2.1. NBUC

The new characterization of the NBUC property is given by the following lemma.

Lemma 2.1: F is NBUC iff

holds for all nonnegative, nondecreasing functions g, any nonnegative, nondecreasing convex function h, and any measurable function

.

Proof: First, we show the sufficiency. Let

, and take α(t) = (a/t) ∧ (a/(a + b)). Note that α(t)t ≥ a iff t ≥ a + b, and, in this case, we have h((1 − α(t))t) = t − a − b. Therefore,

On the other hand, the right-hand side is

Therefore, we obtain the inequality

Replace a with a + ε in the above inequality and let ε → 0+. This yields

that is, F is NBUC.

Next, we show the necessity. Suppose that F is NBUC. We first consider the case

. Among all the functions

, the function defined by α₀(t) = (a/t) ∧ (a/(a + b)) will make the integrand g(α(t))h((1 − α(t))t) the largest possible since it is not zero iff α(t)t ≥ a and (1 − α(t))t > b, and in this case, to make the value the largest possible, α(t) should be as small as possible. From the proof of the sufficiency part, it is easy to see that

for any

. Thus, if we let

where g_i ≥ 0, 0 ≤ a₁ ≤ ··· ≤ a_n < +∞, h_i ≥ 0, and 0 ≤ b₁ ≤ ··· ≤ b_m < +∞, then it follows that

Thus, we have shown that the result is correct for functions of the above form. Since any nonnegative, nondecreasing function is a nondecreasing limit of functions like g and any nonnegative, nondecreasing convex function is a nondecreasing limit of functions like h, it follows from the Lebesgue monotone convergence theorem that the inequality holds. █

Corollary 2.1: If F is NBUC, then

holds for all nonnegative, nondecreasing functions g, any nonnegative, nondecreasing convex function h, and any constant α ∈ [0,1].

Next, we consider the problem of preservation under mixture of the NBUC class. Analogous to Lynch [11], we require two conditions:

Remark: Note, by (NBUC1),

that is, F(t|θ) is an NBUC distribution for each θ ≥ 0.

Theorem 2.1: Let {F(t|θ),θ ≥ 0} and M satisfy (NBUC1) and (NBUC2). Then F_M is NBUC. Conversely, if F_M is NBUC whenever {F(t|θ),θ ≥ 0} satisfies (NBUC1), then M is NBUC.

Proof: The converse part follows by taking F(t|θ) = I_[0,θ+ε)(t) with ε > 0. Then F(t|θ) is increasing in

is convex in θ, and

where α(a,b,θ) = a ∧ θ/θ. Therefore, F(t|θ) satisfies (NBUC1). Since F_M is NBUC, it follows that M(t) = F_M(t + ε) satisfies

for all a ≥ ε (if we define M(t) = 1 for t < 0, this is true for all a ≥ 0). By letting ε → 0+, we conclude by the Lebesgue dominated convergence theorem that

Since

, we have

Now, replace the a above with a + ε and let ε → 0+. Thus,

that is, M is NBUC.

The other part of the proof is a direct application of Lemma 2.1. In fact,

In the second inequality, we applied Lemma 2.1 since M(θ) is NBUC,

is nonnegative, increasing, and convex in θ and F(t|θ) is increasing in θ for each t. Thus, F_M is NBUC. █

As an application of the above mixture preservation result, we rederive the preservation of NBUC under the convolution result of Hu and Xie [7].

Corollary 2.2: Suppose that F₁ and F₂ are two NBUC lifetime distributions. Then their convolution is also NBUC.

Proof: The survival function F of the convolution is given by

Then, by Fubini's theorem,

By the NBUC property of F₁, for s < a,

Let α(a,s) = (s ∧ a)/s. Then, for all s ≥ 0, we have

Note that F₁(t) is decreasing in t, and, therefore,

is an increasing, convex function of s. The desired result now follows as an application of Theorem 2.1. █

2.2. NBU(2)

The new characterization of the NBU(2) property is given by the following lemma.

Lemma 2.2: F is NBU(2) iff

for all nonnegative, nondecreasing functions g, any nonnegative, nondecreasing concave function h, and any measurable function

.

Proof: First, we show the sufficiency. Let g(t) = 1_[a,+∞)(t) and h(t) =

, and take α(t) = (a ∧ t)/t. We note that α(t)t ≥ a iff t ≥ a, and in this case, we have h((1 − α(t))t) = (t − a) ∧ b. Then

On the other hand,

Therefore, we must have

Replace a with a + ε in the above inequality and let ε → 0+ to conclude that

that is, F is NBU(2).

Next, we show the necessity. Suppose that F is NBU(2). First, we consider the case

. Among all the functions

, the function defined by α₀(t) = (a ∧ t)/t will make the integrand g(α(t))h((1 − α(t))t) for the above-defined g and h the largest possible since it is not zero iff α(t)t ≥ a, and in this case, to make the value the largest possible, α(t) should be as small as possible. Then from the proof of the sufficiency part, it is easy to see that

for any

. Thus, if we let

where g_i ≥ 0, 0 ≤ a₁ ≤ ··· ≤ a_n < +∞, and h_i ≥ 0, 0 ≤ b₁ ≤ ··· ≤ b_m < +∞, then it follows that

Thus we have shown that the result is true for functions of the above form. Since any nonnegative, nondecreasing function is a nondecreasing limit of functions like g and any nonnegative, nondecreasing concave function is a nondecreasing limit of functions like h, it follows from the Lebesgue monotone convergence theorem that the result is true. █

Corollary 2.3: If F is NBU(2), then

for all nonnegative, nondecreasing functions g, any nonnegative, nondecreasing concave function h, and any constant α ∈ [0,1].

Next, we consider the mixture preservation problem of NBU(2). For the NBU(2) preservation result, we require the following:

Remark: By (NBU(2)1), it is easy to see that

that is, F(t|θ) is an NBU(2) distribution for each θ ≥ 0.

Theorem 2.2: Let {F(t|θ),θ ≥ 0} and M satisfy (NBU(2)1) and (NBU(2)2). Then F_M is NBU(2). Conversely, if F_M is NBU(2) whenever {F(t|θ),θ ≥ 0} satisfies (NBU(2)1), then M is NBU(2).

Proof: The converse part follows by taking F(t|θ) = 1_[0,θ+ε)(t). Then, F(t|θ) is increasing in θ for each t and

is concave in θ. Moreover, for α(a,b,θ) = a ∧ θ/θ,

Therefore, F(t|θ) satisfies (NBU(2)1). By our assumption, F_M is NBU(2). Since M(t) = F_M(t + ε) in this case, we conclude that for a ≥ ε,

By letting ε → 0+, we have

Since

, we have

Now, replace a in the above inequality with a + ε and let ε → 0+. Thus,

that is, M is NBU(2).

The other part of the proof is a direct application of Lemma 2.2. In fact,

In the second inequality, we have used Lemma 2.2, noting that F(t|θ) is nondecreasing in θ, M(θ) is NBU(2), and

is nonnegative, nondecreasing, and concave in θ. Thus, F_M is NBU(2). █

As an application of the above mixture preservation result, we give another proof of the preservation of NBU(2) under the convolution result. This result appeared previously in Hu and Xie [7].

Corollary 2.4: Suppose that F₁ and F₂ are two NBU(2) lifetime distributions. Then their convolution is also NBU(2).

Proof: Analogous to the proof of Corollary 2.2, we let α(a,s) = (s ∧ a)/s. Consider

Since F₁ is NBU(2), we have

Now, note that F₁(t) is decreasing in t and that

is an increasing, concave function of s. The result then follows as a direct application of Theorem 2.2. █