A comparison between homogeneous and heterogeneous portfolios

A comparison between homogeneous and heterogeneous portfolios

Insurance: Mathematics and Economics 29 (2001) 59–71 A comparison between homogeneous and heterogeneous portfolios Esther Frostig∗ Department of Stat...

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Insurance: Mathematics and Economics 29 (2001) 59–71

A comparison between homogeneous and heterogeneous portfolios Esther Frostig∗ Department of Statistics, University of Haifa, Mount Carmel, 31905 Haifa, Israel Received 1 December 1999; received in revised form 1 March 2001; accepted 27 March 2001

Abstract We compare two portfolios: in the heterogeneous portfolio the individual risks are independent but not identically distributed. In the homogeneous portfolio the risks are independent and identically distributed. We compare the heterogeneous portfolio with two types of homogeneous portfolios. First, we assume that the distribution of each risk in the homogeneous portfolio is a mixture with equal weights of the risks in the heterogeneous portfolio, and get an upper bound for the heterogeneous portfolio. To get a lower bound we assume that the risks in the homogeneous portfolio are the average of the individual risks in the heterogeneous portfolio. © 2001 Elsevier Science B.V. All rights reserved. Keywords: Portfolios; Supermodular ordering; Stochastic ordering; Convex ordering

1. Introduction Consider N insurance businesses, all are reinsured by the same insurer. The N businesses are exposed to N different risks (X1 , . . . , XN ), where, Xi has a distribution function Fi . We consider the evaluation of the portfolio of the N businesses under three assumptions: 1. Business j insures risk Xj , j = 1, . . . , N. In this case the value of the portfolio is Ψ (X1 , . . . , XN ). We assume that X1 , . . . , XN are independent. We call this portfolio heterogeneous portfolio. 2. Business j can insure all risks. Assume that business j insures risk X(i, j ) with probability 1/N , where X(i, j ) has distribution Fi , and X(i, j ), i, j = 1, 2, . . . , N are independent random variables. In this case the value of the portfolio is Ψ (X(·, 1), . . . , X(·, N )), where X(·, j ) = X(i, j )

with probability

1 , i = 1, 2, . . . , N. N

(1)

Note that the risks in the portfolio are independent identically distributed (i.i.d.) random variables. This portfolio is homogeneous portfolio. 3. Let X(i, j ) be as in (2). Business j insures the average risk, i.e. X¯ j , where N X(i, j ) ¯ . (2) Xj = i=1 N ∗

Fax: +972-4-825-3849. E-mail address: [email protected] (E. Frostig). 0167-6687/01/$ – see front matter © 2001 Elsevier Science B.V. All rights reserved. PII: S 0 1 6 7 - 6 6 8 7 ( 0 1 ) 0 0 0 7 3 - 7

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The value of the portfolio is Ψ (X¯ 1 , . . . , X¯ N ). Also, in this case the risks in the portfolio i.i.d. random variables. In this paper, we compare the homogeneous portfolio described in (1) with the two heterogeneous portfolios described in (2) and (3). We show that for symmetric supermodular function Ψ, Ψ : RN → R EΨ (X1 , . . . , XN ) ≤ EΨ (X(·, 1), . . . , X(·, N )).

(3)

When Ψ is symmetric supermodular function, which is convex in each argument, we prove that EΨ (X1 , . . . , XN ) ≥ EΨ (X¯ 1 , . . . , X¯ N ).

(4)

Supermodular functions are defined in Section 2. The motivation for the problem rose from the problem posed in the book of Kaas et al. (1994), and also in the book by Marshall and Olkin (1979), and first solved by Hoeffding (1963). Assume that Xi , i = 1, . . . , N are independent random variables Bernoulli distributed with  parameter pi . Let p = (p1 , . . . , pN ). In this case X(·, j ) are i.i.d., Bernoulli distributed with parameter p¯ = N i=1 pi /N . It is shown in all the references mentioned above that if ϕ is a convex function then Eϕ(X(·, 1) + · · · + X(·, N )) ≥ Eϕ(X1 + · · · + XN ).

(5)

Note that X(·, 1) + · · · + X(·, N) is Binomial distributed. Thus, as stated in Kaas et al. (1994) “The Binomial is riskiest among sums of Bernoulli distributions”. Similarly, let Xi , i = 1, . . . , N be independent random variables, Xi ). Karlin and Novikoff (1963) (see also Marshall Bernoulli distributed with parameter pi , and let p = (p1 , . . . , pN and Olkin, 1979) generalized the above result, they proved that when p, is majorized by p , and ϕ is convex, then Eϕ(X1 + · · · + XN ) ≥ Eϕ(X1 + · · · + XN ).

(6)

Majorization ordering is discussed in Section 2. A more general problem than the last one is formulated in Marshall and Olkin (1979). Let P, P be two N × m matrices, with column vector pcj (pcj ) sum to 1, j = 1, . . . , m. For j = 1, . . . , m, i = 1, . . . , N, let X(i, j ) be independent random variables where X(i, j ) has distribution Fi , j = 1, . . . , m, and Fi are stochastically ordered. Let XP = (Xpc1 , . . . , Xpc c ), where Xpcj = X(i, j ) with probability p(i, j ). Similarly, define XP . Marshall and Olkin m defined majorization ordering between P and P . Assuming that P majorizes P they found conditions on functions Ψ , Ψ : Rm → R such that EΨ XP ≤ EΨ XP . Their problem is more general than ours, however, the condition that they found on Ψ are too restrictive and difficult to check. For m = 2, let  X(1, j ) with probability pj , X(·, j ) = (7) X(2, j ) with probability 1 − pj , similarly define X (·, j ), j = 1, . . . , m, as mixtures of X(1, j ) and X(2, j ) with weights pj and 1−pj , respectively. Marshall and Olkin proved that when p majorizes p, then for symmetric supermodular function Ψ : EΨ (X(·, 1), . . . , X(·, m)) ≥ EΨ (X (·, 1), . . . , X (·, m)).

(8)

Our problem differs from Marshall and Olkin problem since we compare the random vector (X1 , . . . , XN ) in which the Xi are not identically distributed with a vector whose components are i.i.d. random variables with distribution which is a mixture with equal weights of Fi . Another results related to this study are the results derived by Ma (2000). Ma considered the following portfolios: (Ip1 X1 , . . . , Ipm Xm ),where Xj are risks, and Ipj is indicator random variable which is 1 if risk j is in the portfolio and Xj exchangeable then and 0 otherwise, where P (Ip j = 1) = pj . Ma proved that when Ipj are independent m p majorizes p , implies that m I X is smaller in convex ordering than I X p j j =1 j j =1 pj j . Stochastic and convex

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61

ordering are defined in the next section. When Xj are stochastically ordered Ma shows that mwhen (g(p1 ), . . . , g(pm )) )), where g(x) = −log x or g(x) = (1 − x)/x, then majorizes (g(p1 ), . . . , g(pm j =1 Ipj Xj is stochastically  I X . Although, Ma’s problem is related to ours, the two problems are different: we compare smaller than m j =1 pj j a portfolio with different risks to a portfolio with i.i.d. risks, where Ma compares two portfolios with different risks where the difference between portfolios is the probabilities of the appearance of the risks. This paper is organized as follows. In Section 2, we give some definition of stochastic ordering, in Section 3, we compare between the portfolios described in (1) and (2), in Section 4, we compare between portfolios described in (1) and (3), conclusions and related future research are given in Section 5. 2. Definitions and notations Let X = (X1 , . . . , Xn ) and Y = (Y1 , . . . , Yn ) be random vectors. We write X=st Y if X and Y have the same probability law. Definition 1. Let X and Y be random variables with distribution functions FX and FY , respectively. We say that X is smaller than Y in stochastic order, written X≤st Y , if FX (t) ≥ FY (t) for all t. The following proposition is proved in the literature, see for example Ross (1983), Shaked and Shanthikumar (1994) and Kaas et al. (1994). Proposition 2. The following statements are equivalent: 1. X≤st Y . 2. Eg(X) ≤ Eg(Y ) for all non-decreasing functions g. ˜ st X, Y˜ =st Y and 3. There exist a probability space with two random variables X˜ and Y˜ defined on it such that X= ˜ ˜ X ≤ Y with probability 1. For 0 ≤ u ≤ 1, let FX−1 (u) = inf{x : FX (x) ≥ u}. Remark 3. Let U be a random variable uniformly distributed on (0, 1), then X≤st Y implies that FX−1 (U ) ≤ FY−1 (U )

(9)

with probability 1. Definition 4. Let X and Y be random variables with distribution functions FX and FY , respectively. We say that X is smaller than Y in increasing convex ordering or stop-loss ordering written X≤icv Y , if Eg(X) ≤ Eg(Y ) for all convex non-decreasing functions g. Remark 5. Stochastic ordering implies stop-loss ordering. This ordering is applied in the actuarial and economic literature to compare two risks. X≤icv Y means X is less risky than Y in the sense that X is less likely to take extreme values than Y , or that X is more variable than Y , see Kaas et al. (1994). Since X≤icv Y implies that Eg(X) ≤ Eg(Y ) for g non-decreasing and convex, thus any risk averter insurer prefers loss X to a loss Y . Definition 6. X is less than Y in convex ordering X≤cv Y , if X≤icv Y and EX = EY. Remark 7. If X≤cv Y then Eg(X) ≤ Eg(Y ) for all convex functions g, thus convex ordering implies increasing and convex (stop-loss) ordering. Let x = (x1 , . . . , xm ), and let Π x be a permutation of x.

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Definition 8. A function g : Rm → R, is symmetric if g(x) = g(Π x) for all permutations Π . Next we define supermodular (L-superadditive) function. The following definition can be found, for example in Marshall and Olkin (1979). Definition 9. A function g : Rm → R, is supermodular (submodular) if g(x1 , . . . , xi + , . . . , xj + δ, . . . , xm ) + g(x1 , . . . , xi , . . . , xj , . . . , xm ) ≥ (≤)g(x1 , . . . , xi + , . . . , xj , . . . , xm ) + g(x1 , . . . , xi , . . . , xj + δ, . . . , xm ) holds for all x ∈ Rm , 1 ≤ i < j ≤ m, and all , δ ≥ 0. Next, we define the supermodular and the symmetric supermodular ordering between two random vectors. These relations have been considered in the applied probability and statistics literature, see among others Meester and Shanthikumar (1993), Joe (1990), Bäuerle (1997) and Shaked and Shanthikumar (1997). The supermodular ordering was introduced in the actuarial literature by Müller (1997), Bäuerle and Müller (1998), Goovaerts and Dhaene (1999), Hu and Wu (1999) and Dhaene and Denuit (1999). Definition 10. A random vector X = (X1 , . . . , Xm ) is said to be smaller than a random vector Y = (Y1 , . . . , Ym ) in supermodular ordering, written X≤sm Y, if Eg(X) ≤ Eg(Y) for all supermodular functions g, such that the expectation exists. Definition 11. A random vector X = (X1 , . . . , Xm ) is said to be smaller than a random vector Y = (Y1 , . . . , Ym ) in symmetric supermodular ordering, written X≤symsm Y, if Eg(X) ≤ Eg(Y) for all symmetric supermodular functions g such that the expectation exists. One of the main tools in the actuarial literature to compare between risks is the increasing convex and convex ordering. Lately, the supermodular ordering and the symmetric super modular ordering were introduced in the actuarial literature in order to compare between vectors of risks. When X≤sm Y then the components of Y are more dependent then the  components of X Müller, 1997). If X≤sm Y and φi are monotone in the same direction, (see m φ (X )≤ φ i = 1, . . . , m, then m i cv i=1 i (Yi ). Thus, if the insurance company pays φi (Xi ) for loss Xi , we can i=1 i compare the total compensation sums of two portfolios X and Y (see Müller, 1997). Examples for symmetric supermodular functions Ψ : RN → R are: Ψ (x1 , . . . , xN ) = ϕ(x1 + · · · + xN ), where ϕ is convex. Ψ (x1 , . . . , xN ) = min(x 1 , . . . , xN ).  |x Ψ (x1 , . . . , xN ) = N i=1 i |. Thus, the supermodular ordering enables us to compare the minimal risk of two portfolios, and any convex function of the sum of the risks in the two portfolios. The following definition is from Marshall and Olkin (1979). Definition 12. Let x, y ∈ Rm , and denote by x[1] ≥ · · · ≥ x[m] , the components of x in decreasing order, analogously for y. The vector y majorizes the vector x, written x ≺ y if: k 

x[i] ≤

i=1

k  i=1

y[i] ,

k = 1, . . . , m − 1,

m  i=1

x[i] =

m 

y[i] .

i=1

Definition 13. Let N be the non-negative integers, y = (y1 , . . . , yN ), and assume that yi > yj . A transformation T : N m → N m is a simple transfer if: T y = y ,

E. Frostig / Insurance: Mathematics and Economics 29 (2001) 59–71

63

where yi = yi − 1,

yj = yj + 1,

yk = yk ,

k = i, j.

Clearly, y ≺ y. The following lemma can be found in Marshall and Olkin (1979). Lemma 14. Assume that x, y ∈ N m and that x ≺ y, then x can be derived from y by successive applications of finite number of simple transfers. In the proof of the main result in the next section, we apply some ideas from Karlin (1974). Karlin compared between sampling with replacement and sampling without replacement. Consider a finite population Ω = {a1 , . . . , aN }. Let (Z1 , . . . , Zn ) be a sample of size n from Ω. Let F denote sampling with replacement and let PF be the probability measure on Ω n defined by F, i.e. for (x1 , . . . , xn ) ∈ Ω n PF (x1 , . . . , xn ) =

1 . Nn

Similarly, let W denotes sampling without replacement and let PW be the probability measure on Ω n defined by W, i.e. for (x1 , . . . , xn ) ∈ Ω n PW (x1 , . . . , xn ) =

1 . N (N − 1), . . . , (N − n + 1)

Karlin (1974) proved that for supermodular function Ψ : Rn → R EW Ψ (X1 , . . . , Xn ) ≤ EF Ψ (X1 , . . . , Xn ).

(10)

Karlin (1974) proved (10), for more general functions Ψ . In our case we will consider only symmetric supermodular functions.

3. Upper bound In this section, we compare between the heterogeneous portfolio and the homogeneous portfolios described in (1) and (2) in Section 1. The main result of this section will follow from Lemmas 15 and 16. Before stating and proving the lemmas and the main theorem we need some more definitions and notations. Let X1 , . . . , XN be independent random variables, Xi has a distribution Fi . We assume that Xi are stochastically ordered, i.e. X1 ≤X2 ≤ · · · ≤XN . st

st

st

Let X(·, 1), . . . , X(·, N) be i.i.d. random variables as defined in Eq. (1). Let Ψ be a symmetric function, Ψ : Rn → R. We will describe now how to derive EΨ (X(·, 1), . . . , X(·, n)). Let Xjc be the column vector of (X(1, j ), . . . , X(N, j )). Let Xn be an N × n matrix with columns X1c , . . . , Xnc . For ij ∈ {1, . . . , N}, we have Ψ (X(·, 1), . . . , X(·, n)) = Ψ (X(i1 , 1), . . . , X(in , n)) with probability

1 . Nn

(11)

Let Ω = {1, . . . , N}, then EΨ (X(·, 1), . . . , X(·, n)) =

1 Nn

 (i1 ,...,in )∈Ω n

EΨ (X(i1 , 1), . . . , X(in , n)).

(12)

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The expectation in (12) can be calculated by applying the following steps: 1. Determine an integer p, 1 ≤ p ≤ n. 2. Determine a p-dimensional vector k = (k1 , . . . , kp ) with integer components such that k1 ≥ · · · ≥ kp > 0, p 

(13)

ki = n.

(14)

ι=1

3. From the matrix Xn choose p rows (r1 , . . . , rp ) at random without replacement. 4. Define Ω(n, k) as all the random samples of size n from the matrix Xn such that there is exactly one component ˜ from each column and ki components from row ri , i = 1, . . . , p. Thus Ω(n, k) contains all the vectors X(n, k) of size n of the following form: p p ˜ X(n, k) = (X(r1 , (11 ), . . . , X(r1 , (1k1 ), X(r2 , (21 ), . . . , X(r2 , (2k2 ), . . . , X(rp , (1 ), . . . , X(rp , (kp )), j

where (h denotes the index of the column of the h th random variable that was sampled from row rj . Note that there are N !/(N − p)! such vectors. Karlin (1974) denoted the above sampling plan by F[k]. Let X·n = (X(·, 1), . . . , X(·, n)). Define Ek Ψ (X·n ) =

 ˜ X(n,k)∈Ω(n,k)

1 ˜ EΨ (X(n, k)). N !/(N − p)!

Since Ψ is symmetric we have that EΨ (X·n ) =

n! 1  N! Ek Ψ (X·n ). Nn (N − p)! k1 !, . . . , kp ! k

Let gi : (0, 1) → R, i = 1, 2, . . . , N, and u = (u1 , u2 , . . . , un ). Let k = (k1 , . . . , kp ) be a vector such that (13) and (14) hold. Define, k 

k 

k 

(gi1 1 , gi2 2 , . . . , gip p ) = (gi1 (u1 ), . . . , gi1 (uk1 ), gi2 (uk1+1 ), . . . , gi2 (uk1 +k2 ), . . . , gip (uk1 +···+kp−1 +1 ), . . . , gip (un )). Let Θ(p) be all the ordered samples i = (i1 , . . . , ip ) of size p without replacements from Ω. Using the above notation we can write  1  1  1 Ek Ψ (X·n ) = ··· Ψ ((Fi−1 )k1  , . . . , (Fi−1 )kp  ) du1 , . . . , dun . p 1 N!/(N − p)! u1 =0 un =0 i∈Θ(p)

Lemma 15. Assume that Ψ is symmetric and supermodular. For all n ≤ N , and p = 2 the following holds. If k ≺ k then Ek Ψ (X·n ) ≤ Ek Ψ (X·n ).

(15)

Proof. The proof is very similar to the proof of Lemma 2.1 in Karlin (1974). Consider two cases: (1) n = 2m + 1, (2) n = 2m.

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65

Case 1 (n = 2m + 1). E[m+1,m] Ψ (X·n ) =

 i∈Θ(2)

=

 i∈Θ(2)

 ×

1 N (N − 1) 1 N (N − 1)

1

un =0





1 u1 =0



1 u1 =0

···

1 un =0

Ψ ((Fi−1 )m+1 , (Fi−1 )m ) du1 , . . . , dun 1 2

···

Ψ (Fi−1 (u1 ), Fi−1 (u2 ), (Fi−1 )m−1 , (Fi−1 )m ) du1 , . . . , dun . 1 1 1 2

(16)

Similarly, E[m+2,m−1] Ψ (X·n ) =

 i∈Θ(2)

=

 i∈Θ(2)

 ×

1 N (N − 1) 1 N (N − 1)

1

un =0





1 u1 =0



1 u1 =0

···

1 un =0

Ψ ((Fi−1 )m−1 , (Fi−1 )m+2 ) du1 , . . . , dun 1 2

···

Ψ (Fi−1 (u1 ), Fi−1 (u2 ), (Fi−1 )m−1 , (Fi−1 )m ) du1 , . . . , dun . 2 2 1 2

(17)

The stochastic ordering implies that either Fi−1 (uh ) ≤ Fi−1 (uh ), 1 2

h = 1, 2, 0 < uh < 1

Fi−1 (uh ) ≥ Fi−1 (uh ), 1 2

h = 1, 2, 0 < uh < 1.

or

The supermodularity of Ψ and Eqs. (16) and (17) imply that E[m+1,m] Ψ (X·n ) + E[m+2,m−1] Ψ (X·n ) ≥ 2E[m+1,m] Ψ (X·n ). Thus E[m+2,m−1] Ψ (X·n ) ≥ E[m+1,m] Ψ (X·n ).

(18)

Assume that for all 0 ≤ h ≤ k ≤ m − 1, we have E[m+h+1,m−h] Ψ (X·n ) ≥ E[m+h,m−h+1] Ψ (X·n ).

(19)

We shall prove now (19) for h = k + 1. E[m+k+2,m−k−1] Ψ (X·n ) =

 i∈Θ(2)

=

 i∈Θ(2)

 ×

1 N (N − 1) 1 N (N − 1)

1

u2 =0



1 u1 =0



1 u1 =0

 ···

1 u2 =0

Ψ ((Fi−1 )m+k+2 , (Fi−1 )m−k−1 ) du1 , . . . , dun 1 2

···

Ψ (Fi−1 (u1 ), Fi−1 (u2 ), (Fi−1 )m+k , (Fi−1 )m−k−1 ) du1 , . . . , dun . 1 1 1 2

(20)

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Similarly, 

E[m+k,m−k+1] Ψ (X·n ) =

i∈Θ(2)



=

i∈Θ(2)

 ×

1 N (N − 1) 1 N (N − 1)

1

u2 =0



1 u1 =0



1 u1 =0

 ···

1 u2 =0

Ψ ((Fi−1 )m+k , (Fi−1 )m−k+1 ) du1 , . . . , dun 1 2

···

Ψ (Fi−1 (u1 ), Fi−1 (u2 ), (Fi−1 )m+k , (Fi−1 )m−k−1 ) du1 , . . . , dun . 2 2 1 2

(21)

Eqs. (20) and (21) and the supermodularity of Ψ yield that E[m+k+2,m−k−1] Ψ (X·n ) + E[m+k,m−k+1] Ψ (X·n ) ≥ 2E[m+k+1,m−k] Ψ (X·n ). The induction hypothesis (19) yields the result. Next we shall show that E[2m+1,0] Ψ (X·n ) ≥ E[2m,1] Ψ (X·n ).

(22)

The stochastic ordering and the supermodularity of Ψ yield (N − 1)

 1  1 1 ··· Ψ (Fi−1 (u1 ), Fi−1 (u2 ), (Fi−1 )2m−1 ) du1 , . . . , dun 1 1 1 N u1 =0 un =0

i1 ∈Ω

+

 1  1 1 ··· Ψ (Fi−1 (u1 ), Fi−1 (u2 ), (Fi−1 )2m−1 ) du1 , . . . , dun 2 2 1 N u1 =0 un =0

i∈Θ(2)

≥2

 1  1 1 ··· Ψ (Fi−1 (u1 ), (Fi−1 )2m ) du1 , . . . , dun , 2 1 N u1 =0 un =0

i∈Θ(2)

or E[2m−1,2] Ψ (X·n ) + E[2m+1,0] Ψ (X·n ) ≥ 2E[2m,1] Ψ (X·n ).

(23)

Inequality (19) yields E[2m,1] Ψ (X·n ) ≥ E[2m−1,2] Ψ (X·n ), thus (23) and (24) yield (22). Case 2 (n = 2m). In this case

(24)

E. Frostig / Insurance: Mathematics and Economics 29 (2001) 59–71

2E[m+1,m−1] Ψ (X·n ) =

 i∈Θ(2)

 ×

1 N (N − 1)

1

un =0



+

i∈Θ(2)



×

1

un =0



≥2

i∈Θ(2)



×



un =0

u1 =0

···

Ψ (Fi−1 (u1 ), Fi−1 (u2 ), (Fi−1 )m−1 , (Fi−1 )m−1 ) du1 , . . . , dun 1 1 1 2 1 N (N − 1)



1 u1 =0

···

Ψ (Fi−1 (u1 ), Fi−1 (u2 ), (Fi−1 )m−1 , (Fi−1 )m−1 ) du1 , . . . , dun 2 2 1 2

1 N (N − 1)

1

1

67



1 u1 =0

···

Ψ (Fi−1 (u1 ), Fi−1 (u2 ), (Fi−1 )m−1 , (Fi−1 )m−1 ) du1 , . . . , dun . 1 2 1 2

Thus 2E[m+1,m−1] Ψ (X·n ) ≥ 2E[m,m] Ψ (X·n ). The rest of the proof is as in Case 1. In the rest of this section, we will denote by k an N -component vector k = (k1 , . . . , kp , 0, . . . , 0), where the first p components are positive all the rest are 0 and (13) and (14) hold. Lemma 16. Assume that k ≺ k . Then Ek Ψ (X·n ) ≤ Ek Ψ (X·n ).

(25)

Proof. According to Lemma 14 it suffices to consider k ≺ k which differ only in two components. Let k, k be two vectors such that N  i=1

ki =

N  i=1

ki = n,

and assume that k, k differ only in the i th and j th components. Assume that k has p positive components. Let n2 = kι + kj , k2 = [ki , kj ], I = {h1 , . . . , hp−2 } and k˜ = (k1 , . . . , ki−1 , ki+1 , . . . , kj −1 , kj +1 , . . . , kp ). ˜ ∈ Ω(n − n2 , k) ˜ is given, and ˜ − n2 , k) Assume that X(n ˜ = (X(h1 , (1 ), . . . , X(h1 , (1k ), . . . , X(hp−2 , (p−2 ), . . . , X(hp−2 , (p−2 )). ˜ − n2 , k) X(n 1 kp−2 1 h 1

˜ n be the matrix obtained from Xn by deleting the p − 2 rows and n − n2 columns which are the indices of the Let X 2 ˜n ˜ − n2 , k). ˜ Let Ω(n ˜ 2 , k2 ) be all the random samples of size n2 from the matrix X components of the vector X(n 2 such that there is exactly one component from each column and ki components from row ri1 , and kj components

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˜ 2 , k2 ) contains all the from row ri2 , where (ri1 , ri2 ) is ordered sample without replacement from (Ω − I). Thus Ω(n ˜ 2 , k2 ) of size n2 of the following form: vectors X(n ˜ 2 , k2 ) = (X(r1 , (1 ), . . . , X(r1 , (1k ), X(r2 , (2 ), . . . , X(r2 , (2k )). X(n 1 1 i j Define, E[ki ,kj ,X(n−n ˜ ˜ Ψ (X·n ) = 2 ,k)]





˜ 2 ,k2 )∈Ω(n ˜ 2 ,k2 ) X(n

1 N −p−2 2

˜ 2 , k2 ), X(n ˜ − n2 , k)). ˜ 2 EΨ (X(n

Let kh ,...,kh



p−2 F(h11,...,hp−2) = ((Fh−1 )kh1  , . . . , (Fh−1 ) 1 p−2

khp−2 

).

Thus E[ki ,kj ,X(n−n ˜ ˜ Ψ (X·n ) 2 ,k)] =



1 N (N − 1) 2



(i,j )∈(Ω−I) i=j



1 u1 =0

···

1 un =0

kh ,...,kh



p−2 Ψ (Fi−1 )ki  , (Fj−1 )kj  , F(h11,...,hp−2) ) du1 , . . . , dun .

A similar expression holds for E[k k ,X(n−n ˜ ˜ Ψ (X·n ). 2 ,k)] i j

˜ − n2 , k) ˜ ∈ Ω(n − n2 , k): ˜ Lemma 15 yields that for all X(n E[ki ,kj ,X(n−n ˜ ˜ ˜ Ψ (X·n ) ≤ E[k ,k ,X(n−n ˜ Ψ (X·n ). 2 ,k)] 2 ,k)] i

(26)

j

˜ − n2 , k) ˜ yields (25). Taking expectation from (26) with respect to X(n



We are now ready to state and to prove the main theorem of this section. Theorem 17. If X1 ≤X2 ≤ · · · ≤XN ,

(27)

(X1 , . . . , XN ) ≤ (X(·, 1), . . . , X(·, N )).

(28)

st

st

st

then symsm

Proof. Let e be an N -dimensional vector e = (1, . . . , 1). Clearly, e ≺ k, for any other N -dimensional vector k. Lemma 16 yields that for symmetric supermodular function Ψ E[e] Ψ (X·N ) ≤ E[k] Ψ (X·N ). Thus EΨ (X·N ) =

1  N! N! N! 1  N! ψ(X ) ≥ E Ee Ψ (X·N ) k ·N NN (N − p)! k1 !, . . . , kp ! NN (N − p)! k1 !, . . . , kp ! k

= EΨ (X1 , . . . , XN ), (29) proves (28).

k

(29) 

E. Frostig / Insurance: Mathematics and Economics 29 (2001) 59–71

69

Remark 18. Assume n ≤ N . Let en be an n-dimensional vector en = (1, . . . , 1). Assume that we take a sample of size n without replacement out of Ω = {1, . . . , N}. Lemma 16 yields that EΨ (X·n ) ≥ Een Ψ (X·n ). Note that Een Ψ (X·n ) =

 i∈Θ(n)

1 EΨ (Xi1 , . . . , Xin ), N!/(N − n)!

where {i1 , . . . , in } is a sample without replacement of size n from Ω. Example 19. Let Xi , i = 1, . . . , N be independent random variables, Xi is Bernoulli distributed with parameter  pi . In this case X(·, j ) are i.i.d. Bernoulli distributed with parameter p¯ = (1/N ) N i=1 pi . Theorem 1 yields that (X1 , . . . , XN ) ≤ (X(·, 1), . . . , X(·, N )). symsm

(30)

This is a generalization of (5). Example 20. Let a0 < a1 < · · · < aN . Assume that for 1 ≤ i, j ≤ N , ai − ai−1 = aj − aj −1 . Assume that X1 , . . . , XN are independent random variables where Xi is uniformly distributed on (ai−1 , ai ). In this case X(·, j ) are i.i.d. uniformly distributed on (a0 , aN ). Also in this case (30) holds. 4. Lower bound Define N stochastically identical and independent portfolio as follows: Xj = (X(1, j ), X(2, j ), . . . , X(N, j )),

j = 1, . . . , N,

where X(i, j ), j = 1, . . . , N are as defined in Section 1. Define the vectors Yi as follows: Y1

=

(Y (1, 1), . . . , Y (N, 1)) = (X(1, 1), X(2, 1), . . . , X(N − 1, 1), X(N, 1)),

Y2

=

(Y (1, 2), . . . , Y (N, 2)) = (X(2, 2), X(3, 2), . . . , X(N − 1, 2), X(1, 2)), .. .

Yj

=

(Y (1, j ), . . . , Y (N, j )) = (X(j, j ), X(j + 1, j ), . . . , X(N, j ), X(1, j ), . . . , X(j − 1, j )), .. .

YN

=

(Y (1, N ), . . . , Y (N, N)) = (X(N, N ), X(1, N ), . . . , X(N − 1, N )).

Also, define  N N N X(i, j ) X(i, j ) X(i, j ) i=1 i=1 i=1 j ¯ = , ,..., . X N N N Clearly, for every realization of (X(1, j ), X(2, j ), . . . , X(N − 1, j ), X(N, j )), we have that ¯ j. Yj  X The following proposition is stated in Marshall and Olkin (1979).

(31)

70

E. Frostig / Insurance: Mathematics and Economics 29 (2001) 59–71

Proposition 21 (Fan and Lorentz, 1954). Let Φ be a supermodular function that is convex in each argument separately, Φ : R m → R. If x 1 , . . . , x m and y 1 , . . . , y m are N-dimensional vectors such that x i ≺ y i , i = 1, . . . , m then N  j =1

Φ(xj1 , . . . , xjm ) ≤

N  j =1

Φ(yj1 , . . . , yjm ).

From Proposition 21, we conclude (4) as follows. Proposition 22. Let Φ be a symmetric supermodular function that is convex in each argument separately Φ : R N → R. Then  N N N j =1 X(j, 1) j =1 X(j, 2) j =1 X(j, N ) EΦ(X1 , X2 , . . . , XN ) ≥ EΦ , ,..., . (32) N N N ¯ i that for every realization (31) holds. Let Φ function which is Proof. It is clear from the definition of Yi and X symmetric and satisfies the conditions of the proposition. Using the symmetry of Φ, we get N

1 EΦ(Y (i, 1), Y (i, 2), . . . , Y (i, N )) N i=1  N N N N 1 i=1 X(i, 1) i=1 X(i, 2) i=1 X(i, N ) EΦ ≥ , ,..., N N N N i=1    N N N j =1 X(i, 2) j =1 X(i, N ) i=1 X(i, 1) = EΦ , ,..., .  N N N

EΦ(X1 , X2 , . . . , XN ) =

(33)

5. Conclusions In this paper, we derived upper and lower bound on heterogeneous portfolio by two types of homogeneous portfolios. Hoeffding (1963) and Karlin and Novikoff (1963) showed that if X = (X1 , . . . , XN ) is a vector of independent ) is Bernoulli distributed random variables with vector of parameters p = (p1 , . . . , pN ), and X = (X1 , . . . , XN another vector of independent Bernoulli distributed random variables with vector of parameters p then p ≺ p implies that (6) holds. Marshall and Olkin (1979) showed that if X(1, j ), j = 1, . . . , N are i.i.d. and X(2, j ), j = 1, . . . , N are i.i.d. and independent of X(1, j ), X(1, j )>st X(2, j ) and X(·, j ), X (·, j ), pj , pj are as in (7), then p ≺ p yields that (X (·, 1), . . . , X (·, N)) ≤ (X(·, 1), . . . , X(·, N )). symsm

(34)

Our contribution is establishing the last identity in the case where X(·, j ) are a mixture of more then two variables, and i.i.d. with mixing probability vector pj = ((1/N ), . . . , (1/N )), j = 1, . . . , N, and p j = ej , where ej is a vector with the j th component equal to 1 and all other component 0. Thus, for this extreme case we find that symmetric supermodular are the functions that solve Marshall and Olkin problem introduce in Section 1. For future research we try to find other special cases in which “simple” family the functions solves Marshall and Olkin problem.

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