ESTIMATING OF THE PROPORTIONAL HAZARD PREMIUM FOR HEAVY-TAILED CLAIM AMOUNTS
4. PROOF OF THE MAIN RESULT
The following proposition is instrumental for the proof of our result Proposition 4.1. Let F be a distribution function fulfilling (1.3) with ξ ∈(0,1),δ >0 and some realc. Suppose that Lis locally bounded in [x0,+∞) forx0≥0. Then, fornlarge enough, for any un=O(nαξ),α∈(0,1),we have
pn = P X1> un
= c 1 +o(1) n−α, γn2 = var
Z un
0
F(x)1/ρ−1
1{X1≤x}dx
= O n2α(ξ−1/ρ+1) ,
and √
npnu−δn L(un) = O n−α/2−αξ δ+1/2 . Proof of the Theorem 2.1: Let us write
(4.1) √
n Πbρ,n−Πρ
= An+Bn , where
An = √ n
Z un
0
h
Fn(x)1/ρ
− F(x)1/ρi dx , and
Bn = √ n
b
p1/ρn ρβbn 1−ξbnρ −
Z ∞
un
F(x)1/ρ
dx
. We begin by Bn, we may rewriteBnas follows:
Bn =Bn,1+Bn,2 ,
where
Bn,1 = (pbn)1/ρ ρβbn
1−ξbnρ −(pn)1/ρ ρ βn 1−ξ ρ , and
Bn,2 = (pn)1/ρ ρ βn 1−ξρ −
Z ∞
un
F(s)1/ρ
ds . First, observe thatBn,1, may be rewrite into
Bn,1 = ρβbn 1−ξbnρ
√n
(bpn)1/ρ−(pn)1/ρ + (pn)1/ρ ρ
1−ξbnρ
√n βbn−βn
+ ρ2βn(pn)1/ρ (1−ξbnρ) (1−ξ ρ)
√n ξbn−ξ .
From Smith (1987), we have, asn→ ∞ (4.2) βbn/βn−1 = OP u−δn L(un)
and ξbn−ξ = OP u−δn L(un) . On the other hand, by the central limit theorem, we have
(4.3) pbn−pn= OP p pn/n
as n→ ∞. Then, with the delta method, we obtain
Bn,1 = θ1 1 +oP(1)√
n(pbn−pn) +θ2 1 +oP(1)√
n(βbn−βn) +θ3(1 +oP(1))√
n(ξbn−ξ) , where
θ1 = β(pn)1/ρ−1
1−ξ ρ , θ2 = ρ(pn)1/ρ
1−ξ ρ , θ3 = ρ2β(pn)1/ρ (1−ξ ρ)2 . Either, for Bn,2, we have
Bn,2 = (pn)1/ρ ρβn
ξ ρ−1 − Z ∞
un
F(s)1/ρ
ds . We may rewrite
Fun(s) = F(un+s) F(un) =
1 + s
un
−1/ξ1 + (un+s)−δL(un+s) 1 +u−δn L(un) .
This allows us to rewrite Z ∞
0
F(s+un)1/ρ
ds =
= F(un)1/ρZ ∞ 0
Fun(s)1/ρ
ds
= F(un)1/ρZ ∞
0
"
1 + s un
−1/ξ1 + (un+s)−δL(un+s) 1 +u−δn L(un)
#1/ρ
ds
= p1/ρn
1
1 +u−δn L(un) 1/ρ
× Z ∞
0
"
1 + s un
−1/ξ
1 + (un+s)−δL(un+s) #1/ρ
ds
= p1/ρn
1
1 +u−δn L(un) 1/ρ
u1/ξρn Z ∞
un
x−1/ξρ 1 +x−δL(x)1/ρ
dx
= p1/ρn
1
1 +u−δn L(un) 1/ρ
u1/ξρn
×
ξ ρ
1−ξ ρ u1−1/ξρn
+ Z ∞
un
x−1/ξρ−δL(x)1/ρdx
.
Since function L is locally bounded in [x0,∞) for x0 ≥0 and x−δL(x) is non-increasing near infinity, then for all large n, we have
u1/ξρn Z ∞
un
x−1/ξρ−δL(x)1/ρdx = O u−δn ,
and therefore, for all large n Z ∞
un
F(x)1/ρdx = p1/ρn βnρ 1−ξ ρ
1−u−δn L(un) +O u−δn L(un)1/ρ .
Consequently
Bn,2 = O u1−1/ρξ−δ/ρn ,
which means, since 1−1/ρξ−δ/ρ <0, thatBn,2→P 0 as n→ ∞.
For An, we have
(4.4) An = √
n Z un
0
h Fn(x)1/ρ
− F(x)1/ρi dx .
We next show that, the right-hand side of (4.4), converge to 0 in probability, by
the use of the Taylor formula, we have Z un
0
h
Fn(x)1/ρ
− F(x)1/ρi dx =
= 1 ρ
Z un
0
Fn(x)−F(x)
F(x)1/ρ−1
dx
= −1 ρ
Z un
0
Fn(x)−F(x)
F(x)1/ρ−1
dx
= −1 ρ
Z un
0
1 n
X1(Xi≤x)−F(x)
F(x)1/ρ−1
dx
= −1 ρ
1 n
X Z un
0
1(Xi≤x) F(x)1/ρ−1
dx − Z un
0
F(x) F(x)1/ρ−1
dx
= −1 ρ
h
Z−E[Z1] i
,
where
Zi :=
Z un
0
F(x)1/ρ−1
1(Xi≤x)dx . We assume that
γn2 = var(Z1).
We are going to calculateγn. ForF(x) =x−1/ξO(1) andun=nαξO(1), we have E[Zi] =
Z un
0
F(x)1/ρ−1
E
1(Xi≤x) dx
= Z un
0
F(x)1/ρ−1 1−F(x) dx
= Z un
0
F(x)1/ρ−1
dx − Z un
0
F(x)1/ρ
dx
= Z un
0
x−1/ξ(1/ρ−1) dx −
Z un
0
x−1/ξρ dx
O(1)
= ρ ξ u1−1/ξρ+1/ξn
ρ ξ+ξ−1 −ρ ξ u1−1/ξρn ξρ−1
! O(1)
and
E(Zi2) = E Z un
0
F(x)1/ρ−1
1(Xi≤x)dx Z un
0
F(y)1/ρ−1
1(Xi≤y)dy
= Z un
0
Z un
0
F(x)1/ρ−1
F(y)1/ρ−1
Eh
1(Xi≤x)1(Xi≤y) i
dx dy
= Z un
0
Z un
0
F(x)1/ρ−1
F(y)1/ρ−1
min F(x), F(y) dx dy
=
= Z un
0
F(x)1/ρ−1Z x
0
F(y)1/ρ−1
F(y)dy
dx +
Z un
0
F(y)1/ρ−1Z un
x
F(x)1/ρ−1
F(x)dx
dy
= ρ2ξ2 u2(1−1/ξρ+1/ξ) n
(ρ ξ+ρ−1)2 − 2ρ2ξ2 u2−2/ξρ+1/ξn (ξ ρ−1) (2ρ ξ+ρ−2
! O(1), we conclude that
γn= nα(ξ−1/ρ)O(1) . Now, we show that
√n γn
Z−E[Z1] D
→ N(0,1), as n→ ∞.
With Lindeberg–Feller Theorem (see e.g. Chapter 2 in Durrett (1996)), note that
√n
γn Z−E[Z1]
= Xn
k=1
Z un
0
F(x)1/ρ−1
1(Xk≤x)dx − E[Z1] γn
√n
= Xn
k=1
Sk,n ,
where
E(Sk,n) = 0, E(Sk,n2 ) = 1/n and Xn
k=1
E(Sk,n2 ) = 1 for all n≥1. We need to show that
Xn
k=1
Eh
|Sk,n|21 |Sk,n|> ǫi
→ 0, as n→ ∞. Indeed, we have
Xn
k=1
E h
|Sk,n|21 |Sk,n|> ǫi
= 1 γn2 Eh
Zk−E[Z1]2
1Zk−E[Z1]> ǫ γn√ ni
.
Since
Zk−E[Z1]
≤un, then the right side of the previous inequality is less or equal than
u2n γ2n Eh
1
Zk−E[Z1] > ǫ γn
√ni
= u2n
γn2 PhZk−E[Z1] > ǫ γn
√ni . In view of Tchebychev’s inequality, we get
u2n
γn2 PhZk−E[Z1]
> ǫ γn√ ni
≤ u2n γn2
1 ǫ γn√
n2 .
Further, for all α∈(0,1), ξ∈(0,1) andǫ >0, withun=O(nαξ) was used, then u2n
ǫ nγn4 = n−2αξ+4α/ρ−1O(1). We must assume: 4α/ρ−2αξ <1 for that Pn
k=1Eh
|Sk,n|21 |Sk,n|> ǫi
→0 asn→ ∞.
Finally, we obtain that
√n γn
b
Πρ,n−Πρ
→ −1 ρ
√n γn
Z−E[Z1] +θ1
ppn(1−pn) γn
√n(pbn−pn) ppn(1−pn) + θ2βn
√pnγn
√npn βbn/βn−1
+ θ3
√pnγn
√npn ξbn−ξ
+oP(1), This enable us to rewrite into
√n
γn Πbρ,n−Πρ
→ −1
ρW1+θ1
ppn(1−pn) γn W2 +
p2 (1 +ξ)θ2βn
√pnγn W3 + (1 +ξ)θ3
√pnγn W4+oP(1), where (Wi)i=1,4 are standard normal rv’s with E[WiWj] = 0 for every i, j = 1, ...,4, except for
E[W3W4] = E
"
p 1
2 (1 +ξ)
√npn βbn/βn−1 1 (1 +ξ)
√npn ξbn−ξ#
= 1
(1 +ξ)p
2 (1 +ξ) Eh√
npn βbn/βn−1√
npn ξbn−ξi
= − 1
p2 (1 +ξ) .
From Lemma A-2 of Johansson 2003, under the assumptions of Theorem 2.1, we have, for any real numbers, t1, t2, t3 and t4,
E
"
exp (
i t1
√n
γn Z−E[Z1] +i√
npn(t2, t3) βbn/β−1 ξbn−ξ
! +i t4
√n(pbn−pn) ppn(1−pn)
)#
→ exp
−t21 2 −1
2(t2, t3)Q−1 t2
t3
−t24 2
1 +oP(1) asn→ ∞, whereQ−1 is that in (2.6),γ2n= Var(Z1) and i2 =−1.
It follows that, with this result that
√n
γnσn Πbρ,n−Πρ D
→ N(0,1), as n→ ∞ , where
σ2n := 1 ρ2 + θ12
γn2 pn(1−pn) + 2 (1 +ξ)θ22βn2 pnγn2 +(1 +ξ)2θ23
pnγn2 −2(1 +ξ)βnθ2θ3
pnγn2 . This complete the proof of Theorem (2.1).
ACKNOWLEDGMENTS
This work has been supported by the National Agency of the University Development Research of Algeria (ANDRU) PNR project, code: 08/E09/5100.
We are grateful to an anonymous referee for constructive criticism and a number of queries that helped us to produce a substantial revision of the paper. I also thank the Professor Necir who encouraged me to address this work.
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