Theorem 1.B.10 of Shaked and Shanthikumar (2007, p.22) it would follow that �3�0 and hence Y� rhX.

N. Misra et al. / Journal of Statistical Planning and Inference 139 (2009) 2084 -- 2090 2089

For 0�x < y <∞, we have

��y

��21(x, y) =⎛⎝ n∏

i=1

{(1 − Fxi (xi))(1 − F

xi (yi))}

⎞⎠

⎡⎣ n∏i=1

{1 − F

yi (xi ∧ yi)

1 − Fxi (xi ∧ yi)

}

×n∑

k=1

Fyk(xk ∧ yk)(− ln(Fk(xk ∧ yk)))

1 − Fyk(xk ∧ yk)

−n∏

i=1

{1 − F

yi (xi)

1 − Fxi (xi)

} n∑k=1

Fyk(xk)(− ln(Fk(xk)))

1 − Fyk(xk)

⎤⎦ (2.7)

and

��21(x, y) + ��21(y, x) =⎛⎝ n∏

i=1

{(1 − Fxi (xi))(1 − F

xi (yi))}

⎞⎠

⎡⎣ n∏i=1

{1 − F

yi (xi ∧ yi)

1 − Fxi (xi ∧ yi)

}+

n∏i=1

{1 − F

yi (xi ∨ yi)

1 − Fxi (xi ∨ yi)

}

−n∏

i=1

{1 − F

yi (xi)

1 − Fxi (xi)

}−

n∏i=1

{1 − F

yi (yi)

1 − Fxi (yi)

}⎤⎦ .

Using Lemma 2.1, along with the facts that 0 < Fi(xi)�Fi(xi ∧ yi) <1, i = 1, . . . ,n and 0�x < y <∞, we get

1 − Fyi (xi)

1 − Fxi (xi)

�1 − F

yi (xi ∧ yi)

1 − Fxi (xi ∧ yi)

, ∀i ∈ {1, . . . ,n} (2.8)

and

Fyk(xk)(− ln(Fk(xk)))

1 − Fyk(xk)

�Fyk(xk ∧ yk)(− ln(Fk(xk ∧ yk)))

1 − Fyk(xk ∧ yk)

, ∀k ∈ {1, . . . ,n}. (2.9)

Now, on using (2.8) and (2.9) in (2.7), we get

��y

��21(x, y)�0.

To show that ��21(x, y) + ��21(y, x)�0, consider the following three cases:Case I: ��21(x, y) + ��21(y, x) = 0 for xi�yi, i = 1, . . . ,n;Case II: ��21(x, y) + ��21(y, x) = 0 for yi�xi, i = 1, . . . ,n;Case III: For some l ∈ {1, 2, . . . ,n−1}andapermutation (i1, . . . , in) of (1, . . . ,n),−∞ < xij �yij <∞, j=1, . . . , l, and−∞ < yij �xij <∞, j=

l + 1, . . . ,n.In this case,

��21(x, y) + ��21(y, x) =⎛⎝ n∏

i=1

{(1 − Fxi (xi))(1 − F

xi (yi))}

⎞⎠

⎛⎝ l∏

j=1

⎧⎨⎩1 − F

yij (xij )

1 − Fxij (xij )

⎫⎬⎭ −

l∏j=1

⎧⎨⎩1 − F

yij (yij )

1 − Fxij (yij )

⎫⎬⎭

⎞⎠

×⎛⎝ n∏

j=l+1

⎧⎨⎩1 − F

yij (yij )

1 − Fxij (yij )

⎫⎬⎭ −

n∏j=l+1

⎧⎨⎩1 − F

yij (xij )

1 − Fxij (xij )

⎫⎬⎭

⎞⎠ . (2.10)

On using Lemma 2.1(i), along with the facts that 0 < Fij (yij )�Fij (xij ) <1, j = 1, . . . , l; 0 < Fij (xij )�Fij (yij ) <1, j = l + 1, . . . ,n, and0�x < y <∞, it follows that

1 − Fyij (yij )

1 − Fxij (yij )

�1 − F

yij (xij )

1 − Fxij (xij )

, j = 1, . . . , l (2.11)

and

1 − Fyij (xij )

1 − Fxij (xij )

�1 − F

yij (yij )

1 − Fxij (yij )

, j = l + 1, . . . ,n. (2.12)

Now, on using (2.11) and (2.12) in (2.10), we have ��21(x, y) + ��21(y, x)�0. �

In the following theorem, we provide conditions under which Y� stX holds.

2090 N. Misra et al. / Journal of Statistical Planning and Inference 139 (2009) 2084 -- 2090

Theorem 2.4. For frailty models (2.1) and (2.2), if V1� stV2, then Y� stX.

Proof. Let � : Rn → R be an increasing function, such that E(�(X)) and E(�(Y)) exist. Then, using (2.3), (2.4) and the Fubinitheorem, we get

�4 = E(�(X)) − E(�(Y))

=∫ ∞

0vnh1(v)

⎛⎝∫

Rn+�(t)

⎧⎨⎩

n∏i=1

Fv−1i (ti)fi(ti)

⎫⎬⎭ dt

⎞⎠ dv −

∫ ∞

0vnh2(v)

⎛⎝∫

Rn+�(t)

⎧⎨⎩

n∏i=1

Fv−1i (ti)fi(ti)

⎫⎬⎭ dt

⎞⎠ dv

=∫ ∞

0h1(v)

(∫[0,1]n

�(F−11 (u1/v1 ), . . . , F

−1n (u1/vn )) du

)dv −

∫ ∞

0h2(v)

(∫[0,1]n

�(F−11 (u1/v1 ), . . . , F

−1n (u1/vn )) du

)dv

= E(k(V2)) − E(k(V1)),

where dt = dt1, . . . , dtn, du = u1, . . . ,un and

k(v) = −∫[0,1]n

�(F−11 (u1/v1 ), . . . , F

−1n (u1/vn )) du, v�0.

Clearly k(v) is an increasing function of v on Rn+. Since V1� stV2, we conclude that �4 = E(k(V2)) − E(k(V1))�0, and therefore

Y� stX. �

To end this paper, we present a sufficient condition to Y�uoX hold.

Theorem 2.5. For frailty models (2.1) and (2.2), if V1� icvV2, then Y�uoX.

Proof. For fixed x ∈ Rn, consider

�5 = F(x) − G(x) = E(�(V2)) − E(�(V1)),

where�(v)=−(∏n

i=1Fi(xi))v, v ∈ R1

+. For fixed x ∈ Rn, since�(v) is an increasing and concave function of v on R1+, and V1� icvV2,

it follows that �5�0. Therefore F(x)�G(x), ∀x ∈ Rn, and the assertion is proved. �

3. Conclusions

Using theory of stochastic orders, we make stochastic comparisons between frailty models arising from different choices offrailty distribution. Gupta and Kirmani (2006) and Xu and Li (2008) obtained stochastic properties of univariate frailty models,which are the special cases of the multivariate frailty models considered in this paper. Theorems 2.1–2.3 are multivariategeneralization of results proved by Gupta and Kirmani (2006) and Xu and Li (2008).

Acknowledgements

The authors are grateful to the referee and the Associate Editor for their comments on the manuscript, which has improvedthe paper. The second author would like to acknowledge the financial assistance from the C.S.I.R., India, for carrying out thisresearch work.

References

Clayton, D.G., 1978. A model for association in bivariate life tables and its application in epidemiological studies of familial tendency in chronic disease incidence.Biometrika 65, 141–151.

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