THE CM PROPERTY AND COMPARISON OF RISK PROCESSES

OF RISK PROCESSES

ANA MARIA R ˘ADUCAN and GHEORGHIT¸ ˘A ZB ˘AGANU

We generalize a theorem from [3] concerning comparisons of queues and apply it to both queues and risk processes.

AMS 2000 Subject Classification: 60G50, 60J15, 60K10, 60K25.

Key words: risk process, queue, service time, interarrival time, waiting time.

1. QUEUING SYSTEMS AND RISK PROCESSES

The simplest G/G/1 queue is defined by two independent sequences of i.i.d. random variables (Xn)n≥1 and (Yn)n≥1. Xn is theservice time for the nth customer,Yntheinterarrival time. Among the characteristics of a queue we shall focus on the waiting time until entering service of nth customer, Wn. It is well known that the sequence (Wn)n≥1 is given by the recurrence (1.1) W₁= 0, W_n+1= (W_n+X_n−Y_n)₊

(see, for instance, [2]), or, equivalently, with the notation ξi=Xi−Yi,i≥1, (1.2) W_n+1= (W_n+ξ_n)₊ ∀n≥1.

Remark 1.1. Let us denote by G_n the distribution of W_n and by F the distribution of ξ_n. It was proved in [3] that the distributions G_n follow the recurrence G0 = δ0, G1 =F₍₊₎, Gn+1 = (Gn∗F)₍₊₎, ∀n ≥1, where F₍₊₎ = F◦f⁻¹,withf(x) =x₊.Exactly,F₍₊₎=qδ₀+pF_|[0,∞),whereq =F((−∞,0)), p= 1−q, andF_|[0,∞)(A) = ^F_F^{(A∩[0,∞))}_([0,∞)) .

The classic risk proccesshas the form

(1.3) V(t) =S_N(t)−ct,

where S0= 0, Sn=X1+X2+· · ·+Xn,n≥1 andN(t) = max{k|T_k ≤t}, with T₀ = 0, T_n=σ₁+σ₂+· · ·+σ_n,n≥1.

MATH. REPORTS10(60),3 (2008), 277–287

Here Xn is the value of the nth claim, Tn is the arrival moment of the nth claim, and σ_n is the interarrival time. The constant c is the intensity of the cash flow coming to the insurer.

Remark1.2. If we are interested only in the ruin moments, they may only occur at momentsT_n.ButV(T_n) =S_n−c(σ₁+σ₂+· · ·+σ_n) =ξ₁+ξ₂+· · ·+ξ_n, where ξi = Xi −Yi and Yi = cσi. The distribution of ξi is F_X ∗F−Y. The maximum aggregate loss after n transactions is the process (Ln)n≥0 defined by L_n = max(0, ξ₁, ξ₁+ξ₂, . . . , ξ₁ +ξ₂+· · ·+ξ_n). It is well known (see, for instance, [8]) that

(1.4) L₀ = 0, L_n+1= (L^D _n+ξ_n)₊ ∀n≥0.

Therefore, the pair of distribution functionshF_X, FYican be interpreted from both the points of view of a queue and of a risk process.

Among many posibilities of comparison of queues and risk processes we focus on two of them:

Definition 1. Consider the queueshF_X, F_YiandhF_X⁰, F_Y⁰i. We say that – hF_X, F_Yi is better than hF_X⁰, F_Y⁰i (and denote this relation by hF_X, FYi ≺≺ hF_X⁰, F_Y⁰i,orhX, Yi ≺≺ hX⁰, Y⁰i) iff

(1.5) P(W_n> x)≥P(W_n⁰ > x)^not⇔G_n(x)≤G⁰_n(x), ∀n≥1, ∀x≥0, where W_n⁰ is the sequence constructed by reccurence (1.2) with the sequences (X_n⁰)n≥1 of i.i.d. random variables having the distribution FX⁰,and (Y_n⁰)n≥1

with the distribution F_Y⁰.

–hF_X, FYi is asimptotically better thanhF_X⁰, F_Y⁰i iff (1.6) P(W∞> x)≥P(W_∞⁰ > x)^not⇔G∞(x)≤G⁰_∞(x),

where W∞ and W_∞⁰ are the limits of (Wn)n and (W_n⁰)n as n → ∞. It is also known (see, for instance, [2]) that Wn has a weak limit if and only if EX_n<EY_n (which here is equivalent to Eξ₁ <0).

Intuitively, a queue is better that another one if the waiting time of the nth custoner is smaller in the first queue than in the second one, for each n ≥1. If the waiting time in the first queue tends to be smaller as n→ ∞, we say that it is asimptotically safer than the second queue.

From the point of view of an insurer, the safest risk process is one that has the smallest ruin probability after n transactions, for each n≥1. If one risk process has smaller infinite horizon probability than another one, we say that the first process is asimptotically safer than the second one. The reason

is that

hF_X, F_Yi ≺≺ hF_X⁰, F_Y⁰i ⇔P(L_n> u)≥P(L⁰_n> u) (1.7)

⇔ψn(u)≥ψ⁰_n(u), ∀n≥1, ∀u≥0,

where P(Ln > u) = ψn(u) is the ruin probability after n transactions, with initial capital u and

hF_X, F_Yi ≺ hF_X⁰, F_Y⁰i ⇔P(L∞> u)≥P(L⁰_∞> u) (1.8)

⇔ψ(u)≥ψ⁰(u), ∀u≥0, where ψ(u) is the ruin probability.

An alternative way to describe these concepts is as follows: it is known that the random variableX is said to bestochastically dominated by the random variable Y (see, for instance, [5], [6]) iff P(X > x) ≤ P(Y > x)

∀x ≥ 0. Denote that by X ≺_st Y, or, FX ≺_st FY. Let now two sequences of i.i.d. random variables, (ξ_n)n≥1 and (ξ_n⁰)n≥1 such that

(1.9) ξn∼F,(Ln)n≥0, L0= 0, Ln+1 = (Ln+ξn)+, ξ⁰_n∼F⁰,(L⁰_n)n≥0, L⁰₀= 0, L⁰_n+1 = (L⁰_n+ξ_n⁰)+

Notation. We writeF ≺₊₊ F⁰ (orξ ≺₊₊ξ⁰,) iff Ln≺_st L⁰_n ∀n≥0, and F ≺₊F⁰ (or ξ≺₊ξ⁰) iff L∞≺_stL⁰_∞.

Coming back to our previous definitions, we see that

(1.10) hF_X, F_Yi ≺≺ hF_X⁰, F_Y⁰i ⇔ξ≺₊₊ξ⁰⇔G_n≺_st G⁰_n, ∀n≥1, and

(1.11) hF_X, F_Yi ≺ hF_X⁰, F_Y⁰i ⇔ξ≺₊ξ⁰ ⇔G∞≺_st G⁰_∞. In [3] there was proved the result below.

Theorem1.1. Let E_adenote the exponential distribution with parameter a. If X∼Ea,Y ∼Eb,X⁰ ∼E_a⁰,Y⁰∼E_b⁰,then

(1.12) hF_X, F−Yi ≺≺ hF_X⁰, F−Y⁰i ⇔

a≥a⁰ ρ≤ρ⁰, where ρ= ^EX_EY .

Remark 1.3. In terms of queues, the ratio ^EX_EY is called the traffic in- tensity. In ruin theory, the same ratio is denoted _1+θ¹ , and θ is called the loading factor. The meaning of Theorem 1.1 is that in the world of exponential distributions the following assertion holds: if we have to compare two queues and both service time and trafic intensity are smaller in the first one, then the

first one is better. Or, in terms of risk theory: if the claims are smaller in the first risk process and the loading factor is greater, then the first risk process is safer.

Our purpose is to generalize this fact to other distributions.

The proof of Theorem 1 relies on a remarkable property of the exponen- tial distribution, namely,

Ea∗E−b =pEa+qE−b

with p = _EX+EY^EY , q = _EX^EX_+EY. Here, we denote by E−b the distribution function with density e−b(x) = be^bx1_(−∞,0)(x), x ∈ R. The exact meaning of this notation is as follows: if E_b is the distribution function of the random variable X, thenE−b is the distribution function of the random variable−X.

Definition 2. We say that the distribution functionsF andGare conju- gated iff

(1.13) F ∗G=pF + (1−p)G for somep∈(0,1).

Remark 1.4. This property has been studied by D. Dugu´e for the first time in 1939 in terms of characteristic functions. He found two pairs (ϕ₁, ϕ₂) such thatϕ1(t)ϕ2(t) = ^ϕ1(t)+ϕ₂ ^2(t).In later papers (see, for example, [4] or [7]) one can find other examples, though with incomplete or incorrect proofs.

2. GENERALIZATION

We prove the following result

Theorem 2.1. Let X, X⁰ > 0 (a.s.), Y, Y⁰ ≥ 0 such that X ∼ F1, X⁰ ∼F₁⁰,−Y ∼F2,−Y⁰ ∼F₂⁰. Assume that the pairs (F1, F2) and (F₁⁰, F₂⁰) are conjugated and ρ, ρ⁰ <1,where ρ= ^EX_EY, ρ⁰ = ^EX_EY0⁰.Then

(i)ρ≤ρ⁰ and F1≤F₁⁰ ⇒F1∗F2≺₊₊F₁⁰∗F₂⁰ ⇒ρ≤ρ⁰ and F1 ≤ ^p_p⁰F₁⁰; (ii)F1∗F2 ≺₊F₁⁰∗F₂⁰ if and only if

∞

P

n=0

(1−ρ)ρⁿΓn≺_st

∞

P

n=0

(1−ρ⁰)ρ⁰ⁿΓ⁰_n. Here Γn=F₁^∗n and Γ⁰_n=F₁^0∗n.

The proof will be divided into several steps. Let us fix some nota- tion: F = F₁ ∗F₂, F⁰ = F₁⁰ ∗ F₂⁰, G₀ = G⁰₀ = δ₀, G_n+1 = (G_n ∗F)₍₊₎, G⁰_n+1 = (G⁰_n∗F⁰)₍₊₎,Γ is the row vector (Γ0,Γ1,Γ2, . . .),Γ⁰ is the row vector (Γ⁰₀,Γ⁰₁,Γ⁰₂, . . .),e is the column vector (1,0,0, . . .)⁰.

Step 1. Lemma 2.2. If F = F1∗F2 = pF1+qF2, then Gn = ΓQⁿe, where

(2.1) Q=Q(p) =

























q q² q³ . . . qⁿ⁻¹ qⁿ qⁿ⁺¹ . . . p pq pq² . . . pqⁿ⁻² pqⁿ⁻¹ pqⁿ . . . 0 p pq . . . pqⁿ⁻³ pqⁿ⁻² pqⁿ⁻¹ . . . . . . .

0 0 0 . . . p pq pq² . . .

0 0 0 . . . 0 p pq . . .

0 0 0 . . . 0 0 p . . .

. . . .

















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

 .

Proof. For n = 1, we have ΓQe = qΓ0 +pΓ1 = qδ0 +pF1 = (qF2+ pF₁)₍₊₎=F₊=G₁, hence (2.1) is true. Next, it is clear thatG_n has the form (2.2) G_n=α_n,0Γ₀+α_n,1Γ₁+· · ·+α_n,nΓ_n, ∀n≥0,

where α_n,j are some real numbers. To find them notice that

G_n∗F = n

X

k=0

α_n,kΓ_k

∗(qF₂+pF₁) (2.3)

=q

n

X

k=0

αn,k(Γk∗F2) +p

n

X

k=0

αn,kΓk+1.

On the other hand it is easy to check that

(2.4) Γ_k∗F₂ =pΓ_k+pqΓ_k−1+q²pΓ_k−2+· · ·+q^k−1pΓ₁+q^kF₂. Taking this into account, we see that

Gn+1= Γ0(αn,0q+αn,1q²+· · ·+αn,n−1qⁿ+αn,nqⁿ⁺¹)+

+ Γ₁(α_n,0p+α_n,1pq+· · ·+αn,n−1qⁿ⁻¹p+α_n,nqⁿp)+

+ Γ2(αn,1p +αn,2pq+· · ·+αn,n−1qⁿ⁻²p+αn,nqⁿ⁻¹p)+

. . .

+ Γ_n(αn,n−1p+α_n,npq) + Γ_n+1(α_n,npq),

which can be written as Gn+1= (Γ0,Γ1,Γ2, . . .)·

·



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q q² q³ . . . qⁿ⁻¹ qⁿ qⁿ⁺¹ . . . p pq pq² . . . pqⁿ⁻² pqⁿ⁻¹ pqⁿ . . . 0 p pq . . . pqⁿ⁻³ pqⁿ⁻² pqⁿ⁻¹ . . . . . . .

0 0 0 . . . p pq pq² . . .

0 0 0 . . . 0 p pq . . .

0 0 0 . . . 0 0 p . . .

. . . .



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























 αn,0

α_n,1 αn,2

. . . αn,n

0 0 . . .



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

.

Step 2. LetP = (Pn)n≥0 be a column-stochastic matrix such that the columns are stochastically increasing, which means that P

i≥k

p_i,n ≤ P

i≥k

p_i,n+1,

∀k≥1,∀n≥0 (or,Pn≺_stPn+1,∀n≥0).We call such a matrixmonotonous.

We say that P is dominated byQand write P ≺_st QifP_n≺_st Q_n,∀n≥0.

Remark 2.1. IfP, P⁰, Q, Q⁰ are monotonous matrices such thatP ≺_st Q and P⁰ ≺_st Q⁰ then P P0 ≺_st QQ⁰. Indeed, the nth column of P P⁰ is Un = P

k≥1

p_1,kp⁰_k,n,P

k≥1

p_2,kp⁰_k,n, . . .0

= (u_1,n, u_2,n, . . .)⁰, the nth column of QQ⁰ is V_n =

P

k≥1

q_1,kq_k,n⁰ , P

k≥1

q_2,kq⁰_k,n, . . .0

= (v_1,n, v_2,n, . . .)⁰ and is easy to prove that P

j≥0

u_l+j,n ≤ P

j≥0

v_l+j,n, ∀l ≥ 1, ∀n ≥ 1. Let now s_l,k = p_l,k +p_l+1,k+

· · ·, tl,k = ql,k +ql+1,k+· · ·. Then S1 = P

j≥0

ul+j,n = P

k≥1

p⁰_k,nsl,k and S2 = P

j≥0

vl+j,n = P

k≥1

q_k,n⁰ tl,k. From the hypothesis, sl,k ≤ tl,k and s⁰_k,n ≤ t⁰_k,n and if P

k≥0

x_k ≥ 0, ∀k ≥ 1 then x_k ≥ 0, ∀k ≥ 1. It follows that S₁ ≤ S₂ hence P P⁰ ≺QQ⁰.

We intend to prove that G_n ≺ G⁰_n, i.e., Γ(x)Qⁿ(p)e⁰ ≺ Γ⁰(x)Qⁿ(p⁰)e⁰. First, we notice that Q(p) ≺Q(p⁰). Indeed, a comparison of thenth columns of these matrices shows that the above domination condition is equivalent to (2.5) q^j+1 ≥q^0j+1, ∀j= 1, . . . , n.

But, from the hypothesisρ≤ρ⁰,this is obvious. Then, according to Remark 4, Qⁿ(p) ≺Qⁿ(p⁰), ∀n≥1. It is also true that Γ(x)≺Γ⁰(x). Therefore Gn≺_st G⁰_n, ∀n ≥ 1, which is equivalent by definition to F ≺₊₊ F⁰. Thus, the first implication from (i) is proved.

To prove the second implication, remark that F ≺₊₊ F⁰ ⇒ G1 ≺_st G⁰₁. Moreover,G₁= (F₁∗F₂)₍₊₎= (pF₁+qF₂)₍₊₎=pF₁+qδ₀andG⁰₁=p⁰F₁⁰+q⁰δ₀.

To findp and q notice that E(X−Y) is the expectation of F1∗F2.The expectation ofpF₁+qF₂is equal topEX−qEY, hence E(X−Y) =pEX−qEY. Then

(2.6) p= EX

EX+ EY = ρ

ρ+ 1, q= E Y

EX+ EY = 1 ρ+ 1.

But G₁ ≺_st G⁰₁ means that G₁(x)≥G⁰₁(x) for any x >0⇔pF₁(x)≤p⁰F₁⁰(x)

∀x >0.Lettingx→0 and taking into account thatF1(0 + 0) =F₁⁰(0 + 0) = 1, we see thatp≤p⁰⇒ρ≤ρ⁰.Moreover, F1(x)≤ ^p_p⁰F₁⁰(x).

(ii)Step 3. Let us consider a Markov chain Z such that Z_n+1= (Z_n+ ηn)+, whereηn are i.i.d. random variables with

η1 ∼

. . . −2 −1 0 1 . . . pq³ pq² pq p

.

We can write ξn = 1−ηn ∼Negbin(1, p), n≥1. In our case the ruin is not certain if Eη_n<0,which is equivalent toq > p.

The transpose of the transition matrix ofZ is

Q=



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

q p 0 0 0 0 . . .

q² pq p 0 0 0 . . . q³ pq² pq p 0 0 . . . q⁴ pq³ pq² pq p 0 . . . . . . .











 .

It is known that if Zn → Z∞ as n → ∞, then the distribution of Z∞

is a distribution Π = (π₀, π₁, . . .) which satisfies the equation ΠQ⁰ = Π, or explicitely,πn=qⁿ⁺¹π0+pqⁿπ1+· · ·+pqπn+pπn+1,∀n≥0.It appears that pπ_n+2=π_n+1−qπ_nfor any n≥1, henceπ_n= (1−ρ)ρⁿ,∀n≥0,withρ= _p^q. Otherwise written Π = Negbin(1,1−ρ).In this way, we proved

Lemma2.3. We have

(2.7) G∞=

∞

X

n=0

(1−ρ)ρⁿΓn.

Now, the proof of (ii) from Theorem 2.1 is obvious.

Here are some particular cases of (2.7).

Lemma2.4. Let us denote α= 1−α for any α∈(0,1). Then a)If X ∼exp(a) then G∞=ρδ₀+ρexp(aρ) =ρδ₀+ρexp(a−b).

b) If X ∼Negbin(1, α) thenG∞=ρδ₀+ρNegbin(1,_1−αρ^αρ ).

c)If X ∼Geometric(α) thenG∞=ρδ0+ρGeometric(αρ).

d) If X ∼ B(1, α) then G∞ = Negbin(1,_1−ρα^ρ ), where B(1, α) =αδ0+ αδ1.

Proof. Since Γn = F_X^∗n, if mF is the moment generating function cor- responding to the distribution function F then, according to Lemma 2.3, m_G∞(t) = (1−ρ)

∞

P

n=0

ρⁿ(m_FX(t))ⁿ, hencem_G∞(t) = _1−ρm^1−ρ

FX(t). a)mG∞(t) = (1−ρ)_1−ρ¹a

a−t = (1−ρ)_a−t−ρa^a−t = (1−ρ)(1 + _{a(1−ρ)−t}^ρa ) = ρ+ρ_aρ−t^aρ ⇒G∞=ρδ₀+ρexp(aρ).

b)mG∞(t) = (1−ρ)_1−ρ¹α

1−αet = (1−ρ)_{(1−αρ)−αe}^1−αet t = (1−ρ)(1+ρ_{(1−αρ)−αe}^α t)

= ρ +ρ ^αρ

(1−αρ)(1−_1−αρ^αet ) = ρ +ρ_1−βe^β _t ⇒ G∞ = ρδ0 + ρNegbin(1, β) with β = _1−αρ^αρ .

c)m_G∞(t) = (1−ρ) ¹

1−ρ ^αet

1−αet

= (1−ρ)_{1−(αρ+α)e}^1−αet t = (1−ρ)(1+ρ_{1−(αρ+α)e}^αet t)

=ρ+ρ_1−βe^β _t,withβ =ρα,hence G∞=ρδ0+ρGeometric(αρ).

d)m_G∞(t) = (1−ρ)_{1−ρ(α+αe}¹ _t) = (1−ρ)_{(1−ρα)−αρe}¹ t =

1−ρ 1−ρα

1−_1−ρα^αρ e^t ⇒G∞= Negbin 1,_1−ρα^ρ

.

Remark 2.2. Formulae b), c), d) from Lemma 2.4. can be slightly gen- eralized as follows. Let h >0 be arbitrary. Denote by

– Geometric(α, h) the distribution of hX ifX ∼Geometric(α);

– Negbin(1, α, h) the distribution of hX ifX ∼Negbin(1, α);

– B (k, α, h) the distribution ofhX ifX ∼B(k, α).

Then b), c), d) from Lemma 2.4 become

b⁰) IfX ∼Negbin(1, α, h) thenG∞=ρδ₀+ρNegbin(1,_1−αρ^αρ , h).

c⁰) IfX ∼Geometric(α, h) then G∞=ρδ₀+ρGeometric(αρ, h).

d⁰) IfX ∼B(1, α) thenG∞= Negbin(1,_1−ρα^ρ , h).

Remark 2.3. We have seen that, for X, X⁰ > 0, ρ ≤ ρ⁰ and X ≺_st X⁰ imply G1 ≺_st G⁰₁.But is still true thatG1 ≺_st G⁰₁ does imply thatρ≤ρ⁰ and X ≺_st X⁰? The answer is no. For instance in Examples 1) and 2) below this is true but in Example 3) just the opposite X⁰ ≺_st X holds. In Example 4, there is no stochastical domination at all.

Example 1. X∼Geometric(α, h), Y ∼Negbin(1, β, h).It is known (see [7]), that (FX,F−Y) are conjugated withp= _αβ+α^αβ .We have

G1 =pF_X+qδ0, G⁰₁=p⁰F_X⁰+q⁰δ0 ⇒G1 ≺_st G⁰₁

⇔pαⁿ≤p⁰α⁰ⁿ⇒

α≥α⁰ p≤p⁰.

We notice that p ≤ p⁰ ⇔ ρ ≤ ρ⁰. In this case, FX ≺_st FX⁰ ⇔ FX(x) ≤ FX⁰(x)⇔α≥α⁰.Now, it is clear thatG1≺_st G⁰₁⇒ρ≤ρ⁰ and X≺_stX⁰.

Example 2. X ∼ Negbin(1, α, h), Y ∼ Geometric(β, h). In this case (F_X,F−Y) are conjugated with p = _αβ+α^β ; the computations are similar to those from Example 1.

Example 3. X ∼

0 h α α

, −Y ∼

. . . −3h −2h −h 0 . . . γββ² γββ γβ γ

. It was proved in [7] that (F_X,F−Y) are conjugated if an only if γ = 1−αβ. In this case, F =FX−Y =pF_X +qF−Y with p=γ =αβ.

The reader can check that G1 = (1−α²β)δ0 +α²βδ_h = B (1, αp) and ρ = ^EX_EY = ^αβ_γ = _1−αβ^αβ = _1−p^p .If X⁰ and Y⁰ are of the same type as X and Y, but with parametersα⁰, β⁰, γ⁰, h⁰ instead, we have

G₁ ≺_stG⁰₁ ⇔

h≤h⁰

α²β ≤α⁰²β⁰ ⇔

h≤h⁰ αp≤α⁰p⁰ . On the other hand, X ≺_st X⁰ ⇔

h≤h⁰

α≤α⁰ . As ρ ≤ ρ⁰ ⇔ p ≤ p⁰, this fact confirms once more the implication “FX ≺_st FX⁰,ρ≤ρ⁰ ⇒G1≺_st G⁰₁”.

To find a counterexample, let us takeα= ¹₂,α⁰ = ¹₃,β= ¹₄,β⁰= ³₄,h= h⁰.In this caseρ= ¹₇,ρ⁰= ¹₃,p= ¹₈,p⁰ = ¹₄. Notice thatX⁰ ≺_st X, but it is not true thatX ≺_stX⁰.Moreover Γ_n= B n,¹₂

, Γ⁰_n= B n,¹₃

.The reader can see thatG1= ¹⁵₁₆δ0+₁₆¹δh= B 1,₁₆¹, h

, G⁰₁= ¹¹₁₂δ0+₁₂¹δh= B 1,₁₂¹, h

.Of course G₁ ≺_stG⁰₁.More than that: not onlyG₁ ≺_stG⁰₁, but alsoG∞≺_stG⁰_∞. Indeed, according to Lemma 2.4 d), G∞= (1−ρ) P

n≥0

ρⁿB(n, α) = Negbin(1,_1−ρα^ρ , h) and G⁰_∞ = Negbin(1,_1−ρ^ρ0₀

α⁰, h). Forh =h⁰ = 1, we get in our particular case G∞= Negbin(1,¹²₁₃), G⁰_∞= Negbin(1,¹²₁₄).Obviously,G∞≺_st G⁰_∞.

Remark 2.4. We believe that, at least for distribution of this typeG₁≺_st G⁰₁, G∞ ≺_st G⁰_∞ does imply Gn ≺_st G⁰_n,∀n≥1. If that were true, we would have an example in which X−Y ≺₊₊X⁰−Y⁰ but it would not be true that X ≺_stX⁰!

So, we know thatG1 = B(1, αp),G⁰₁ = B(1, α⁰p⁰),G∞= Negbin(1,_1−ρα^1−ρ )

= Negbin(1,_1−2p+αp^1−2p ) and G⁰_∞ = Negbin(1,_1−2p^1−2p0+α⁰⁰p⁰), and G₁ ≺_st G⁰₁ ⇔ αp≤α⁰p⁰,G∞≺_st G⁰_∞⇔ _1−2p^αp ≤ _1−2p^α0p00. We conjecture that

(2.8) αp≤α⁰p⁰, αp

1−2p ≤ α⁰p⁰ 1−2p⁰ implies

(2.9) Gn≺_stG⁰_n ∀n≥1.

We were unable to infirm it on the computer. We tried tens of randomly chosen pairs (α, p), (α⁰, p⁰) which satisfy (2.8) for nfrom 1 to 100. However, we are able to prove that (2.9) holds at least for n= 2.

Let us thus prove that G1 ≺_st G⁰₁ and G∞≺_st G⁰_∞ impliesG2 ≺_st G⁰₂.This is equivalent to (2.8) ⇒ (2.10), where

(2.10) αp(1−p²+ 2p−αp)≤α⁰p⁰(1−p⁰²+ 2p⁰−α⁰p⁰).

Let us consider the function f : (0,¹₂)×(0,1)→D, f(p, α) = (αp,_1−2p^αp ) (our conditions are 0 < p, p⁰ < ¹₂, 0 < α, α⁰ < 1). The system

x=αp y= _1−2p^αp has the unique solution p = ¹⁻

x y

2 = ^y−x_2y , α = ^x_p = _y−x^2xy. If we denote by h x, y

, αp(1−p² + 2p−αp), the expression we are interested in, we have h(x, y) =x 2− ^x_y − ¹⁻

x y

2

2

−x

, wherex, y satisfy (2.11) 0< x < y, y−x−2xy >0.

A more convenient expression of his

(2.12) 4h(x, y) = 7x−4x²−2x² y −x³

y².

We prove that h is increasing by verifying that its partial derivatives are positive. These are

4∂h

∂x = 7−8x−4x y−3x²

y² = 3(y²−x²) + 4y(y−x−2xy)

y² , ∂h

∂y = 2x² y² +2x³

y³ . Now, applying (2.11), it is clear that ^∂h_∂x ≥ 0, ^∂h_∂y ≥ 0. Therefore G1 ≺_st G⁰₁ and G∞≺_st G⁰_∞ ⇒ G2 ≺_st G⁰₂.

Example 4. X ∼ Geometric(1−e⁻²,2), Y ∼ Negbin(1,e⁻⁴,2), X ∼ Geometric(1−e⁻³,3), Y ∼ Negbin(1,e⁻³,3).In this case

ρ = EX EY =

e² e²−1

e⁴−1 = e²

(e²−1)(e⁴−1), p= ρ

ρ+ 1 = e²

e²+ (e²−1)(e⁴−1), ρ⁰ = EX⁰

EY⁰ =

e³ e³−1

e⁴−1 = e³

(e³−1)(e⁴−1), p⁰= ρ⁰

ρ⁰+ 1 = e³

e³+ (e³−1)(e⁴−1).

Moreover, for x > 0 we have G1(x) = pFX(x) = pe⁻²[^x₂] and G⁰₁(x) = p⁰F_X⁰(x) = p⁰e⁻³[^x₃]. Consequently, G₁ ≺_st G⁰₁ is equivalent to pe⁻²[^x₂] ≤ p⁰e⁻³[^x₃] ∀x > 0 ⇔ sup(e³[^x₃]⁻²[^x₂]) ≤ ^p_p⁰ ⇔ e ≤ ^p_p⁰. As the last inequality holds, we see that G1≺_st G⁰₁.

AsFX ≤FX⁰ ⇔2_x

2

≥3_x

3

and this is not always true (one can see a graphic representation), there is no stochastic domination between X and X⁰! In this case, G₁ ≺_st G⁰₁ ⇒ρ≤ρ⁰,butX⁰ ⊀stX and X⊀stX⁰.

Acknowledgements.This research was supported by The Hungarian-Romanian In- tergovernmental Cooperation for Science and Technology under Grant RO-40/2005.

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[6] R. Szekli, Stochastic Ordering and Dependence in Applied Probability. Springer, Berlin, 1995.

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Received 21 May 2007 “Gheorghe Mihoc-Caius Iacob”

Institute of Mathematical Statistics and Applied Mathematics Casa Academiei Romˆane Calea 13 Septembrie nr. 13 050711 Bucure¸sti 5, Romania

anaraducan@yahoo.ca and

University of Bucharest

Faculty of Mathematics and Computer Science Str. Academiei 14

010014 Bucharest, Romania zbagang@fmi.unibuc.com