THE PUBLISHING HOUSE
OF THE ROMANIAN ACADEMY
PROCEEDINGS OF THE ROMANIAN ACADEMY, Series A,
Volume 17, Number 4/2016, pp. 287–292
UPER AND LOWER BOUNDS FOR A FINITE-TYPE RUIN PROBABILITY
IN A NONHOMOGENEOUS RISK PROCESS
Anişoara Maria RĂDUCAN1, Raluca VERNIC1,2 , Gheorghiţă ZBĂGANU1,3
1
Institute for Mathematical Statistics and Applied Mathematics “Gh. Mihoc- C. Iacob”
2
Ovidius University of Constanţa
3
University of Bucharest
Correspondent author: Gheorghiţă Zbaganu, E-mail: [email protected]
Abstract. Based on many numerical examples, Răducan et al [8] stated a conjecture that relates the
order in which some nonhomogeneous claims arrive - provided that all the claims are comparable - to
the magnitude of the corresponding ruin probability. In that conjecture the usual stochastic order has
been considered for the claims. Now we know that the conjecture was wrong [10] and we prove it for
a stronger order, namely the likelihood ratio order. Being stronger, the likelihood order implies the
usual stochastic one, but for many usual distributions the two types of ordering are equivalent. That
explains our initial conjecture suggested by computer.
Key words: ruin probability, non-homogeneous claims, stochastic orders, risk process, likelihood ratio
domination.
1. STATING OF THE PROBLEM
Among various kinds of stochastic ordering on the positive half-line, the strongest one seems to be the
likelihood ratio (or, shortly, likelihood). If F1 and F2 are two distributions absolutely continuous with
respect to some basic measure μ, we say that F1 ≤like F2 if we can choose the densities fi for Fi , i = 1,2, such
f
that the ratio 1 is decreasing, with the convention that the ratio makes sense only on the union of the
f2
f1
= ∞ . We shall use the same notation for random variables: X1 ≤like X 2
0
means that if F1 ≤like F2 if F1 is the distribution of X1 and F2 is the distribution of X2.
In general, this ordering implies the usual stochastic ordering, denoted F1 ≤ st F2 and defined by
supports of F1 and F2 , and that
F1 ( x ) ≥ F2 ( x ) for all x or, equivalently, by F 1 ( x ) ≤ F2 ( x ) ,∀x; here F = 1 − F is the right tail of F.
However, in some cases, the likelihood ordering is the same with the stochastic one, as we shall see in the
following (for more details on stochastic orderings see, e.g., [1, 11]). For this purpose, let Exp(a) denote the
exponential distribution with parameter a > 0, Gamma (a, b) the gamma distribution with parameters a,b > 0,
U(a,b) the uniform distribution on (a,b), a < b, δx the Dirac point measure, and denote by ∗ the convolution
operator. Then, for example, in the exponential case, if F1 = Exp (a) and F2 = Exp (b), a, b >0, it is easy to
see that F1 ≤like F2 ⇔ F1 ≤ st F2 (⇔a ≥ b). The same equivalence F1 ≤like F2 ⇔ F1 ≤ st F2 holds if
F1 = δα ∗ Exp ( a ) , F2 = δβ ∗ Exp ( b ) ,
being
F1 = Gamma ( m, a ) , F2 = Gamma ( n, b ) ,
also
being
equivalent
equivalent
with
with
(α ≤ β
(m ≤ n
and
and
a ≥ b);
or,
if
a ≥ b);
or,
if
F1 = U ( a, b ) , F2 = U ( a ', b ') which is equivalent with (a ≤ a′ and b ≤ b′).
This is why in [8], after many numerical examples (see, also, [7]), we were led by the computer to the
following:
288
Anişoara Maria Răducan, Raluca Vernic, Gheorghiţă Zbăganu
2
CONJECTURE. Let X = ( X k )1≤k ≤n ,Y = (Yk )1≤k ≤n be independent random vectors valued in \ n+ with
independent components, representing the claim sizes and, respectively, the inter-claim revenues of an
insurance risk process. Suppose that the components of Y are identically distributed and that
X 1 ≤ st X 2 ≤ st ... ≤ st X n . Then Ψ e ≤ Ψ σ ≤ Ψ ϖ , where Ψ e , Ψ σ , Ψ ϖ are the ruin probabilities at or before the
(
nth claim when the claims come in the order ( X k )1≤k ≤n , X σ( k )
)
1≤ k ≤ n
and respectively, ( X n−k +1 )1≤k ≤n , with σ
being an arbitrary permutation of {1,2,…,n}.
We discovered [10] that the conjecture is wrong, though it holds for exponentially distributed claims
with all parameters distinct. We were misled by the fact that the likelihood ordering is the same with the
stochastic one for the distributions we used. In this paper, we prove the correct assertion: the claims should
be ordered in the likelihood order. It follows that these claims can follow different types of distributions,
which was not the case with the claims considered in [7–9] and [14]; there, the claims were assumed to
follow the same type of distribution, but with possible different parameters which implies the non homogeneity of the risk process. Non-homogeneous claims have already been considered in the actuarial
literature in order to capture the fluctuations of the economic environment, see, e.g., De Kok [4], Lefevre and
Picard [5], Paulsen [6], Blazevicius et al. [2], Castaner et al. [3], or Stanford et al. [12] and [13].
The structure of the paper is as follows: in Section 2, we present two preliminary results which maybe
are interesting in themselves and in Section 3 we prove the conjecture if X 1 ≤ st X 2 ≤ st ... ≤ st X n is replaced
by X 1 ≤like X 2 ≤like ... ≤like X n .
2. PRELIMINARY RESULTS
The following lemmas are needed to prove the main results:
LEMMA 2.1. Let f : \ 2+ → \ be an integrable function satisfying the following properties:
i) f ( x, y ) + f ( y, x ) = 0 for any x,y ≥ 0,
ii) if x≤y then f(x,y) ≥0.
α α+β− x
Let α, β > 0 and Dα,β ( f ) =
∫ ∫
0
f ( x, y ) dydx. Then Dα ,β ( f ) ≥ 0.
0
Proof. There are two cases, depending if α > β or otherwise.
Case 1. α > β : we rewrite Dα,β ( f ) = 1B fdλ, , where λ is the Lebesgue measure on R, and 1B is the
∫
indicator function of B. Note that B = B1 ∪ B2 , where
B₁= {( x, y ) ∈ \ 2+ x ≤ α, y ≤ α, x + y ≤ α + β},
B2 = {( x, y ) ∈ \ 2+ x ≤ α, y > α, x + y ≤ α + β}.
Then Dα,β ( f ) = 1B1 fdλ + 1B2 fdλ.
∫
∫
The set B1 is symmetrical, meaning that
B2 = {( x, y )
∈ \ 2+
( x, y ) ∈ B1 ⇔ ( y, x ) ∈ B1 ,
while the set B2 can be rewritten as
x ≤ β, y > α, x + y ≤ α + β}. Therefore, interchanging x and y, applying the assumption (i)
and the symmetry of B1 , followed by Fubini's theorem, we obtain
∫
∞∞
1B1 f dλ =
0 0
∞∞
=−
∫∫
1B1 ( x, y ) f ( x, y ) dxdy =
∫ ∫1
B1
0 0
∞∞
∫ ∫1
B1
( y , x ) f ( y , x ) d yd x
0 0
∞∞
( y, x ) f ( x, y ) dxdy = − ∫ ∫ 1B1 ( x, y ) f ( x, y ) dydx = − ∫ 1B1 fdλ,
0 0
3
Upper and lower bounds for a finite-type ruin probability in a nonhomogeneous risk process
from where
∫
1B1 fdλ = 0. On the other hand,
∫
β α+β− x
1B2 fdλ =
∫ ∫
0
289
f ( x, y ) dydx ≥ 0, since in this case
α
y ≥ α > β ≥ x and according to assumption ii), when x ≤ y then f ( x, y ) ≥ 0. It follows that Dα ,β ( f ) ≥ 0.
Case 2. α ≤ β : now we write Dα,β ( f ) =
αα
∫∫
f ( x , y ) dy dx +
0 0
α α+β− x
∫ ∫
0
f ( x, y ) dydx. The first integral is
0
again equal to 0 by symmetry, while the second one is nonnegative since x ≤ α ≤ y (note that in the second
integral the inequality between the limits α ≤ α + β − x holds since x ≤ α ≤ β ). Q.E.D.
Notation. Let n ≥ 2 be a positive integer. Let X = ( X k )1≤k ≤n be a random vector valued in \ n+ with
independent components. For any i ≤ j denote by X i: j the random vector X i: j := ( X k )i≤k ≤ j .
Let a = ( ak )1≤k ≤n ∈ \ n+ and let
PX ( a ) = P ( X 1 ≤ a1 , X 1 + X 2 ≤ a1 + a2 ,..., X 1 + X 2 + ... + X n ≤ a1 + a2 + ... + an ) .
(1)
We also denote
S k = X 1 +...+X k , sk = a1 +...+ak , ∀k ≥ 1
so that we can rewrite (1) as PX ( a ) = P ( S1 ≤ s1 , S2 ≤ s2 ,..., Sn ≤ sn ) .
If Y is another random vector, we denote
PX ( a Y ) = P ( X 1 ≤ a1 , X 1 + X 2 ≤ a1 + a2 ,..., X 1 + X 2 + ... + X n ≤ a1 + a2 + ... + an Y ) .
With this notation the following result holds:
LEMMA 2.2. Let X be a random vector valued in \ n+ with independent components and a ∈ \ n+ .
Suppose that X i ≤like X i+1 for some 1 ≤ i ≤ n − 1. Let
⎧( X 1 ,.., X i −1 , X i +1 , X i , X i + 2 ,.., X n ) , 2 ≤ i ≤ n − 2
⎪
.
X * = ⎨( X 2 , X 1 , X 3 ,..., X n ) ,
i =1
⎪( X ,.., X , X , X ) ,
i = n −1
n−2
n
n −1
⎩ 1
(2)
Then PX ( a ) ≥ PX * ( a ) or, equivalently, Δi ( a ) = PX ( a ) − PX * ( a ) ≥ 0 .
Proof. Let fi , fi +1 be the densities of X i and X i +1 respectively. Let
δi ( x, y ) = fi ( x ) f i +1 ( y ) − fi ( y ) fi +1 ( x ) .
(3)
As X i ≤like X i+1 , it holds that 0 ≤ x ≤ y ⇒ δi ( x, y ) ≥ 0 and , moreover, it is obvious that
δi ( x , y ) + δi ( y , x ) = 0
for all x, y ≥ 0. We start with i = 1 and we note that
PX ( a ) = E ⎡⎣ P ( S1 ≤ s1 , S2 ≤ s2 ,..., Sn ≤ sn X 1 , X 2 ) ⎤⎦ =
= E ⎡⎣1{ X1 ≤ a1 , X1 + X 2 ≤ a1 + a2 } P ( X 3 ≤ s3 − X 1 − X 2 ,.., X 3 + ... + X n ≤ sn − X 1 − X 2 X 1 , X 2 ) ⎤⎦ =
= E ⎡⎣1{ X1 ≤ a1 , X1 + X 2 ≤ a1 + a2 } PX 3:n ( s3 − X 1 − X 2 , a4 ,.., an X1:2 ) ⎤⎦ =
a1 a1 + a2 − x1
=
∫ ∫
0
0
f1 ( x1 ) f 2 ( x2 ) PX 3:n ( s3 − x1 − x2 , a4 ,.., an ) dx2 dx1 .
(4)
290
Anişoara Maria Răducan, Raluca Vernic, Gheorghiţă Zbăganu
Similarly, we get PX * ( a ) =
a1 a1 + a2 − x1
f1 ( x2 ) f 2 ( x1 ) PX3:n ( s3 − x1 − x2 , a4 ,.., an ) dx2 dx1 .
∫ ∫
0
4
0
Thus
Δ1 ( a ) =
a1 a1 + a2 − x1
δ1 ( x1 , x2 ) PX 3:n ( s3 − x1 − x2 , a4 ,.., an ) dx2dx1 = Da1 ,a2 ( h1 ) ,
∫ ∫
0
(5)
0
Da1 ,a2 ( h ) is defined in Lemma 2.1. and h1 ( x, y ) = δ1 ( x, y ) PX 3:n ( s3 − x1 − x2 , a4 ,.., an ) . Applying
where
Lemma 2.1. yields Δ1 ( a ) ≥ 0. The same reasoning holds in general.
For an arbitrary 2 ≤ i ≤ n − 2 , we have
PX ( a ) = E ⎡ P S1 ≤ s1 , S 2 ≤ s2 ,..., S n ≤ sn X 1:( i +1) ⎤ =
⎢⎣
⎥⎦
= E ⎡1{S1≤s1 ,S2 ≤s2 ,..,Si +1≤si +1} P X i + 2 ≤ si + 2 − Si +1 ,.., X i + 2 + ... + X n ≤ sn − Si +1 X 1:( i +1) ⎤ =
⎢⎣
⎥⎦
= E ⎡1{S1≤s1 ,S2 ≤s2 ,..,Si +1≤si +1} PX i + 2 :n si +2 − Si +1 , ai +3 ,.., an X 1:( i +1) ⎤ .
( )
⎣⎢
⎦⎥
On the other hand,
PX * ( a ) = E ⎡1{S1≤s1 ,...,Si −1≤si −1 ,Si −1+ X i +1≤si ,Si +1≤si +1} P X i + 2 ≤ si + 2 − Si +1 ,.., X i + 2 + ... + X n ≤ sn − Si +1 X 1:( i +1) ⎤
⎢⎣
⎦⎥
= E ⎡1{S1≤s1 ,...,Si −1≤si −1 ,Si −1+ X i +1≤ si ,Si +1≤ si +1} PX i + 2 :n si + 2 − Si +1 , ai +3 ,..., an X 1:( i +1) ⎤
( )
⎢⎣
⎥⎦
(
)
(
(
)
)
(
(
)
)
and
Δi ( a ) = PX ( a ) − PX * ( a ) =
(
(
)
)
= E ⎡ 1{S1≤ s1 ,S2 ≤s2 ,...,Si +1≤si +1} − 1{S1≤s1 ,...,Si −1≤si −1 ,Si −1+ X i +1≤si ,Si +1≤ si +1} PX i + 2 :n si + 2 − Si +1 , ai +3 ,..., an X 1:( i +1) ⎤ =
( )
⎢⎣
⎦⎥
si −1 −
s1 s2 − x1
=
∫ ∫
0
0
i−2
∑ xj
j =1
...
∫
0
⎛
⎜
⎜
⎝
i −1
∏
j =1
⎞
f j (x j )⎟
⎟
⎠
si −
i −1
∑ xj
si +1 −
j =1
i
∑ xj
j =1
∫
∫
0
0
⎛
δi ( xi , xi +1 ) PX i + 2 :n ⎜ si + 2 −
( ) ⎜
⎝
i +1
∑ x ,a
j
j =1
i + 3 ,.., a n
⎞
⎟ dxi +1dxi ..dx1
⎟
⎠
Hence we obtain
Δi ( a ) = PX ( a ) − PX * ( a ) =
s1 s2 − x1
∫ ∫
0
⎛
with hi ( x, y ) = δi ( x, y ) PX i + 2 :n ⎜ si + 2 −
( ) ⎜
⎝
According to Lemma 2.1. D⎛ i −1 ⎞
∑
si −1 −
0
i−2
∑ xj
j =1
...
∫
0
⎛
⎜
⎜
⎝
⎞
i −1
∏ f ( x ) ⎟⎟D
j
j
j =1
⎠
i −1
⎛
⎞
⎜s −
x j ⎟ , ai +1
⎜ i
⎟
j =1 ⎠
⎝
∑
( hi ) dxi −1dxi ..dx1 ,
⎞
− x − y, ai +3 ,..., an ⎟ .
⎟
j =1
⎠
( hi ) ≥ 0 from where Δi ( a ) ≥ 0.
i −1
∑x
j
⎜ s − x ⎟,a
j ⎟ i +1
⎜ i
j =1 ⎠
⎝
Finally, the case i = n − 1 results in a similar way and the proof is completed.
3. MAIN RESULTS
PROPOSITION 3.1. Let X = ( X k )1≤k ≤n ,Y = (Yk )1≤k ≤n , be independent random vectors valued in \ n+
with independent components. Assume that the components of Y are identically distributed, let
ξk = X k − Yk and let L ( ξ ) = max ( 0, ξ1 , ξ1 + ξ 2 ,..., ξ1 + ξ 2 + .. + ξn ) . Let i ∈ {1,2,…,n-1} be fixed and
5
Upper and lower bounds for a finite-type ruin probability in a nonhomogeneous risk process
291
⎧( ξ1 ,.., ξi −1 , ξi +1 , ξi , ξi + 2 ,.., ξ n ) , 2 ≤ i ≤ n − 2
⎪
ξ = ⎨( ξ 2 , ξ1 , ξ3 ,..., ξ n ) ,
i =1
.
⎪( ξ ,.., ξ , ξ , ξ ) ,
i = n −1
n−2 n
n −1
⎩ 1
*
(6)
Suppose that X i ≤like X i+1 .
Then L ( ξ ) ≤ st L ( ξ* ) .
(
)
Proof. We have to prove that P ( L ( ξ ) ≤ t ) ≥ P L ( ξ* ) ≤ t for all t ≥ 0 , or, equivalently, that
P ( X 1 ≤ Y1 + t ,..., X 1 + ... + X n ≤ Y1 + ... + Yn + t ) ≥ P ( X 1* ≤ Y1* + t ,..., X 1* + ... + X n* ≤ Y1* + ... + Yn* + t ) ,
where X * is defined in (2) and Y * is a random vector independent of X * and having the same distribution
as Y . Let H be the common distribution of the Yks. Then, taking into account the fact that the measure H n is
invariant in respect to permutations, we see that
P ( X 1 ≤ Y1 + t ,..., X 1 + ... + X n ≤ Y1 + ... + Yn + t ) =
∫P
X
( a1 + t, a2 ,.., an ) dH n ( a ) ,
\ n+
P ( X 1* ≤ Y1* + t ,..., X 1* + ... + X n* ≤ Y1* + ... + Yn* + t ) =
∫P
X*
( a1 + t, a2 ,.., an ) dH n ( a ) ,
\ n+
hence the difference between the two integrals is
∫ Δ (a
i
1
+ t , a2 ,..., an )dH n ( a ) ≥ 0 according to lemma 2.2.
\ n+
This completes the proof. QED.
(
Let now σ be a permutation of {1,2,…,n} and let σ(ξ) be the vector with the components ξσ( k )
)
1≤ k ≤ n
.
Assuming that, in an insurance context, ξ k represents the loss between two consecutive claims (i.e. claims
number k–1 and k), we let
(
Lσ = max 0, ξσ(1) , ξσ(1) + ξσ( 2 ) ,..., ξσ(1) + ξσ( 2 ) + .. + ξσ( n )
be the maximum aggregate loss given by
)
σ(ξ). It is known that Ψ σ ( t ) = P ( Lσ > t ) defines the ruin
th
probability at or before the n claim (see, for instance [1]) . Let e be the identical permutation and ϖ be the
... n ⎞
⎛1 2
inverse permutation, i.e. ϖ = ⎜
⎟.
⎝ n n − 1 ... 1 ⎠
Then the following result holds.
PROPOSITION 3.2. Let X = ( X k )1≤k ≤n ,Y = (Yk )1≤k ≤n , be independent random vectors valued in \ n+
with independent components, representing the claim sizes and, respectively, the inter-claim revenues of an
insurance surplus process. Assume that the components of Y are identically distributed and that
X 1 ≤like X 2 ≤like ... ≤like X n . Then Ψ e ≤ Ψ σ ≤ Ψ ϖ .
Proof. We have to prove that Le ≤ st Lσ ≤ st Lϖ for any permutation σ. Suppose that σ ≠ e, σ ≠ ϖ and let
X * = σ ( X ) . Then there exist some i, j ∈ {1, 2,.., n − 1} such that σ ( i ) < σ ( i + 1) and σ ( j ) > σ ( j + 1) . Then
(
)
X σ( i ) ≤like X σ( i +1) ⇔ X i* ≤like X i*+1 . According to proposition 3.1. L ( ξ* ) ≤ st L τi ( ξ* ) where τi is the
transposition between i and i+1. Regarding the index j, according to the same proposition,
(
)
L τ j ( ξ* ) ≤ st L ( ξ* ) , where τ j is the transposition between j and j+1. In other words, if σ ≠ e, σ ≠ ϖ , we
can both increase and decrease the value of Ψ σ ( t ) .
The only two cases when we cannot do that are when σ = e (here we can only increase) and σ = ϖ
(when we can only decrease). And noting that any permutation can be built from e or ϖ by successively
permuting two consecutive numbers, the proof is complete.
Example. In the context of Proposition 3.2., for n = 4, we have the following inequalities chains:
292
Anişoara Maria Răducan, Raluca Vernic, Gheorghiţă Zbăganu
6
Ψ1234 ≤ Ψ 2134 ≤ Ψ 2314 ≤ Ψ 2341 ≤ Ψ 3241 ≤ Ψ 3421 ≤ Ψ 4321 ,
Ψ1234 ≤ Ψ1243 ≤ Ψ1423 ≤ Ψ 4123 ≤ Ψ 4132 ≤ Ψ 4312 ≤ Ψ 4321 ,
Ψ1234 ≤ Ψ1324 ≤ Ψ1342 ≤ Ψ1432 ≤ Ψ 4132 ≤ Ψ 4312 ≤ Ψ 4321.
As a thumb rule, if we can proceed from σ to σ’ using inversion of type “if X σ( i ) ≤like X σ( i +1) , we interchange
them”, then Ψ σ ≤ Ψ σ ' .
COROLLARY 3.3. Let
X = ( X k )1≤k ≤n ,Y = (Yk )1≤k ≤n , be independent random vectors valued in \ n+
with independent components. Suppose that the components of Y are identically distributed and the
components of X are Gamma distributed: X i ∼ Gamma( νi , λ i ) .
If ν1 ≤ ν 2 ≤ ... ≤ ν n and λ1 ≥ λ 2 ≥ ... ≥ λ n then Ψ e ≤ Ψ σ ≤ Ψ ϖ .
Proof. It is easy to see that Gamma( ν, λ ) ≤like Gamma( ν ', λ ') if and only if ν ≤ ν ' and λ ≥ λ ' . Indeed,
in this case the ratio
f1
( x) = x ν−ν 'e − x( λ−λ ') is decreasing if and only ν ≤ ν ' and λ ≥ λ ' .
f2
Or, otherwise written Gamma( ν, λ ) ≤ like Gamma( ν ', λ ') iff ν ≤ ν ' and λ ≥ λ ' and the result is
immediate from Proposition 3.2.
ACKNOWLEDGEMENTS
Gheorghiţă Zbăganu acknowledges the support from a research grant of the Romanian National
Authority for Scientific Research CNCS-UEFISCDI, project number PN-II-ID-PCE-2011-3-0908.
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Received October 11, 2015