# Further Results on Dynamic Additive Hazard Rate Model

Authors: Zhengcheng Zhang; Limin Zhang
Journal: Mathematical Problems in Engineering
 (2014)

---

## Abstract

In the past, the proportional and additive hazard rate models have been investigated in the works. Nanda and Das (2011) introduced and studied the dynamic proportional (reversed) hazard rate model. In this paper we study the dynamic additive hazard rate model, and investigate its aging properties for different aging classes. The closure of the model under some stochastic orders has also been investigated. Some examples are also given to illustrate different aging properties and stochastic comparisons of the model.


---

## Body

## 1. Introduction

It is common practice in statistical analysis that covariates are often introduced to account for factors that increase the heterogeneity of a population. When the effect of a factor under study has a multiplicative (or additive) effect on the baseline hazard function, we have a proportional (or an additive) hazard model. The latter category of model is preferred in any situation. For example, in tumorigenicity cases, where the dose effect on tumor risk is of interest, the excess risk becomes an important factor. Clinical trials that seek the effectiveness of treatments often experience lag times of treatment effectiveness after which treatment procedures will be in full effect.In reliability and survival analysis, devices or systems always operate in a changing environment. The conditions under which systems operate can be harsher or gentler in modeling lifetime of the devices or systems. The most known Cox [1] model is that the changing conditions are assumed to act multiplicatively on the baseline hazard rate. This model has been widely used in many experiments where the time to systems' failure depends on a group of covariates, which may be regarded as different treatments, operating conditions, heterogeneous environments, and so forth. P. L. Gupta and R. C. Gupta [2] studied the relation between the conditional and unconditional failure rates in mixtures when the distributions in the mixture follow the proportional hazard rate. For further research, one may see Cox and Oakes [3], Kumar and Westberg [4], Dupuy [5], Lau [6], Zhao and Zhou [7], X. Li and Z. Li [8], and Yu [9].R. C. Gupta and R. D. Gupta [10] proposed and studied the proportional reversed hazard model to analyze failure time data. For more details on this model, see Gupta and Wu [11], X. Li and Z. Li [12], and so forth.Recently, Nanda and Das [13] introduced the dynamic proportional hazard rate (DPHR) model and the dynamic proportional reversed hazard rate (DPRHR) model and studied their properties for different aging classes. The closure of the models under different stochastic orders has also been studied.Aranda-Ordaz [14] first dealt with an additive hazard model
(1)
h
(
t
∣
Z
(
s
)
,
s
≤
t
)
=
β
′
z
(
t
)
+
h
0
(
t
)
,
for
t
≥
0
,
where h
0
(
t
) is a baseline hazard rate and a time-dependent covariate vector Z, representing the changes in the operating conditions, and β is a vector of parameters. For more details, one may see Cox and Oakes [3], Thomas [15], Breslow and Day [16], Finkelstein and Esaulova [17], Lim and Zhang [18], and so forth.Assume thatX and Y are the lifetimes of two systems with corresponding hazard rate functions h
X
(
t
) and h
Y
(
t
) for t
≥
0. Let c
(
t
)
=
β
′
z
(
t
); the model (with time-dependent covariates) in (1) would reduce to the form
(2)
h
Y
(
t
)
=
c
(
t
)
+
h
X
(
t
)
,
∀
t
≥
0
,
which is named as dynamic additive hazard rate (DAHR) model.Sometimes the hazard rate functions ofX and Y may not be additive over the whole interval [
0
,
∞
), but they may be additive differently from different intervals. Specifically, they may be related as
(3)
h
Y
(
t
)
=
c
i
+
h
X
(
t
)
,
t
i
-
1
≤
t
≤
t
i
for i
=
1,2
,
…, and t
0
=
0, where c
i
(
i
=
1,2
,
…
) are some constants. When the intervals [
t
i
-
1
,
t
i
)
(
i
=
1,2
,
…
) become smaller and smaller, a model as in (2) will be naturally obtained.In order to guarantee thath
Y
(
t
) is a hazard rate function of a nonnegative random variable Y, the following lemma is given.Lemma 1.
Assume thatc
(
t
) and h
X
(
t
) are defined before. Then, for t
≥
0, h
Y
(
t
)
=
c
(
t
)
+
h
X
(
t
) is a hazard rate function if and only if the following conditions hold:(i)
c
(
t
)
+
h
X
(
t
)
≥
0, for all t
≥
0;
(ii)
∫
0
∞
‍
(
c
(
t
)
+
h
X
(
t
)
)
d
t
=
∞;
(iii)
if∫
0
t
0
‍
h
X
(
t
)
d
t
=
∞, then
(4)
∫
0
t
0
‍
(
c
(
t
)
+
h
X
(
t
)
)
d
t
=
∞
,for some t
0
<
∞.In Section2 of the paper, we discuss some aging properties of the DAHR model. In Section 3, the closure of DAHR model under different stochastic orderings is studied. Some examples are given to illustrate the results concerned in Sections 2 and 3.Throughout the paper, assume that all random variables under consideration have0 as the common left end point of their supports, and the terms increasing and decreasing stand for monotone nondecreasing and monotone nonincreasing, respectively.


## 2. Aging Properties of DAHR Model

At first we introduce some concepts of aging notions that will be useful in the section. Recall that a random variableX is said to be (a) increasing in failure rate (IFR) [decreasing in failure rate (DFR)] if h
X
(
t
) is increasing [decreasing] in t
≥
0; (b) increasing in failure rate in average (IFRA) [decreasing in failure rate in average (DFRA)] if ∫
0
t
‍
h
X
(
u
)
d
u
/
t is increasing [decreasing] in t
≥
0; (c) new better than used (NBU) [new worse than used (NWU)] if F
-
(
x
+
t
)
≤
[
≥
]
F
-
(
t
)
F
-
(
x
), for all t
,
x
≥
0; (d) new better than used in failure rate (NBUFR) [new worse than used in failure rate (NWUFR)] if h
X
(
t
)
≥
(
≤
)
h
X
(
0
), for all t
≥
0; (e) new better than used in failure rate average (NBAFR) [new worse than used in failure rate average (NWAFR)] if ∫
0
t
‍
h
X
(
u
)
d
u
/
t
≥
[
≤
]
h
X
(
0
), for all t
≥
0. For more discussions on properties of aging notions, readers may refer to Barlow and Proschan [19], Müller and Styan [20], and so forth.In the following we give some aging closure properties between the random variablesX and Y under some conditions of c
(
t
). Some results are obvious and hence their proofs are omitted.Proposition 2.
If the random variableX is IFR (DFR) and, for t
≥
0, c
(
t
) is increasing (decreasing), then the random variable Y is IFR (DFR).In the following, we give two examples related to this proposition. Example3 is an application of the proposition. Example 4 indicates that the condition of c
(
t
) is sufficient but not a necessary one.Example 3.
LetX be a random variable having Weibull distribution with hazard rate function h
X
(
t
)
=
2
t, t
≥
0. Take c
(
t
)
=
t for t
≥
0. It is obvious that c
(
t
) satisfies all the conditions of Lemma 1. Obviously, if X is IFR and c
(
t
) is increasing in t, hence Y is IFR.Example 4.
LetX be a random variable having Weibull distribution with hazard rate function h
X
(
t
)
=
2
t, t
≥
0. Let c
(
t
)
=
(
2
+
t
2
)
/
(
1
+
t
) for t
≥
0. It can be verified that h
X
(
t
)
+
c
(
t
) is increasing in t
≥
0, and hence Y is IFR. However, c
(
t
) is decreasing in t
∈
[
0
,
3
-
1
) but increasing in t
∈
[
3
-
1
,
+
∞
).Proposition 5.
If the random variableX is IFRA (DFRA) and c
(
t
) is increasing (decreasing) in t
≥
0, then the random variable Y is IFRA (DFRA).Proof.
Fort
≥
0, let
(5)
q
(
t
)
=
∫
0
t
‍
h
Y
(
x
)
d
x
t
=
∫
0
t
‍
(
c
(
x
)
+
h
X
(
x
)
)
d
x
t
.
Note that X is IFRA (DFRA) and c
(
t
) is increasing (decreasing) implying that
(6)
q
′
(
t
)
=
c
(
t
)
+
h
X
(
t
)
t
-
∫
0
t
‍
(
c
(
x
)
+
h
X
(
x
)
)
d
x
t
2
=
∫
0
t
‍
(
c
(
t
)
-
c
(
x
)
)
d
x
t
2
+
t
h
X
(
t
)
-
∫
0
t
‍
h
X
(
x
)
d
x
t
2
≥
0
(
≤
0
)
.
Hence the desired result follows directly.Example3 can be regarded as an application of the above proposition. Example 6 below indicates that the condition of c
(
t
) is sufficient but not a necessary one for the monotone property of Y.Example 6.
LetX be a random variable having Weibull distribution with hazard rate function h
X
(
t
)
=
2
t, t
≥
0. Take c
(
t
)
=
-
t for t
≥
0. It is obvious that c
(
t
) satisfies all the conditions of Lemma 1. Obviously, X is IFRA and Y is IFRA. However, c
(
t
) is decreasing in t
≥
0.Proposition 7.
If the random variableX is NBU (NWU) and c
(
t
) is increasing (decreasing) in t
≥
0, then the random variable Y is NBU (NWU).Proof.
We only give the proof for the case of NBU. In order to prove thatY is NBU, it is sufficient to prove that, for all t
≥
0 and x
≥
0,
(7)
e
-
∫
0
x
+
t
‍
(
c
(
u
)
+
h
X
(
u
)
)
d
u
≤
e
-
∫
0
x
‍
(
c
(
u
)
+
h
X
(
u
)
)
d
u
×
e
-
∫
0
t
‍
(
c
(
u
)
+
h
X
(
u
)
)
d
u
.
It is equivalent to
(8)
e
-
∫
t
x
+
t
‍
(
c
(
u
)
+
h
X
(
u
)
)
d
u
≤
e
-
∫
0
x
‍
(
c
(
u
)
+
h
X
(
u
)
)
d
u
.
That is,
(9)
e
-
∫
0
x
‍
(
c
(
u
+
t
)
+
h
X
(
u
+
t
)
)
d
u
≤
e
-
∫
0
x
‍
(
c
(
u
)
+
h
X
(
u
)
)
d
u
.
Note thatX is NBU which implies that
(10)
e
-
∫
0
x
+
t
‍
h
X
(
u
)
d
u
≤
e
-
∫
0
x
‍
h
X
(
u
)
d
u
·
e
-
∫
0
t
‍
h
X
(
u
)
d
u
.
That is,
(11)
e
-
∫
0
x
‍
h
X
(
u
+
t
)
d
u
≤
e
-
∫
0
x
‍
h
X
(
u
)
d
u
.
From the fact thatc
(
t
) is increasing and (11), (9) holds, and hence the desired result follows.Example3 is an application of the above proposition. The following example indicates that the condition of c
(
t
) is sufficient but not a necessary one for the NBU property of Y.Example 8.
Assume thatX is a random variable having exponential distribution with mean 1/2. It is clear that X is NBU. Let c
(
t
)
=
(
1
+
t
)
/
(
1
+
t
2
) for t
≥
0. By some computations, we have
(12)
a
(
t
,
x
)
=
∫
0
x
‍
(
c
(
t
+
u
)
-
c
(
u
)
+
h
X
(
t
+
u
)
-
h
X
(
u
)
)
d
u
=
arctan
(
t
+
x
)
+
1
2
ln
⁡
(
1
+
(
t
+
x
)
2
)
-
arctan
x
+
1
2
ln
⁡
(
1
+
x
2
)
+
2
x
t
.
It can be verified that a
(
t
,
x
) is nonnegative for t
,
x
≥
0 (see also Figure 1). From (9), we conclude that Y is NBU. However, it is easily obtained that c
(
t
) is increasing in [
0
,
2
-
1
) but decreasing in (
2
-
1
,
+
∞
).Figure 1
Plot of thea
(
t
,
x
) for (
x
,
t
)
∈
[
0,100
]
×
[
0,80
].Proposition 9.
If the random variableX is NBUFR (NWUFR) and c
(
t
)
≥
0
(
≤
0
) for t
≥
0, then the random variable Y is NBUFR (NWUFR).Proposition 10.
If the random variableX is NBAFR (NWAFR) and ∫
0
t
‍
c
(
u
)
d
u
≥
(
≤
)
t
c
(
0
) for t
≥
0, then the random variable Y is NBAFR (NWAFR).Proof.
We only give the proof for the case of NBAFR. It is noted thatY is NBAFR which is equivalent to that, for all t
≥
0, (
∫
0
t
‍
(
c
(
u
)
+
h
X
(
u
)
)
d
u
)
/
t
=
(
∫
0
t
‍
h
Y
(
u
)
d
u
)
/
t
≥
h
Y
(
0
)
=
c
(
0
)
+
h
X
(
0
). Note that X is NBAFR if and only if ∫
0
t
‍
h
X
(
u
)
d
u
/
t
≥
h
X
(
0
). Hence the desired result follows from the condition ∫
0
t
‍
c
(
u
)
d
u
≥
t
c
(
0
).Remark 11.
Example3 is an application of Propositions 9 and 10. Example 6 can be regarded as a counterexample, which shows that the condition c
(
t
)
≥
0 is a sufficient but not a necessary one in Propositions 9 and 10.


## 3. Stochastic Comparisons of DAHR Model

Firstly let us recall the concepts of some stochastic orders that are closely related to the main results in this section. A random variableX is said to be larger than another random variable Y in (a) aging intensity ordering (denoted by X
≥
a
i
Y), if
(13)
h
X
(
t
)
∫
0
t
‍
h
X
(
u
)
d
u
≤
h
Y
(
t
)
∫
0
t
‍
h
Y
(
u
)
d
u
,
for all t
≥
0; (b) usual stochastic order (denoted by X
≤
s
t
Y) if F
-
X
(
t
)
≤
F
-
Y
(
t
), for all t
≥
0; (c) hazard rate order (denoted by X
≤
h
r
Y) if h
X
(
t
)
≥
h
Y
(
t
), for all t
≥
0; (d) up hazard rate order (denoted by X
≤
h
r
↑
Y ) if X
-
t
≤
h
r
Y, for all t
≥
0; (e) down hazard rate order (denoted by X
≤
h
r
↓
Y) if X
≤
h
r
[
Y
-
t
∣
Y
>
t
], for all t
≥
0. For more details about stochastic orders, please refer to Shaked and Shanthikumar [21].In the following we give some sufficient (and necessary) conditions of stochastic ordering between random variablesX and Y. Some results are obvious and hence their proofs are omitted.Proposition 12.
SupposeX and Y are two nonnegative random variables satisfying (2). Then, X
≥
a
i
(
≤
a
i
)
Y if c
(
t
)
/
h
X
(
t
) is increasing (decreasing ) in t
≥
0.Proof.
Note thatX
≥
a
i
Y if and only if, for all t
≥
0,
(14)
h
X
(
t
)
∫
0
t
‍
h
X
(
u
)
d
u
≤
c
(
t
)
+
h
X
(
t
)
∫
0
t
‍
(
c
(
u
)
+
h
X
(
u
)
)
d
u
.
It is equivalent to that ∫
0
t
‍
(
c
(
t
)
h
X
(
u
)
-
c
(
u
)
h
X
(
t
)
)
d
u
≥
0. It holds if c
(
t
)
/
h
X
(
t
) is increasing in t
≥
0. The proof of the parenthetical statement is similar.The following example indicates that the condition of the monotone property of thec
(
t
)
/
h
X
(
t
) is sufficient but not a necessary one for the aging intensity ordering between X and Y.Example 13.
Assume thatX is a random variable having exponential distribution with mean 1/2. Let c
(
t
)
=
(
1
+
t
2
)
/
(
1
+
t
) for t
≥
0. By some computations, we have
(15)
a
(
t
)
=
∫
0
t
‍
(
c
(
t
)
h
X
(
u
)
-
c
(
u
)
h
X
(
t
)
)
d
u
=
2
t
(
1
+
t
2
)
1
+
t
-
t
2
+
2
t
+
4
ln
⁡
(
1
+
t
)
.
It can be verified that a
′
(
t
)
≥
0 for t
≥
0, and hence a
(
t
) is increasing in t
≥
0 (see also Figure 2). Note that a
(
0
)
=
0. Thus a
(
t
)
≥
0, for all t
≥
0, and hence X
≥
a
i
Y. However, it is easily obtained that c
(
t
)
/
h
X
(
t
) is decreasing in [
0
,
2
-
1
) but increasing in (
2
-
1
,
+
∞
).Figure 2
Plot of thea
(
t
) for t
∈
[
0,2
].Proposition 14.
SupposeX and Y are two nonnegative random variables satisfying (2). Then, X
≥
s
t
(
≤
s
t
)
Y if and only if ∫
0
t
‍
c
(
u
)
d
u
≥
(
≤
)
0, for all t
≥
0.The following corollary follows immediately from the proposition above.Corollary 15.
Ifc
(
t
)
≥
(
≤
)
0, for all t
≥
0, then X
≥
s
t
(
≤
s
t
)
Y.Proposition 16.
SupposeX and Y are two nonnegative random variables satisfying (2). Then, X
≥
h
r
(
≤
h
r
)
Y if and only if c
(
t
)
≥
(
≤
)
0, for all t
≥
0.Proposition 17.
Suppose thatX and Y are two nonnegative random variables satisfying (2). Then, X
≤
h
r
↑
(
≥
h
r
↑
)
Y if and only if h
X
(
y
+
t
)
-
h
X
(
t
)
-
c
(
t
)
≥
(
≤
)
0, for all y
≥
0 and t
≥
0.Proof.
Note thatX
≤
h
r
↑
Y if and only if
(16)
exp
⁡
[
-
∫
0
x
‍
h
Y
(
u
)
d
u
]
exp
⁡
[
-
∫
0
x
+
t
‍
h
X
(
u
)
d
u
]
is increasing in x, for all t
≥
0. It is equivalent to the fact that
(17)
exp
⁡
[
∫
0
x
+
t
‍
h
X
(
u
)
-
∫
0
x
‍
h
Y
(
u
)
d
u
]
is increasing in x, which is equivalent to the fact that its derivative is nonnegative; that is, h
X
(
y
+
t
)
-
h
X
(
t
)
-
c
(
t
)
≥
0, for all y
≥
0 and t
≥
0. It follows from the condition. The proof of the parenthetical statement is similar.Proposition 18.
Suppose thatX and Y are two nonnegative continuous random variables satisfying (2). Then, X
≤
h
r
↓
(
≥
h
r
↓
)
Y if and only if h
X
(
y
)
-
h
X
(
t
+
y
)
-
c
(
t
+
y
)
≥
(
≤
)
0, for all y
≥
0 and t
≥
0.Its proof is similar to that of Proposition17 and hence is omitted.



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*Source: 752831-2014-07-21.xml*