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328 R. M. Mnatsakanov and F. H. Ruymgaart<br />

Corollary 4.1. Let us assume that (2.2) is valid. Consider � f ∗ α(x) defined in (3.15)<br />

with α(x) given by (3.14). Then<br />

(4.4)<br />

n1/2 α(x) 1/4{ � f ∗ �<br />

α(x)−f(x)}→d Normal [ W f(x)<br />

as n→∞ and f ′′ (x)�= 0.<br />

2 x2√π ]1/2 ,<br />

W f(x)<br />

2 x2√ �<br />

,<br />

π<br />

Proof. From (4.1) and (3.11) with α = α(x) defined in (3.13) it is easy to see that<br />

(4.5)<br />

n1/2 α(x) 1/4{ � f ∗ α(x)−E � f ∗ �<br />

α(x)} = Normal 0,<br />

W f(x)<br />

2 x2√ �<br />

π<br />

+ oP( 1<br />

),<br />

n2/5 as n→∞. Application of (3.4) where α = α(x) is defined by (3.14) yields (4.4).<br />

Corollary 4.2. Let us assume that (2.2) is valid. Consider � f ∗ α∗(x) defined in (3.15)<br />

with α∗ (x) given by<br />

(4.6) α ∗ (x) = n δ ·{<br />

π<br />

4·W 2}1/5<br />

�<br />

x3 · f ′′ �4/5 (x)<br />

� ,<br />

f(x)<br />

2<br />

5<br />

Then when f ′′ (x)�= 0, and letting n→∞, it follows that<br />

(4.7)<br />

n1/2 α∗ (x) 1/4{ � f ∗ α∗(x)−f(x)}→d �<br />

Normal 0,<br />

n1/2 α∗ (x) 1/4{ � f ∗ α∗(x)−E � f ∗ �<br />

α∗(x)} = Normal 0,<br />

< δ < 2.<br />

W f(x)<br />

2 x2√ �<br />

.<br />

π<br />

Proof. Again from (4.1) and (3.11) with α = α∗ (x) defined in (4.6) it is easy to see<br />

that<br />

W f(x)<br />

(4.8)<br />

2 x2√ �<br />

+ oP(1),<br />

π<br />

as n→∞. On the other hand application of (3.4) where α = α ∗ (x) is defined by<br />

(4.6) yields<br />

(4.9)<br />

n1/2 α∗ (x) 1/4{E � f ∗ �<br />

C(x)<br />

α∗(x)−f(x)} = O<br />

n (5δ−2)/4<br />

�<br />

,<br />

W f(x)<br />

as n→∞. Here C(x) ={ 2 x2 √ π }1/2 . Combining (4.8) and (4.9) yields (4.7).<br />

5. An application to the excess life distribution<br />

Assume that the random variable X has cdf F and pdf f defined on [0,∞) with<br />

F(0) = 0. Denote the hazard rate function hF = f/S, where S = 1−F is the<br />

corresponding survival function of X. Assume also that the sampled density g<br />

satisties (1.1) and (1.7). It follows that<br />

(5.1) g(y) = 1<br />

W<br />

{1−F(y)} , y≥ 0 .<br />

It is also immediate that W = 1/g(0) and, f(y) =−W g ′ (y) =− g′ (y)<br />

g(0)<br />

that<br />

hF(y) =− g′ (y)<br />

g(y)<br />

, y≥ 0 .<br />

, y≥0 , so

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