Random sums of random variables and vectors by E. Omey and R. Vesilo

8/14/2019 Random sums of random variables and vectors by E. Omey and R. Vesilo

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Random sums of random variables and vectorsVersion May 6, 2009

E. Omey (*) and R. Vasilo (**)

(*) HUB, Stormstraat 2, 1000 Brussels - [email protected]

(**) Macquarie University, Sydney - [email protected]

Abstract

Let fX; Xi; i = 1; 2;:::g denote independent positive random variables hav-ing a common distribution function F(x) and, independent of X, let N denote

an integer valued random variable. Using S(0) = 0 and S(n) = S(n 1) + Xn,the random sum S(N) has distribution function

G(x) =1X

i=0

P(N = i)P(S(i) x).

and tail distribution G(x) = 1 G(x). The distribution function G is calledsubordinated to F with subordinator N. Under suitable conditions, it can beproved that G(x) s E(N)F(x) as x ! 1. here many results. In this paper weextend some of the existing results. In the place of i.id. random variables, weuse variables that are independentor variables that are asymptotic independent.We also consider multivariate subordinated distribution functions.

Keywords: Subexponential distributions, regular variation, O-regular vari-ation, subordination

AMS 2000 Subject Classication:Primary: 60G50Secondary: 60F10, 60E15, 26A12, 60K99

1 Introduction

Let fX; Xi; i = 1; 2;:::g denote independent, nonnegative random variables (r.vs)having a common distribution function (d.f.) F(x). Independent of X, let Ndenote an integer valued r.v. with pdf pn = P(N = n). Partial sums are givenby S(0) = 0 and S(n) = X1 + X2 + ::: + Xn, n 1. For n 1, the d.f. ofS(n) is

given by P(S(n) x) = Fn(x), where Fn(x) denotes the n-fold convolutionof F with itself. Replacing the index n by the random index N, we obtain therandom sum S(N) =

PNi=0 Xi . The d.f. ofS(N) is given by

G(x) =1X

n=0

pnFn(x).

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The tail distribution is given by

G(x) = 1 G(x) =1X

n=1

pnFn(x).

If p0 = 0 and F has a density f, then G also has a density g given by

g(x) =1X

n=1

pnfn(x),

where fn(x) = f f::: f(x) and where a b(x) =Rx0 a(x y)b(y)dy. Many

papers have been devoted to the asymptotic behaviour of the tail G(x) and ofthe density g(x).

To formulate our results, we recall some of the basic denitions. A positiveand measurable real function g(x) is regularly varying with real index if

limx!1

g(tx)

g(x)= t, 8t > 0.

Notation: g 2 RV(). The function g(x) is in the class L if it satises

limx!1

g(t + x)

g(x)= 1, 8t > 0.

The function g(x) is in the class ORV of O-regularly varying functions if

lim supx!1

g(tx)

g(x)= g(t) < 1, 8t > 0.

For a survey of denitions, properties and applications ofRV and ORV, we referto Bingham et al. (1987), Geluk and de Haan (1987), Seneta (1976), Resnick(1987).

A d.f. F is in the class S of subexponential distributions if

limx!1

1 F2(x)

1 F(x)= 2.

Notation F 2 S. A density function f(x) is in the class SD of subexponentialdensities if it satises

limx!1

f2(x)

f(x)= 2.

Notation f 2 SD. It is well known that ifF(x) 2 L \ ORV, then F 2 S andif f(x) 2 L \ ORV, then f 2 SD. The class S was introduced by Chistyakov(1964), Teugels (1975) and studied by Chover et al (1973a, 1973b), Cline (1987),Embrechts et al. (1979, 1980, 1982, 1985, 1997).

If F 2 S, it is well known that

liminf1 Fn(x)

1 F(x) n, as x ! 1.

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An application of Fatous lemma yields the result that

liminf 1 G(x)1 F(x)

E(N), as x ! 1.

If E(N) < 1, it makes sense to look for an upper bound for the ratioG(x)=F(x). The following result is well known, see Embrechts et. al. (1979,1982), Chover et al. (1973a,b), Stam (1973), Shneer (2004), or Daley et al.(2006).

Proposition 1 (1) (a) Suppose that E((1 + ")N) < 1 for some " > 0.(i) If F 2 S, then G 2 S and G(x) s E(N)F(x).(ii) If f 2 SD, then g 2 SD and g(x) s E(N)f(x).(b) If F 2 RV(), > 1 and if E(N+1+") < 1, then G 2 RV() and

G(x) s E(N)F(x).(c) If X 0 is an - stable (0 < < 1) r.v., then Fn(x) nF(x) and

G(x) E(N)F(x).

Result (b) shows that if we have weaker assumptions on N, we have toassume more about F.

The main contributions of the paper are to extend Proposition 1 in a numberofways. In Section 2 we discuss the case in which we assume that the r.vs Xiare independent and not necessarily i.i.d.. Secondly, we consider Proposition 1(b) in the case where the mean = E(X) = 1. Also we formulate some newresults in the case where the Xi are dependent and asymptotically independent.In Section 3 of the paper, we state and prove a bivariate analogue of Proposition1.

In the results below, limits are always limits as x ! 1 or t ! 1 or

min(x; y) ! 1. The notation a(x)t

b(x) means that ua(x) b(x) vb(x)for some u ; v > 0 and all x x. The notation a(x) s b(x) means thata(x)=b(x) ! 1. We use similar notations for bivariate functions.

2 Univariate results

Our aim here is to discuss the tail distributions P(S(n) > x) and G(x) Moreprecisely, we want to obtain universal inequalities, asymptotic inequalities andasymptotic equalities. If the Xi are i.i.d. with a nite mean , many results areknown. In the literature, much less is known in the innite means case.

2.1 Upper Bounds

For further use, we dene the integrated tail mF(x) as mF(x) = Rx0 F(t)dt.Clearly, we have xF(x) mF(x). Recall that the Laplace-Stieltjes Transform(LST) of K(x) is given by

bK(s) = Z10

exp(sx)dK(x).

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Clearly the LST of F(x) = P(X x) is given by f(s)

bF(s) = E(exp(sX)).

Moreover, we have bmF(s) = (1 f(s))=s:Now suppose that Xi has d.f. Fi(x), LST fi(s) and integrated tail mi(x).For the sum S(n), n 1, we set

mS(n)(x) =

Zx0

P(S(n) > z )dz.

The following result slightly extends Lemma 5(i) of Daley et al (2007).

Lemma 2 (2) (i) We have mS(n)(x) Pn

i=1 mi(x) and P(S(n) > x) Pni=1 mi(x)=x.

(ii) In the i.i.d. case, we have mS(n)(x) nmF(x) and P(S(n) > x) nmF(x)=x.

Proof. The LST of mS(n)(x) is given by

bmS(n)(s) = 1 f1(s)f2(s):::fn(s)s .Repeated use of the equality 1 ab = (1 a)b + (1 b) shows that

bmS(n)(s) = nXi=1

bmi(s)ai(s),where for each i, ai(s) is the product of one or more of the fj(s). So we ndthat

mS(n)(x) =n

Xi=1mi Ai(x),

where Ai(x) is the convolution product of one or more of the Fj (x). It followsthat

mS(n)(x) nX

i=1

mi(x).

Using the inequality xP(S(n) > x) mS(n)(x), we have (i). The second resultfollows from the rst result.

Remark. This Lemma makes only sense if the means i = E(Xi) areinnite.

2.2 Lower Bounds

In general, it is hard to obtain lower bounds valid for all x > 0. We prove thefollowing liminf-result.

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Lemma 3 (3) (i) For n = 1; 2;::: we have

liminf P(S(n) > x)Pni=1 Fi(x)

1. (1)

(ii) In the i.i.d. case, we have

liminfP(S(n) > x)

F(x) n. (2)

Proof. (i) We have

P(S(n) > x) = 1

Zx0

P(S(n 1) x z)dFn(z)

= Zx

0

P(S(n 1) > x z)dFn(z) + (1 Fn(x))

P(S(n 1) > x)Fn(x) + Fn(x).

It follows that

P(S(n) > x) P(S(n 1) > x) + Fn(x) P(S(n 1) > x)Fn(x).

Using similar arguments for S(n 1); S(n 2);:::, we obtain that

P(S(n) > x) P(S(n 2) > x) + Fn1(x) + Fn(x)

P(S(n 2) > x)Fn1(x) P(S(n 1) > x)Fn(x)

and then

P(S(n) > x) F1(x) + ::: + Fn(x)

n1Xi=1

P(S(i) > x)Fi+1(x).

Using

0 P(S(i) > x)Fi+1(x)

F1(x) + ::: + Fn(x) P(S(i) > x),

we obtain that

P(S(n) > x)

F1(x) + ::: + Fn(x) 1

n1Xi=1

P(S(i) > x),

and (1) follows. The second result follows from the rst result.

2.3 Subordination in the i.i.d. case

For the subordinated tail, in the i.i.d. case, we obtain the following result.

Lemma 4 (4) (i) We have xG(x) mG(x) E(N)mF(x).(ii) If = E(X) = 1, then mG(x) s E(N)mF(x).

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Proof. Daley et al. (2007, Lemma 5)Note that we have xG(x) E(N)mF(x) without any extra conditions. In

the innite means case this is a new result.To relate mF(x) and F(x), we need an extra assumption about F(x).

Lemma 5 (5) (i) For 0 < 1, we have F(x) 2 RV() if and only ifmF(x) 2 RV(1 ) and both statements imply that mF(x) s xF(x)=(1 ).

(ii) If F(x) 2 OR with (F) > 1, then mF(x) t xF(x)

Proof. (i) This is a standard result in regular variation theory, (e.g. Binghamet al., Theorem 1.6.4).

(ii) This is a standard result in the ORV-theory, (e.g. Bingham et al. Corol-lary 2.6.2.

The main result of this section is the following.

Theorem 6 (6) (i) For 0 < 1, we have F(x) 2 RV() if and only if

G(x) 2 RV(), and both statements imply that G(x)s

E(N)F(x).(ii) F(x) 2 OR with (F) > 1 if and only if G(x) 2 OR with (G) > 1,and both statements imply that G(x) t F(x).

Proof. (i) Since (cf. Lemma 4 (ii)) mG(x) s E(N)mF(x), the result followsfrom Lemma 5 (i).

(ii) First assume that F(x) 2 OR with (F) > 1. Using mF(x) t xF(x),we obtain that mG(x) t xF(x) and hence

xG(x)

xF(x)

mG(x)

xF(x) A.

Using liminfG(x)=F(x) E(N), we nd that G(x) t F(x). But then it followsthat G(x) 2 OR with (G) > 1.

Now assume that G(x) 2 OR with (G) > 1. Using mF(x) t mG(x) txG(x), we can nd constants 0 < a < b and x so that

axG(x) mF(x) bxG(x); x x.

Now take t > 1 and observe on the one hand that

mF(xt) mF(x) =

Zxtx

F(z)dz F(x)x(t 1).

On the other hand, we have mF(xt) mF(x) axtG(xt) bxG(x) and we getthat

F(x)(t 1) atG(xt) bG(x)

orF(x)

G(x)

(t 1) atG(xt)

G(x)

b.

Because G(x) 2 OR with (G) > 1, for t suciently large we nd that

liminfF(x)

G(x)(t 1) > 0.

Since we always have the lim sup-result, the lemma follows.

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2.4 Subordination in the independent case

If we have only independent components, we proceed in a dierent way. In thenext results we have to make extra assumptions about the asymptotic behaviourof the tails Fi. The assumptions that we use are similar to those made inSkucait (2004) and Maejima (1972). We have the following result.

Lemma 7 (7) Suppose that for i 1 we have

liminfFi(x)

(x) d(i).

Then

liminfG(x)

(x) E(D(N)),

where D(n) = Pn1 d(i).Proof. First note that

G(x) =1X

n=1

pnP(S(n) > x)Pn

i=1 Fi(x)

nXi=1

Fi(x).

Under the assumptions of the Lemma, for each n 1, we have that

liminf

Pni=1 Fi(x)

(x) D(n),

where D(n) = Pn1 d(i). Using Fatou s lemma and (1), we obtain the desiredresult.To nd also an upper bound, we use stronger assumptions and proceed as

follows.

Lemma 8 (8) Suppose that for each n 1 we have Fn(x)=(x) ! d(n) 0,and(x) =

Rx0 (z)dz " 1 as x " 1. Also suppose that for some constant c 0

we havenX

i=1

mi(x) (x)A(n); 8x c,

and E(A(N)) < 1. Then supxc mG(x)=(x) E(A(N)), and mG(x) s(x)E(D(N)), where D(n) =

Pn1 d(i).

Proof. Using Fi(x)=(x) ! d(i) 0 and (x) ! 1, we have mi(x)=(x) !d(i). From here it follows that

nXi=1

Fi(x)=(x) ! D(n),

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andn

Xi=1

mi(x)=(x) ! D(n).

Using Lemma 2 (i), we have

mG(x)

(x)=

1X1

pnmS(n)(x)

(x)

1X1

pn

Pni=1 mi(x)

(x),

and it follows that

supxc

mG(x)

(x)

1X1

pnA(n) = E(A(N)).

For the second result, ifE(D(N)) < 1, there is nothing to prove. So we assumethat E(D(N)) < 1. To prove the second result, we use Pratts extension ofFatous lemma, cf. Pratt (1960), Johns (1957). Observe the following facts. Wehave

(1)Pn

i=1 mi(x)=(x) ! D(n), as x ! 1;(2) 0 A(n)

Pni=1 mi(x)=(x), for x > c;

(3) A(n) Pn

i=1 mi(x)=(x) ! A(n) D(n), as x ! 1.Using Fatous lemma, we obtain that

liminf1X

n=1

p(n)

"A(n)

nXi=1

mi(x)=(x)

#

Xp(n) [A(n) D(n)] = E(A(N)) E(D(N)).

Since P1

n=1p(n)A(n) = E(A(N)), we nd that

limsupX

p(n)nX

i=1

mi(x)=(x) E(D(N)).

We conclude that

lim supmG(x)

(x) E(D(N)).

On the other hand, using

1

(x)P(S(n) > x) =

P(S(n) > x)Pni=1 Fi(x)

Pni=1 Fi(x)

(x),

and (1), we get that

liminf1

(x)P(S(n) > x) D(n).

It follows that

liminfG(x)

(x) E(D(N)).

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Using Fatous lemma again, we obtain that

liminf mG(x)(x)

E(D(N))

and we conclude that mG(x) s E(D(N))(x).As before, we need an extra condition on (x) or (x), to get a result for

G(x), cf. Lemma 5.

2.4.1 Example 1

Take Fi(x) = F(x) = (x), for all i 1. Here we have d(i) = 1 and D(n) = n.

2.4.2 Example 2

Let a(i) > 0, h(i) > 0 and Fi(x) = Fh(i)(a(i)x). As before, we use the integrated

tails and for i 1, we set mi(x) = mFi(x) where

mi(x) =

Zx0

(1 Fh(i)(a(i)z))dz =1

a(i)

Za(i)x0

(1 Fh(i)(z))dz.

Clearly we have 1 Fh(i)(z) h(i)(1 F(z)), and then we obtain that

mi(x) h(i)

a(i)mF(a(i)x); i 1,

andn

Xi=1 m

i(x)

n

Xi=1

h(i)

a(i) mF(a(i)x).

Multiplying by pn = P(N = n) and taking sums, we obtain the universal boundthat

mG(x) E(NX

i=1

h(i)

a(i)mF(a(i)x)).

2.4.3 Example 3

Take Example 2 again and assume that F(x) 2 RV(), 0 < < 1. For xedi we have, as x ! 1,

1 Fh(i)(a(i)x) s h(i)(1 F(a(i)x) s h(i)a(i)F(x).

Hence, we obtain that mi(x) s h(i)a(i)mF(x), so that

nXi=1

mi(x) s D(n)mF(x),

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where D(n) =

Pni=1 h(i)a

(i). On the other hand, we have

Fi(x)F(x)

h(i) F(a(i)x)F(x)

.

Since F(x) 2 RV(), Potters bounds give

F(a(i)x)

F(x) A(a(i))+; a(i) 1; x x,

andF(a(i)x)

F(x) B(a(i)); a(i) 1; a(i)x x; x x.

For the remaining case of a(i) 1; a(i)x x and x x, we have

F(a(i)x)F(x)

1F(x)

.

Also, since x x=a(i), we have F(x=a(i)) F(x). Since 1=a(i) 1, theusual bounds gives

B(1

a(i))

F(x=a(i))

F(x) A(

1

a(i))+,

and then it follows that

1

F(x)

1

F(x=a(i))

(a(i))

BF(x)= C(a(i)).

We conclude that

Fi(x)

F(x) Ah(i)(a(i))+; a(i) 1; x x,

andFi(x)

F(x) Dh(i)(a(i)); a(i) 1; x x.

It follows that

Fi(x)

F(x) Ch(i) max((a(i))+; (a(i))); x x.

This implies that

supxx

Pni=1 Fi(x)F(x)

A(n),

where

A(n) =nX

i=1

h(i)max(a(i))+; (a(i))).

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For G(x) we nd that

supxx

G(x)

F(x) E(A(N)).

A lower bound is easy to nd. We have

Fi(x)

F(x)! h(i)a(i).

Using Fatous lemma, we obtain that

liminfG(x)

F(x)X

pih(i)a(i).

Now we proceed to nd bounds for the mfunctions. We have

mi(x) h(i)a(i)

mF(a(i)x); i 1.

In the regularly varying case F(x) 2 RV(), 0 < < 1, we have mF(x) 2RV(1 ). Potters bounds show that

mF(a(i)x)

mF(x) Aa1+"(i), for x x; a(i) 1,

mF(a(i)x)

mF(x) 1, for x x; a(i) 1.

But then

supxx

Pni=1 mi(x)mF(x)

= supxx

nXi=1

h(i)a(i)

mF(a(i)x)mF(x)

A(n)

where

A(n) = AnX

i=1;a(i)1

h(i)a+"(i) +nX

i=1;a(i)1

h(i)

a(i).

If E(A(N)) < 1, we can proceed as before, and we nd that

mG(x)

mF(x)! E(D(N)).

In this example, we also have that mF(x) s xF(x)=(1). Using the monotone

density theorem, we nd that

G(x) s E(D(N))F(x).

Special cases are h(i) = 1 and h(i) = i.

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2.4.4 Example 4

Let a(i) > 0 and Fi(x) = Fi

(a(i)x) and assume that F 2 RV(), 0 < < 1.First note that1 Fi(x) i(1 F(a(i)x=i).

As in Example 3, it follows that

Fi(x)

F(x) Ah(i)(

a(i)

i)+; a(i)=i 1; x x,

andFi(x)

F(x) Dh(i)(

a(i)

i); a(i)=i 1; x x.

It follows that

Fi(x)

F(x) Ch(i) max((

a(i)i

)+; (a(i)

i)); x x.

For mi(x) we nd that

mi(x) =

Zx0

(1 Fi(a(i)z)dz i

a(i)mF(a(i)x).

Example 4 can be completed as in Example 3.

2.5 Subordination in a dependent case

In this section we assume that the Xi are dependent, but asymptotically inde-

pendent. We assume that for each i 6= j we have

P(Xi > x; Xj > x)

Fi(x) + Fj (x)! 0. (3)

In the next result we formulate additional assumptions to prove that P(S(n) >x) asymptotically equals

Pn1 Fi(x).

Proposition 9 (9) Assume that (3) holds and that for each i, Fi(x) 2 ORVwith i(t) ! 1 as t " 1. Then

P(S(n) > x) sn

X1Fi(x).

Proof. Part 1. First we prove the result for n = 2. Choose , 0 < < 1 andwrite

P(X1 + X2 > x) = I+ II II I+ IV

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where

I = P(X1 + X2 > x; X2 > (1 )x),II = P(X1 + X2 > x; X1 > (1 )x),

II I = P(X1 > (1 )x; X2 > (1 )x),

IV = P(X1 + X2 > x; x X1 (1 )x;x X2 (1 )x).

For I, (and similarly for II) we have F2(x) I F2((1 )x), and then itfollows that

1 lim

sup

inf

I

F2(x) 2(1 ).

For II I, we write

II I =II I

II I(a) II I(a),

where II I(a) = P(X1 > (1 )z) + P(X2 > (1 )z). Now observe that

P(X1 > (1 )x) =F1((1 )x)

F1(x)F1(x)

and similarly for P(X2 > (1 )z). It follows that

II I(a) max(F1((1 )x)

F1(x);

F2((1 )x)

F2(x))(F1(x) + F2(x))

and hence also that

limsupII I(a)

F1(x) + F2(x)

max(1(1 ); 2(1 ))

Using (3) we obtain that

II I

F1(x) + F2(x)=

II I

II I(a)

II I(a)

F1(x) + F2(x)! 0.

Now we investigate IV . We have IV P(X1 > x; X2 > x). As in the caseof II I, we get that

IV

F1(x) + F2(x)! 0.

Combining all terms, we nd that

1 limsupinfP(X1 + X2 > x)F1(x) + F2(x) max(A(1 ); B(1 )).Now we take # 0, to obtain that

P(S(2) > x)

F1(x) + F2(x)=

P(X1 + X2 > x)

F1(x) + F2(x)! 1.

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This proves the result for n = 2.Part 2. Assume that the result holds for S(2), S(3), ..., S(n 1). to prove

the result for S(n) we consider S(n 1) and Xn. By the induction hypothesiswe have P(S(n 1) > x) s

Pn1i=1 Fi(t). It is straightforward to prove that

limsupP(S(n 1) > tx)

P(S(n 1) > x)= n1(t) max(1(t); 2(t);:::;n1(t)).

This shows that P(S(n 1) > x) 2 ORV and n1(t) ! 1 as t " 1.Now we prove that S(n 1) and Xn are asymptotically independent. Using

fX1 x=(n 1); X2 x=(n 1);:::;Xn1 x=(n 1)g fS(n 1) xg ,

it follows that

P(S(n 1) > x; X n > x) P(S(n 1) > x; X n > x=(n 1))

n1Xi=1

P(Xi > x=(n 1); Xn > x=(n 1))

Using (3), it follows that P(S(n 1) > x; Xn > x) = o(1)(Pn

i=1 Fi(x)) or that

P(S(n 1) > x; Xn > x) = o(1)(P(S(n 1) > x) + Fn(x)).

We can proceed as in Part 1 to prove that P(S(n) > x) s P(S(n 1) >x) + Fn(x). This proves the result.

Now let N denote an integer variable, independent of all the Xi aqnd considerthe random sum S(N). We have

P(S(N) > x) =1X

i=1

pnP(S(n) > x)

We prove the following result.

Theorem 10 (10) Assume that the conditions of Proposition 9 hold.(i) If there is a d.f. F(x) such that

liminfFi(x)

F(x) d(i),

then

liminfP(S(N) > x)

(x) E(D(N)),

where D(n) =Pn

i=1 di.

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(ii) If there is a d.f. F(x) such that F(x) 2 RV() and such that

supx0

Fi(x)F(x)

a(i)

andFi(x)

F(x)! d(i).

IfE(N+"A(N)) < 1, whereA(n) =Pn

i=1 a(i), thenP(S(N) > x) s E(D(N))F(x).

Proof. (i) Under the assumptions of the theorem we get that

liminf

Pni=1 Fi(x)

(x) D(n)

where D(n) = Pni=1 d(i). Fatous lemma yields thatliminf

P(S(N) > x)

(x) E(D(N)).

(ii) Using P(S(n) > x) Pn

i=1 Fi(x=n), for x 0, we have

1

F(x)P(S(n) > x)

F(x=n)

F(x)

nXi=1

Fi(x=n)

F(x=n)

F(x=n)

F(x)A(n),

where A(n) =Pn

i=1 a(i). Using F(x) 2 RV(), we obtain that

F(x=n)

F(x)

Cn+", for x=n x and x x.

For x=n x and x x, we have x nx and F(nx) F(x). Aso we have(x=n) bounded. We nd that

F(x=n)

F(x)

1

F(nx).

For small n 1, this is bounded. For large n, we can use F(x) 2 RV() tond that

n+"F(n) ! 1, as n ! 1.

We nd that

F(x=n)

F(x) Cn+"

, for x=n x

and x; n x

.

As a conclusion, we have that

F(x=n)

F(x) Cn+", for n 1 and x x,

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and1

F(x)P(S(n) > x) Cn+"A(n), for n 1 and x x.

We can use Lebesgues theorem to nd that .

P(S(N) > x)

F(x)! E(D(N)).

This proves the result.

3 Multivariate case

3.1 Introduction and notation

For convenience and without loss of generality we only discuss the two-dimensionalcase. Let F(x; y) = P(X x; Y y) denote a bivariate d.f. with marginalsF1(x) = FX (x) and F2(x) = FY (x) and suppose that X 0, Y 0. Partial

sums will be denoted by (S(1)n ; S(2)n ) and we use the notation Fn(x; y) = 1

Fn(x; y), where Fn(x; y) is the d.f. of(S(1)n ; S

(2)n ). We consider random indices

N or (N; M) independent of the Xi; Yj and as before we set S(1)0 = S

(2)0 = 0.

For convenience, we also dene Fn;m(x; y) = P(S(1)n x; S

(2)m y), n; m 1.

The d.f. of the random vector of random sums (S(1)N ; S

(2)N ) is given by

G(x; y) =P1

n=0pnFn(x; y), where pn = P(N = n). The tail is given by

G(x; y) =1

Xn=1pnFn(x; y).

If we have dierent random indices for each of the components, we study the

random vector of random sums (S(1)N ; S0(2)M ). In this case we have H(x; y) =P1

n=0

P1m=0pn;mFn;m(x; y), where pn;m = P(N = n; M = m). The tail is

given by

H(x; y) =1X

n=0

1Xm=0

pn;mFn;m(x; y).

As in the univariate case we use integrated tails as follows. For F we dene mFas

mF(x; y) =

Zx0

Zy0

F(u; v)dudv, and

mi(x) = Zx0

Fi(x)dx; i = 1; 2.

Using F(x; y) F1(x)+F2(y) and Fi(xi) F(x1; x2), we obtain that mF(x; y)satises

mF(x; y) ym1(x) + xm2(y),

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and thatym1(x) mF(x; y), xm2(y) mF(x; y).

As before, we always have xyF(x; y) mF(x; y).

3.2 General inequalities

To obtain some general inequalitities, rst observe that

max(1 P(S(1)n x); 1 P(S(2)m x)) 1 Fn;m(x; y),

and that1 Fn;m(x; y) 1 P(S

(1)n x) + 1 P(S

(2)m x).

Among others, it follows that

G(x; y) G1(x) + G2(y),

where G1(x) = P(S(1)N x) and G2(y) = P(S

(2)N y). For H(x; y), we nd

thatH(x; y) H1(x) + H2(y),

where H1(x) = P(S(1)N x) and H2(y) = P(S

(2)M y).

Now we can use the univariate upper bounds Gi or Hi of the previous sectionto nd bivariate upper bounds for G or H.

Using Lemma 4 and Theorem 6 we obtain the following result.

Lemma 11 (11) (i) xyG(x; y) mG(x; y) 2E(N)mF(x; y)(ii) xyH(x; y) mH(x; y) (E(N) + E(M))mF(x; y)(iii) Suppose that Fi 2 ORV with (Fi) > 1, or that Fi 2 RV(i) with

0 i < 1. Then, as min(x; y) ! 1, we have G(x; y) = O(1)F(x; y) andH(x; y) = O(1)F(x; y).

Proof. (i) We have

1 Fn;m(x; y) 1 P(S(1)n x) + 1 P(S

(2)m x)

nm1(x)=x + mm2(y)=y.

From this it follows that

mG(x; y) ymG1(x) + xmG2(y)

E(N)(ym1(x) + xm2(y))

2E(N)mF(x; y).

(ii) In a similar way, for H we have

mH(x; y) ymH1(x) + xmH2(y)

E(N)ym1(x) + E(M)xm2(y)

(E(N) + E(M))mF(x; y).

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(iii) For G and H we have that

xyG(x; y) 2E(N)mF(x; y),xyH(x; y) (E(N) + E(M))mF(x; y).

Now, if Fi 2 ORV with (Fi) > 1, or if Fi 2 RV(i) with 0 i < 1, wehave mi(x) t xFi(x) and then it follows that

xyF(x; y) mF(x; y) ym1(x) + xm2(y)

C1yxF1(x) + C2xyF2(y)

(C1 + C2)xyF(x; y).

As min(x; y) ! 1, we nd that xyF(x; y) t mF(x; y). But then, as min(x; y) !1, we have G(x; y) = O(1)F(x; y). In a similar way we nd that H(x; y) =O(1)F(x; y).

In Lemma 9 (iii) we found that G(x; y) and H(x; y) are bounded above byF(x; y). To obtain a lower bound, we prove the following general result, cf.Lemma 3(ii).

Lemma 12 (12) For all n; m 1 we have

lim inf min(x;y)!1

1 P(S(1)n x; S

(2)m y)

1 F(x; y) min(n; m).

Proof. First take n = m. We will prove the result by induction on n. Firstnote that for n = 1 the result holds. Assume the result holds for n = 1; 2;:::;kand consider the case where n = k + 1. We have

1 Fk+1;k+1(x; y) = 1 Zxu=0

Zyv=0

Fk;k(x u; y v)dF(u; v)

=

Zxu=0

Zyv=0

(1 Fk;k(x u; y v))dF(u; v) + 1 F(x; y)

(1 Fk;k(x; y))F(x; y) + (1 F(x; y)).

By the induction step, we nd that

liminf1 Fk+1;k+1(x; y)

1 F(x; y) k + 1.

Hence the result follows.Now take m = n + k, k 1 and write S

(2)m = S

(2)n + Rk. We have

1 Fn;m(x; y) = 1 P(S(1)n x; S

(2)n + Rk y)

= 1

Zy0

Fn;n(x; y z)dFk(z),

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where Fk(z) = P(Rk z). We nd that

1 Fn;m(x; y) = Zy0

(1 Fn;n(x; y z))dFk(z) + 1 Fk(y)

(1 Fn;n(x; y))Fk(y) + 1 Fk(y)

(1 Fn;n(x; y))Fk(y).

Using the rst result, we obtain that

liminf1 Fn;m(x; y)

1 F(x; y) n.

This proves the result.Going to random sums, we have the following Corollary.

Corollary 13 (13) We have

lim inf min(x;y)!1

G(x; y)

F(x; y) E(N),

and

lim inf min(x;y)!1

H(x; y)

F(x; y) E(min(N; M)).

The next result is the bivariate analogue of Lemma 4.

Lemma 14 (14) (i) If E(X) = E(Y) = 1, then, as min(x; y) ! 1, we havemG(x; y) t mF(x; y) and mH(x; y) t mF(x; y).

(ii) If Fi 2 ORV with (Fi) > 1, or if Fi 2 RV(i) with 0 i < 1,then, as min(x; y) ! 1, we have G(x; y) t F(x; y) and H(x; y) t F(x; y).

Proof. (i) In Lemma 11 we proved that mG(x; y) = O(1)mF(x; y) and mH(x; y) =O(1)mF(x; y). To prove (i), choose c and x

such that

H(x; y) cF(x; y), x; y x.

Taking integrals, we obtain that

mH(x; y) c

Zxx

Zyx

F(u; v)dudv = c(mF(x; y) R),

where

R =Zx

0

Zyx

+Zx

x

Zx0

+Zx0

Zx0

!F(u; v)dudv

= R(1) + R(2) + R(3).

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For the rst term, we have

R(1) Zx0

Zyx

(F1(u) + F2(v))dudv

m1(x)y + xm2(y),

and it follows that

R(1)

mF(x; y) m1(x

)y

mF(x; y)+ x

m2(y)

mF(x; y)

m1(x)

1

m1(x)+ x

1

x! 0,

as min(x; y) ! 1. In a similar way, we have R(2)=mF(x; y) ! 0. Since R(3),is bounded, we nally have R(3)=mF(x; y) ! 0. We conclude that

liminf

mH(x; y)

mF(x; y) > 0.

This proves the result.(ii) This follows from Lemma 11 and Lemma 13.Remarks.

1) We need extra conditions to nd exact asymptotic results for G(x; y) andH(x; y). Such results will be discussed below.

2) If in the place of mF, we can also consider kF-functions as follows:

kF(x; y) =

Zx0

Zy0

P(X > u; Y > v)dudv.

Clearly we have P(X > u; Y > v) P(X > u) and P(X > u; Y > v) P(Y > v). It follows that

kF(x; y) ym1(x) and kF(x; y) xm2(y).

For (Sn; Tm), we obtain that

kn;m(x; y) ynm1(x) and kn;m(x; y) xmm2(y),

and then also

kn;m(x; y) nmF(x; y) and kn;m(x; y) mmF(x; y),

or kn;m(x; y) min(n; m)m(x; y). Taking random sums, we obtain that kH(s; y) E(min(N; M))m(x; y).

3.3 Subexponential marginals

In the next result we start from subexponential marginals F1(x) and F2(x).Then automatically F1; F2 2 L. In the next result we prove that we the jointd.f. is a multivariate subexponential d.f.. Multivariate subexponential d.f. havebeen studied by Cline and Resnick (1992), Mallor et al (2006). See also Mallorand Omey (2006) and Omey (2006). The next result extends Proposition 11 ofBaltrunas et al. (2006).

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Theorem 15 (15) Suppose that F1(x) 2 S, F2(x) 2 S.Then for all n; m 1and as min(x; y) ! 1 we have

(i)

1 P(S(1)n x; S(2)m y) = (n m)

+F1(x) + (m n)+F2(y) (4)

+ min(n; m)F(x; y) + o(1)F(x; y),

and(ii)

P(S(1)n > x; S(2)m > y ) = min(n; m)P(X > x; Y > y) + o(1)F(x; y). (5)

Proof. We consider Fn;m(x; y) and rst assume that m = n+k, where n; k 1.Now consider the partial maxima

M(1)n = max(X1; X2;:::;Xn),

M(2)m = max(Y1; Y2;:::;Ym).

We write 1 Fn;m(x; y) = An;m(x; y) + Bn;m(x; y), where

An;m(x; y) = P(M(1)n x; M

(2)m y) P(S

(1)n x; S

(2)m y),

Cn;m(x; y) = 1 P(M(1)n x; M

(2)m y).

First consider An;m(x; y). Writing An;m(x; 1) = A1;n(x) and An;m(1; y) =A2;m(y) we have

0 An;m(x; y) A1;n(x) + A2;m(y).

For A1;n(x),we have

A1;n(x) = Fn1 (x) F

n1 (x) = R1;n(x) + B1;n(x),

where

R1;n(x) = 1 Fn1 (x) n(1 F1(x)), and

B1;n(x) = 1 Fn1 (x) n(1 F1(x)).

For B1;n(x) we use the inequality

j1 xn n(1 x)j

n

2

(1 x)2, 0 x 1, (6)

to obtain that B1;n(x) = O(1)F21(x). Since F1(x) 2 S, we have R1;n(x) =

o(1)F1(x). We conclude that

A1;n(x) = o(1)F1(x) + O(1)F21(x).

In a similar way, we can treat A2;m(y). Using F1(x) F(x; y) and F2(y) F(x; y), we nd that

An;m(x; y) = o(1)F(x; y) + O(1)F2

(x; y).

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Now consider Cn;m(x; y). Since m = n + k, we nd that Cn;m(x; y) =1 Fn(x; y)Fk2 (y). Using (6) twice, we nd that

Cn;m(x; y) = 1 (1 nF(x; y) + O(1)F2

(x; y))(1 kF2(y) + O(1)F22(y)),

and then it follows that

Cn;m(x; y) = nF(x; y) + kF2(y) nkF(x; y)F2(y)

+O(1)F2

(x; y) + O(1)F22(y).

Again using F1(x) F(x; y) and F2(y) F(x; y), this gives

Cn;m(x; y) = nF(x; y) + kF2(y) + O(1)F2

(x; y).

We conclude that

1 Fn;m(x; y) = nF(x; y) + kF2(y) + o(1)F(x; y) + O(1)F2

(x; y).

In a similar way, for n = m + k, m; k 1, we get that

1 Fn;m(x; y) = mF(x; y) + kF1(x) + o(1)F(x; y) + O(1)F2

(x; y).

This proves (4). To prove (5), we use identity

P(X > x; Y > y) = P(X > x) + P(Y > y ) (1 P(X x; Y y)). (7)

Without assuming more, it is not clear which of the terms is dominant inthese expressions. It should be noted that this expression holds as min(x; y) !1. In the next section we will assume that min(x; y) ! 1 in a more preciseway.

3.4 Regular variation

Now assume that there exist functions a(t) and b(t) such that as t ! 1, wehave a(t) " 1 and b(t) " 1 and such that

t(1 F(a(t)x; b(t)y)) ! (x; y) < 1, (8)

for all x; y > 0 and min(x; y) < 1. Taking 0 < x < 1 and y = 1, we have

t(1 F1(a(t)x) ! (x; 1).

Replacing t by ai(t), the inverse of a(t),we have

ai(t)(1 F1(tx)) ! (x; 1).

If (x; 1) > 0, we nd that (x; 1) is of the form (x; 1) = cx, where

c > 0, and then it follows that F1(x) 2 RV(). Moreover, also ai

(t) andconsequently also a(t) are regularly varying functions.

Relations of the type (8) were studied among others by de Haan et al (1983,1984) and Omey (1982, 1989, 1990).

Using (8) and Theorem 15, we have the following result. It generalizes aresult of Omey (1990, Corollary 2.3).

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Theorem 16 (16) Suppose that F1(x); F2(y) 2 S and suppose that (8) holds.Then

(i)

t(1 P(S(1)n a(t)x; S(2)m b(t)y))

! (n m)+(x; 1) + (m n)+(1; y) + min(n; m)(x; y),

(ii)tP(S(1)n > a(t)x; S

(2)m > b(t)y) ! min(n; m)(x; y),

for allx;y > 0 and x + y > 0.

Remark. If F1(x) 2 RV, then automatically F1(x) 2 S.

Now consider the subordinated process and P(S(1)N x; S

(2)M y). For the

marginals, there are many situations (cf. Proposition 1) under which we have

P(S(1)N > x) s E(N)F1(x), as x ! 1, (9)

P(S(2)M > y) s E(M)F2(y), as y ! 1. (10)

We prove the following result.

Theorem 17 (17) Suppose that F1(x); F2(x) 2 S and suppose that (8), (9)and (10) hold. Then for all x;y > 0, we have

(i) tP(S(1)N > a(t)x; S(2)M > b(t)y) ! E(min(N; M))(x; y),

(ii)

t(1 P(S(1)N a(t)x; S

(2)M b(t)y) !

E(N M)+(x; 1) + E(M N)+(1; y)

+Emin(N; M)(x; y).

Proof. We have

P(S(1)N > x; S(2)M > y) =

XXpn;mP(S(1)n > x;S(2)m > y)).Now observe the following facts:

(1) pn;mtP(S(1)n > a(t)x; S

(2)m > b(t)y)) ! pn;m min(n; m)(x; y);

(2) pn;mtP(S(1)n > a(t)x; S

(2)m > b(t)y)) pn;mtP(S

(1)n > a(t)x);

(3) pn;mtP(S(1)n > a(t)x) ! pn;mn(x; 1);

(4)PP

pn;mtP(S(1)n > a(t)x) = tP(S

(1)N > a(t)x) ! E(N)(x; 1);

(5)PP

pn;mn(x; 1) = E(N)(x; 1).Using Pratts extension of Lebesgues theorem, we get that

tP(S(1)N > a(t)x; S

(2)M > b(t)y) ! E(min(N; M))(x; y).

The second result follows from the rst result and (7), (9), (10).

Remarks1) IfX and Y are independent and if (8) holds, we have

tP(X > a(t)x; Y > b(t)y) = tP(X > a(t)x)P(Y > b(t)y) ! (x; 1) 0 = 0.

In this case, using (7), we obtain that (x; y) = (x; 1) + (1; y).2) IfF(x; y) = min(F1(x); F2(y)) and (8) holds, then (x; y) = min((x; 1); (1; y)).

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3.5 Second order behaviour

In the univariate theory, it is possible to prove rate of convergence results inProposition 1. As an example, we mention some results of Omey and Willekens(1986, 1987), Willekens (1986). Related results are in Omey (1994), Baltrunasand Omey (1998, 2002), Baltrunas et al. (2006), Omey and Teugels (2002), andthe references given there. In the rst result we assume that = E(X) < 1and we use the notation

R1;N(x) = P(S(1)N > x) E(N)F1(x).

Proposition 18 (18) (Omey and Willekens, 1987) Assume that E((1+ )N) x; Y > y) = Axy (max(x; y)), x; y 1.

In this example we have

F1(x) = Ax, x 1,

F2(y) = Ay, y 1,

andF(x; y) = F1(x) + F2(y) Ax

y(max(x; y)).

2) As a second example, take F such that

F(x; y) = max(F1(x); F2(y)) + (1 )(F1(x) + F2(y)),

where 0 < < 1. ???

To be continued

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Ph.D. Thesis, K.U.Leuven (in Dutch).

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Random sums of random variables and vectors by E. Omey and R. Vesilo

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