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Intuitive approximations for the renewal function
Statistics and Probability Letters 84 (2014) 72–80
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Intuitive approximations for the renewal function Kosto V. Mitov a,∗ , Edward Omey b a
Aviation Faculty - NMU, 5856 D. Mitropolia, Pleven, Bulgaria
b
HUB-EHSAL, Stormstraat 2, 1000, Brussels, Belgium
article
info
Article history: Received 30 April 2013 Received in revised form 24 September 2013 Accepted 25 September 2013 Available online 29 September 2013
abstract ∗n It is hard to find explicit expressions for the renewal function U (x) = n=0 F (x). Many researchers have made attempts to find suitable approximations for U (x). In this paper we present simple approximations and show that they cover many of the known results. © 2013 Elsevier B.V. All rights reserved.
∞
MSC: 60K05 60E10 26A12 Keywords: Renewal function Approximations Regular variation The class gamma
1. Introduction Suppose that X , X1 , X2 , . . . are i.i.d. and nonnegative r.v. with distribution function F (x) = P (X ≤ x) and F (0+) = 0. If F a density, we denote it by f (x). The tail of F is denoted by F (x) = 1 − F (x). If X has a finite mean, we denote it by µ = EX = has ∞ ¯ F (x)dx. We assume that F is non lattice. Let F (s) = E (e−sX ) denote the Laplace transform of F . Let Sn denote the partial 0 sums Sn = X1 + X2 + · · · + Xn , n ≥ 1, and let S0 = 0. Clearly we have P (Sn ≤ x) = F ∗n (x), n ≥ 0. As F ∗ (.) denote the ∞usual ∗0 ∗1 ∗n ∗(n−1) ∗(n−1) n-fold convolution of F (.) with itself, that is F (x) = 1, F (x) = F (x), F (x) = F ∗ F ( x) = 0 F (x − y)dF (y), n = 2, 3, . . .. In what follows we will use the same notation in more general sense. If G(.) is a measure on the positive half ∞ line and g (.) is an arbitrary measurable function we will denote G ∗ g (x) = 0 g (x − y)dG(y). ∞ ∗n The renewal function U (x) is given by U (x) = n=0 F (x). It is well known that the renewal function satisfies the following renewal equation U (x) = 1+U ∗F (x). It is well known that U (x) < ∞, for all x ≥ 0. If µ < ∞, the renewal theorem states that U (x)/x → 1/µ as x → ∞ and Blackwell’s theorem shows that for any fixed y > 0, U (x + y) − U (x) → y/µ as x → ∞. For these results we refer to Blackwell (1948) or Feller (1971). There have been many successful attempts to obtain the rate of convergence in these results. Some authors such as Rogozin (1972) or Frenk (1983, 1987) use Banach algebra techniques. Alsmeyer (1991), Carlson (1983) or Stone (1965) used Fourier analysis. Alsmeyer (1991) and Ney (1981) used coupling methods. In Dohli et al. (2002), the authors give an overview of numerical approximations in the renewal theorems. In this paper we propose a simple and intuitive approach to approximate U (x). It turns out that our approximations cover all the results
∗
Corresponding author. E-mail addresses: [email protected], [email protected] (K.V. Mitov), [email protected] (E. Omey).
0167-7152/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.spl.2013.09.030
K.V. Mitov, E. Omey / Statistics and Probability Letters 84 (2014) 72–80
73
that we have found in the literature. To define our approximations, recall that the Laplace transform of U (x) is given by
U (s) =
∞
e−sx dU (x) = (1 − F (s))−1 ,
s > 0.
0
When µ = EX < ∞, we can define the equilibrium distribution function Fe (x) as Fe (x) = µ
−1
x
F (y)dy
for x ≥ 0.
0
If EX 2 < ∞ then the mean of the equilibrium distribution is finite. We will denote µe = Fe (x) has the density fe (x) = µ−1 F (x), x ≥ 0. Clearly we have Fe (s) = (1 − F (s))/(µs), s > 0 and it follows that 1
1
∞ 0
(1 − Fe (x))dx < ∞. Clearly
1
, s > 0. µs 1 − ( 1 − Fe (s)) ∞ ∞ n Using a Taylor expansion, we obtain that U (s) = µ1s n=0 Tn (s). Using Newton’s binomial theorem, we n=0 (1 − Fe (s)) = U (s) =
µs Fe (s)
=
obtain that
n 1 1 k (1 − Fe (s))n = C (−1)k Fek (s). Tn (s) = µs µs k=0 n
It follows that U (x) =
∞
Tn (x),
for almost all x ≥ 0,
(1)
n=0
x
k k where Tn (x) = µ1 F ∗k (y)dy. We denote by Fe∗k (x) the kfold convolution of Fe (x) with itself. It is tempting to k=0 Cn (−1) 0 e use formula (1) to obtain consecutive approximations for U (x). Based on (1) we propose the following approximations for the renewal function: for fixed k ≥ 0, we set
n
Uk (x) =
k
Tn (x).
n =0
In the paper we consider the cases 0 ≤ k ≤ 3 and show that our approximation corresponds to the approximations that have been published in many papers before. 2. The functions Tn (x) 2.1. Some alternative expressions Lemma 1. Let x ≥ 0. We have T0 (x) = x/µ and for n ≥ 1, we have Tn (x) = −
n 1
µ
Cnk (−1)k
x
(1 − Fe∗k (y))dy.
(2)
0
k=1
Moreover, if µe < ∞, for n ≥ 2 we have Tn (x) =
n 1
µ
Cnk (−1)k
∞
(1 − Fe∗k (y))dy.
x
k=1
Proof. The result for T0 (x) is clear. The result for Tn (x) follows because we have n
Cnk (−1)k = (1 − 1)n = 0.
k=0
µ
Now assume that µe < ∞. Clearly we have Tn (∞) = − µe then take the first derivative to find:
(x − 1)n =
n
Cnk xk (−1)n−k
and
n(x − 1)n−1 =
k=0
n
n
k=1
Cnk (−1)k k. Now we use Newton’s binomial formula and
Cnk kxk−1 (−1)n−k ,
x ∈ R.
k=1
k n−k Taking x = 1 this is 0 = = (−1)n k=1 Cnk k(−1)k and so we find that Tn (∞) = 0. The result follows. k=1 Cn k(−1) Now we consider into more details the terms Tk (x), k = 1, 2, 3.
n
n
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Lemma 2. For n ≥ 2, let Re,n (x) = 1 − Fe∗n (x) − n(1 − Fe (x)). We have T1 (x) = µ−1 T3 (x) = µ−1
x
F e (y)dy,
0 x
T2 (x) = −µ−1
x
Re,2 (y)dy, 0
(Re,3 (y) − 3Re,2 (y))dy.
0
If µe < ∞, we have T2 (x) = µ−1
∞
Re,2 (y)dy,
T3 (x) = −µ−1
∞
(Re,3 (y) − 3Re,2 (y))dy.
x
x
Proof. Using (2) with n = 1 we have the result for T1 (x). Using (2) with n = 2, we have T2 (x) = −µ
−1
x
−2(1 − Fe (y)) + (1 − Fe∗2 (y)) dy = −µ−1
x
Re,2 (y)dy.
0
0
Using (2) with n = 3, we have T3 (x) = µ−1
x
3(1 − Fe (y)) − 3(1 − Fe∗2 (y)) + (1 − Fe∗3 (y))dy 0
= µ−1
x
(Re,3 (y) − 3Re,2 (y))dy. 0
If µe < ∞, we have
∞ 0
Re,k (y)dy = 0, k ≥ 2, and the second result follows.
2.2. Asymptotic behavior: regular variation In this section we study the asymptotic behavior of Tk (x), 1 ≤ k ≤ 3. More precisely we study the regularly varying behavior of the functions involved. Recall that a positive and measurable function g is regularly varying with index α, α ∈ R if it satisfies lim g (xt )/g (t ) = xα ,
∀x > 0.
t →∞
Notation: g ∈ RV (α). Typical examples are g (x) = xα or g (x) = xα log(x). For a standard book about regular variation and its applications, we refer to Bingham et al. (1987) or Geluk and de Haan (1981). In the first result we discuss the asymptotic behavior of Re,n (x). We need the following results of Omey and Willekens (1987). Lemma 3 (Omey and Willekens (1987, Corollary 2.4)). Suppose that G(x) = P (Y ≤ x) has a density g ∈ RV (−β) with β > 2. For n ≥ 2, let Rn (x) = 1 − G∗n (x)− n(1 − G(x)). Then λ = EY < ∞ and for all n ≥ 2, we have Rn (x) ∼ λn(n − 1)g (x), x → ∞. In our case, we have g (x) = (1 − F (x))/µ and G(x) = Fe (x). The lemma gives the following result. Proposition 1. Suppose that F (x) ∈ RV (−α) and α > 2. Then µe < ∞ and for all n ≥ 2 we have Re,n (x) ∼ n(n − 1)(µe /µ)F (x),
x → ∞.
Now we discuss the asymptotic behavior of Re,3 (x) − 3Re,2 (x). In the result we assume that F has a density f . Proposition 2. Suppose that f (x) ∈ RV (−β) and β > 3. Then µe < ∞ and Re,3 (x) − 3Re,2 (x) ∼ f (x)2µ2e /µ, x → ∞. Proof. First note that Re,3 (x) = Re,2 ∗ Fe (x) + 2Re,2 (x). It follows that Re,3 (x) − 3Re,2 (x) = Re,2 ∗ Fe (x) − Re,2 (x). Clearly we have Re,3 (x) − 3Re,2 (x) = µ−1
x
(Re,2 (x − y) − Re,2 (x))F (y)dy − Re,2 (x)F e (x) 0
x
(Re,2 (x − y) − Re,2 (x))fe (y)dy − Re,2 (x)F e (x),
= 0
where fe (x) = F (x)/µ is the density of Fe . Here and later we denote by fe⊗2 the usual convolution of the density fe with itself, that is fe⊗2 is the density of Fe∗2 . We have Re,2 (x − y) − Re,2 (x) = Fe∗2 (x) − Fe∗2 (x − y) − 2(Fe (x) − Fe (x − y))
x
(fe⊗2 (z ) − 2fe (z ))dz .
= x −y
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75
It follows that Re,3 (x) − 3Re,2 (x) =
x
x
(fe⊗2 (z ) − 2fe (z ))fe (y)dzdy − Re,2 (x)F e (x).
(3)
x −y
y=0
First we analyze r2 (x) = fe⊗2 (x) − 2fe (x). We have r 2 ( x) = 2
x/ 2
fe (x − y)fe (y)dy − 2fe (x)
0 x/ 2
=2
(fe (x − y) − fe (x))fe (y)dy − 2fe (x)F e (x/2)
0
=
2
µ
x/ 2
x
f (z )dzfe (y)dy − 2fe (x)F e (x/2) = I − II . z =x −y
y=0
By assumption we have f ∈ RV (−β). Then for 0 ≤ y ≤ x/2, we have x/2 ≤ z ≤ x. For fixed y, it follows that 1
f (x)
x
f (z )dzfe (y) → yfe (y),
x → ∞.
z =x −y
On the other hand, since f ∈ RV (−β), one obtains 1
f (x)
f (z )
x
f (z )dz ≤ z =x −y
sup
x/2≤z ≤x
f ( x)
y = O(1)y.
Using Lebesgue’s theorem, we obtain that I /f (x) → (2/µ)
∞
yfe (y)dy = 2µe /µ,
x → ∞.
y=0
Because β > 3, we also have II = O(1)xf (x)x2 f (x) = o(1)f (x). We conclude that r2 (x) ∼ f (x)2µe /µ ∈ RV (−β). Now we consider Re,3 (x) − 3Re,2 (x). Using (3), we have Re,3 (x) − 3Re,2 (x) =
x y=0
x
r2 (z )dzfe (y)dy − Re,2 (x)F e (x).
z =x−y
Since r2 (x) ∈ RV (−β), we have
x y=0
x
r2 (z )dzfe (y)dy ∼ r2 (x)µe ∼ 2(µ2e /µ)f (x),
x → ∞.
z =x −y
Since Re,2 (x) = O(1)fe (x)F e (x) and f (x) ∈ RV (−β), β > 3, we obtain that Re,2 (x) = o(1)f (x), x → ∞. We conclude that Re,3 (x) − 3Re,2 (x) ∼ 2(µ2e /µ)f (x),
x → ∞.
In the case of Tk (x), 1 ≤ k ≤ 3, we have the following results. Theorem 1. (i) If F e (x) ∈ RV (−γ ), γ > 1 then µe /µ − T1 (x) ∼ (µ(γ − 1))−1 xF e (x), x → ∞. (ii) If F (x) ∈ RV (−α) with α > 2 then µe < ∞ and T2 (x) ∼ 2µe F e (x), x → ∞. (iii) If there exists the density f (x) of F (x) and f (x) ∈ RV (−β) with β > 3 then µe < ∞ and T3 (x) ∼ −2µ2e µ−2 F (x), x → ∞. 2.3. Asymptotic behavior: the class Γ The regularly varying case does not include examples such as the exponential distribution or the gamma distribution. In extreme value theory, the following class Γ (g ) appears in a natural way. A positive and measurable function h belongs to the class Γ (g ) with auxiliary function g if and only if for every fixed y ∈ R, lim h(x + yg (x))/h(x) = e−y .
x→∞
It is well known that the auxiliary function g should satisfy g (x) = o(x), x → ∞ and for every fixed y, g (x + yg (x)) ∼ g (x), x → ∞. For properties and representation theorems about the class Γ , we refer to Geluk and de Haan (1981). For distribution functions F , we consider the tail h(x) = F (x). It is well known that F ∈ Γ (g ) implies that µ < ∞ and that F e (x) = µ−1
x
∞
F (t )dt ∼ µ−1 g (x)F (x) ∈ Γ (g ),
x → ∞.
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As a consequence we have µe < ∞ and also that as x → ∞, ∞
F e (t )dt ∼ g (x)F e (x) ∼ µ−1 g 2 (x)F (x).
(4)
x
Recall that µe /µ − T1 (x) = µ−1
∞
F e (t )dt. The following result follows from (4).
x
Lemma 4. If F ∈ Γ (g ), then µe /µ − T1 (x) ∼ µ−2 g 2 (x)F (x), x → ∞. Now consider T2 (x). Recall that we have T2 (x) = µ−1 2fe (z ))dz. We have the following result.
∞ x
Re,2 (t )dt. Using densities, we can write Re,2 (x) =
∞ x
(fe⊗2 (z ) −
∞
Lemma 5. (i) If fe (x) = o(1)fe⊗2 (x), then T2 (x) ∼ µ−1 x (1 − Fe∗2 (z ))dz , x → ∞. (ii) If also there exists the density f of F such that f ∈ Γ (1), f (x) = o(1)f ⊗2 (x) and f 2 (x/2) = o(1)f ⊗2 (x), then T2 (x) ∼ µ−3 (1 − F ∗2 (x)), x → ∞. Proof. (i) If fe (x) = o(1)fe⊗2 (x) then r2 (x) = fe⊗2 (x) − 2fe (x) ∼ fe⊗2 (x), x → ∞ and the result follows from Re,2 (x) ∼
∞
fe⊗2 (z )dz = 1 − Fe∗2 (x),
x → ∞.
x
(ii) Clearly we have 1 − Fe∗2 (x) = 2
x/ 2
2
F e (x − y)dFe (y) + F e (x/2) = I + II ,
0
where I = (2/µ)
x/2 0
2
F e (x − y)F (y)dy, and II = F e (x/2).
Since f ∈ Γ (1) we have µF e (x) ∼ F (x) ∼ f (x), x → ∞, cf. (4). Now choose ε > 0 and then x◦ so that
(1 − ε)f (x) ≤ µF e (x) ≤ (1 + ε)f (x),
x ≥ x◦
and
(1 − ε)f (x) ≤ F (x) ≤ (1 + ε)f (x),
x ≥ x◦ .
Now take x ≥ 2x◦ and write I = (2/µ)
x◦
F e (x − y)F (y)dy +
x/2
x◦
0
F e (x − y)F (y)dy
= (2/µ) (I (A) + I (B)) .
For I (B), we have
(1 − ε)2 2µ−2
x/2
x◦
f (x − y)f (y)dy ≤ I (B) ≤ (1 + ε)2 2µ−2
x/2
x◦
f (x − y)f (y)dy.
Note that
x/ 2
2 x◦
f (x − y)f (y)dy = f ⊗2 (x) − 2
x◦
f (x − y)f (y)dy. 0
Since f ∈ Γ (1), we have 1 f (x)
x◦
f (x − y)f (y)dy → 0
x◦
ey f (y)dy,
x → ∞.
0
Using f (x) = o(1)f ⊗2 (x), it follows that
x◦ 0
f (x − y)f (y)dy = o(1)f ⊗2 (x). We conclude that
(1 − ε)2 /µ2 ≤ lim inf(I (B)/f ⊗2 (x)) ≤ lim sup(I (B)/f ⊗2 (x)) ≤ (1 + ε)2 /µ2 . x→∞
x→∞
Now consider I (A). Using F (x) ∼ f (x) ∈ Γ (1) and then f (x) = o(1)f ⊗2 (x), we obtain that I (A) = O(1)f (x)
x◦
ey F (y)dy = o(1)f ⊗2 (x). 0
The overall conclusion is that I ∼ f ⊗2 (x)/µ2 . We have II = o(1)f ⊗2 (x) by assumption. It follows that 1 − Fe∗2 (x) ∼ µ−2 f ⊗2 (x), x → ∞, and as a consequence also that T2 (x) ∼ µ−3 (1 − F ∗2 (x)),
x → ∞.
K.V. Mitov, E. Omey / Statistics and Probability Letters 84 (2014) 72–80
77
Some examples will show the use of the previous results. Note that in our case we have fe (x) = (1 − F (x))/µ so that fe⊗2 (x)/fe (x) = µ−1
x
F (x − y) F (x)
0
F (y)dy
holds.
• WEIBULL: If F (x) = C exp(−λxα ) with α > 0, λ > 0, x ≥ 0 then F ∈ Γ (g ) where g (x) = x1−α /λ. If α > 1, we have that as x → ∞, F (x − y)/F (x) ∼ exp(−λ((x − y)α − xα )) = exp(α y(1 + o(1))λxα−1 ) and it follows that F (x − y)/F (x) → ∞, ∀y > 0. Using the lemma of Fatou, we obtain that limx→∞ fe⊗2 (x)/fe (x) = ∞. • GAMMA-TYPE: If f (x) = Cxβ−1 e−x , β > 0, x ≥ 0 then f ∈ Γ (1) and F (x) ∼ f (x) ∈ Γ (1). In this case we have F (x − y)/ ∞ F (x) → ey , x → ∞ and again it follows that lim infx→∞ fe⊗2 (x)/fe (x) ≥ 0 ey F (y)dy = ∞. In the gamma-case, we have ⊗2 that f (x) is again gamma with parameter 2β . We clearly find that f (x) = o(1)f ⊗2 (x) and that f 2 (x/2) = o(1)f ⊗2 (x). Lemma 5(ii) can be applied.
• LOGISTIC: If F (x) = C /(1 + ex ), then F ∈ Γ (1). Now we have F (x − y)/F (x) ∼ (1 + ex )/(1 + ex−y ) ∼ e−x /(e−x + e−y ) → ey ,
∞
x → ∞,
and lim infx→∞ fe (x)/fe (x) ≥ 0 e F (y)dy = ∞. For the density f we have f (x) = Cex /(1 + ex )2 ∼ Ce−x . As in the gamma-case, Lemma 5 is applicable. • EXTREME VALUE DISTRIBUTION: If f (x) = e−x exp(−e−x ), then f ∈ Γ (1) and F (x) ∼ f (x) ∼ e−x . As in the gamma case, we find that f (x) = o(1)f ⊗2 (x) and as before we can apply Lemma 5. ⊗2
y
3. Consecutive approximations for U (x) 3.1. The regularly varying case Now we use the partial sums of Tn (x) as consecutive approximations for U (x). In each case we relate our result to known results. 3.1.1. Case 1: U0 (x) = T0 (x) = x/µ We have U0 (x) = x/µ and U0 (x + y)− U0 (x) = y/µ. The famous renewal theorem states that U (x) ∼ x/µ and Blackwell’s result states that U (x + y) − U (x) → y/µ, cf. Blackwell (1948) or Feller (1971). 3.1.2. Case 2: U1 (x) = T0 (x) + T1 (x) Clearly we have that U1 (x) − x/µ = T1 (x). If F ∈ RV (−α), α > 2, we have F e ∈ RV (1 − α), µe < ∞ and from Theorem 1 we have
µe /µ − U1 (x) + x/µ ∼ xF e (x)/(µ(α − 2)),
x → ∞.
To formulate a Blackwell type of result, note that U1 (x + y) − U1 (x) − y/µ = µ−1
y
F e (x + z )dz . 0
Since F e (x) → 0, as x → ∞, we have U1 (x + y) − U1 (x) → y/µ, x → ∞. Moreover, if F e ∈ RV (−γ ), γ > 1, we obtain that U1 (x + y) − U1 (x) − y/µ ∼ yF e (x)/µ,
x → ∞.
x • In the literature, the estimate U (x) − x/µ ∼ µ−1 0 F e (y)dy, x → ∞ appears in many places. • In his paper, Ney (1981) used a coupling argument to prove that |U (x + y) − U (x) − y/µ| = O(1)F e (x). • In his Ph.D. dissertation Frenk (1983, Theorem 4.1.9) states conditions under which we have U (x + y) − U (x) − y/µ ∼ (y/µ)F e (x),
x → ∞,
see also Frenk (1987, Theorem 3.1.11). 3.1.3. Case 3: U2 (x) = T0 (x) + T1 (x) + T2 (x) If F (x) ∈ RV (−α), α > 2, from Theorem 1 we have that T2 (x) ∼ 2µe F e (x), x → ∞, and it follows that U2 (x) − x/µ − µ−1
0
x
F e (y)dy ∼ 2µe F e (x),
x → ∞.
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Our approach suggests the following approximation for U (x): U (x) = x/µ + µ−1
x
F e (y)dy + 2µe F e (x).
0
As to a Blackwell type of result, note that U2 (x + y) − U2 (x) − y/µ −
x +y
F e (z )dz = −µ
x+y
−1
x
Re,2 (z )dz , x
or U2 (x + y) − U2 (x) − y/µ − (y/µ)F e (x) = µ−1
y
F e (x + z )dz − yF e (x) −
0
y
Re,2 (x + z )dz . 0
Using Proposition 1, we obtain that
µ
−1
y
Re,2 (x + z )dz ∼ 2yF (x)µe /µ2 ,
x → ∞.
0
For the other term, we have y
F e (x + z )dz − yF e (x) =
y
(Fe (x) − Fe (x + z ))dz y z y2 −1 = −µ F (x + u)dudz ∼ −µ−1 F (x) ,
0
0
0
2
0
x → ∞.
We conclude that U2 (x + y) − U2 (x) − y/µ − (y/µ)F e (x) ∼ −µ−2 F (x) 2µe y + y2 /2 ,
x → ∞.
• Baltrunas and Omey (2002) provided conditions under which x U (x) − x/µ − µ−1 F e (y)dy = O(1)F e (x) 0
holds. This result extends Frenk (1983, Lemma 4.1.12) and Embrechts and Omey (1983). • Frenk (1983, Lemma 4.1.13) showed that if EX β < ∞, β ≥ 2, then xβ−1
U (x) − x/µ − µ−1
x
F e (y)dy
→ 0, x → ∞.
0
This was also proved by Stone (1965) for strongly non lattice distributions. See also Carlson (1983).
• In Frenk (1983, Theorem 4.1.16), the author proves a result of the form (cf. also Rogozin (1972)), x +y 1 U (x + y) − U (x) − y/µ − µ−1 F e (z )dz = O(1) F e (x). x
x
• Also Frenk (1983, Theorem 4.1.21) and Frenk (1987, Theorem 3.1.23) obtains a result of the form x +y F e (z )dz ∼ cF (x), x → ∞, for some constant c . U (x + y) − U (x) − y/µ − µ−1 x
3.1.4. Case 4: U3 (x) = T0 (x) + T1 (x) + T2 (x) + T3 (x) If f ∈ RV (−β), β > 3, we have µe < ∞, Re,3 (x) − 3Re,2 (x) ∼ 2µ2e f (x)/µ, x → ∞, and T3 (x) ∼ −2µ2e µ−2 F (x), x → ∞. We obtain that U3 (x) − x/µ − µ−1
x
F e (y)dy − µ−1
∞
0
Re,2 (y)dy ∼ −2µ2e F (x)/µ2 ,
x
This suggests and approximation for U (x) of the form U (x) = x/µ + µ−1
x
F e (y)dy + µ−1
∞
0
Re,2 (y)dy − 2µ2e F (x)/µ2 . x
For a Blackwell type of result, note that we have U3 (x + y) − U3 (x) − y/µ − µ
−1
x +y
F e (z )dz + µ x
= T3 (x + y) − T3 (x) ∼ 2yµ2e f (x)/µ2 ,
−1
x +y
Re,2 (y)dy x
x → ∞.
We have found no corresponding result in the literature.
x → ∞.
K.V. Mitov, E. Omey / Statistics and Probability Letters 84 (2014) 72–80
79
3.2. The case of Γ (g ) First we consider U1 (x) = x/µ + T1 (x). If F ∈ Γ (g ), then Lemma 4 shows that
µe /µ − U1 (x) + x/µ ∼ µ−2 g 2 (x)F (x),
x → ∞.
Our approximation for the renewal function is given by U (x) = x/µ + µe /µ − µ−2 g 2 (x)F (x). Also, we have U1 (x + y) − U1 (x) − y/µ = µ−1
y
F e (x + z )dz . 0
If F ∈ Γ (1), we have F e (x) ∼ µ−1 F (x) ∈ Γ (1), x → ∞, and it follows that U1 (x + y) − U1 (x) − y/µ ∼ µ−1 F (x)
y
e−z dz = µ−1 F (x)(1 − e−y ),
x → ∞.
0
Now we consider U2 (x). Assume that the conditions of Lemma 5(ii) are satisfied. Then we find that U2 (x) − x/µ − T1 (x) ∼ µ−3 (1 − F ∗2 (x)),
x → ∞.
Our approximation for the renewal function is given by U (x) = x/µ + T1 (x) + µ−3 (1 − F ∗2 (x)). As an example, let f (x) = xe−x denote a gamma density of order 2. In this case we have µ = 2 and F (x) ∼ f (x) ∈ Γ (1). We have F (s) = 1/(1 + s)2 . For the renewal function, we find
U (s) =
1
=1+
1 − F (s)
1
=1+
s(2 + s)
1 2s
−
1 4(1 + s/2)
so that x 1 3 x 1 − (1 − e−2x ) = + + e−2x . 2 4 4 2 4 Now we calculate T1 (x). Clearly we have U (x) = 1 +
1 − F (s) Fe (s) = =
2+s 2(1 + s)2
2s
so that 1 2s
(1 − Fe (s)) =
3 + 2s 4(1 + s)2
=
1 2(1 + s)
+
1 4(1 + s)2
,
and hence 1 1 F1 (x) + F (x), 2 4 where F1 (x) = 1 − e−x . It follows that T1 (∞) = µe /µ = 3/4 and T 1 ( x) =
3
1
− T 1 ( x) =
4 For U1 (x) we find U1 (x) =
x 2
2
+
F 1 (x) +
3
1 4
F (x) ∼
1
1
2
4
1 4
F (x).
− e−x − F (x).
4
Since F (x) ∼ xe−x , x → ∞, we find for example that U (x) − U1 (x) =
1 −2x 1 1 1 xe−x e + e−x + F (x) ∼ F (x) ∼ , 4 2 4 4 4
x → ∞.
Remark 1. In Alsmeyer (1991, page 152) the author obtains the following formula in the case of X ∼ GAMMA(m, 1). He finds U (x) = 1 +
x m
−
m−1 2m
−
m−1 1
m j =1
α(m, j) (α(m,j)−1)x e , 1 − α(m, j)
where α(m, j) are the (complex) m-unit roots given by α(m, j) = exp(i2π j/m). In the case of m = 2, we have α(2, 1) = exp(iπ) = −1 and we find back U (x) = 1 +
x 2
−
1 4
1
1
4
4
+ e−2x =
3 + 2x + e−2x .
80
K.V. Mitov, E. Omey / Statistics and Probability Letters 84 (2014) 72–80
Remark 2. In Frenk (1987, Chapter 3.4) the author also considers distributions for which Fˆ (s) can be written as P (s)/Q (s) where P (s) and Q (s) are polynomials without common factors. His Theorems 3.2.2 and 3.2.4 extend the formula of Remark 1. 4. Concluding remarks
• In this paper we only considered the case where µe < ∞. If µe = ∞, we can use the analogue of Lemma 3 that was proved in Omey and Willekens (1986). A sufficient condition for µe < ∞ is EX 2 < ∞. • For discrete random variables, it is not hard to obtain similar results for the renewal sequence un . • In this paper we restricted to regularly varying functions and obtained asymptotic equalities. It is possible to obtain O(.) types of results by using dominated varying functions. Acknowledgment The authors are very grateful to the referee for the very detailed comments and suggestions which improved the paper significantly. References Alsmeyer, G., 1991. Erneurerungstheorie, B.G. Teubner, Stuttgart. Baltrunas, A., Omey, E., 2002. Second order renewal theorem in the finite means case. Theory Probab. Appl. 47, 127–132. Bingham, N.H., Goldie, C.M., Teugels, J.L., 1987. Regular Variation. Encyclopedia of Math. and its Appl. Cambridge Universtiy Press, Cambridge. Blackwell, D., 1948. A renewal theorem. Duke Math. J. 15, 145–151. Carlson, H., 1983. Remainder term estimates of the renewal function. Ann. Probab. 11, 143–157. Dohli, T., Kaio, N., Osaki, S., 2002. Renewal processes and their computational aspects. In: Osaki, S. (Ed.), Stochastic Models in Reliability and Maintenance. Springer-Verlag, Berlin, pp. 1–24. Embrechts, P., Omey, E., 1983. A property of longtailed distributions. J. Appl. Probab. 21, 80–87. Feller, W., 1971. An Introduction to Probability Theory and its Applications, Vol. 2, second ed. Wiley, New York. Frenk, J.B.G., 1983. On renewal theory, Banach algebras and functions of bounded increase. Ph.D Thesis, Mathematisch Centrum, Amsterdam. Frenk, J.B.G., 1987. On Banach Algebras, Renewal Measures and Regenerative Processes. Centre for Mathematics and Computer Science, Amsterdam, CWI tract 38. Geluk, J.L., de Haan, L., 1981. Regular variation, extensions and Tauberian theorems. CWI Tract 40, Amsterdam. Ney, P., 1981. A refinement of the coupling method in renewal theory. Stochastic Process. Appl. 11, 11–26. Omey, E., Willekens, E., 1986. Second-order behaviour of the tail of a subordinated probability distribution. Stochastic Process. Appl. 21, 339–353. Omey, E., Willekens, E., 1987. Second-order behaviour of distributions subordinate to a distribution with finite mean. Comm. Statist. Stoch. Models 3 (3), 311–342. Rogozin, B.A., 1972. Asymptotics of renewal functions. Theory Probab. Appl. 21, 669–686. Stone, C., 1965. On characteristic functions and renewal theory. Trans. Amer. Math. Soc. 120, 327–342.
Contents lists available at ScienceDirect
Statistics and Probability Letters journal homepage: www.elsevier.com/locate/stapro
Intuitive approximations for the renewal function Kosto V. Mitov a,∗ , Edward Omey b a
Aviation Faculty - NMU, 5856 D. Mitropolia, Pleven, Bulgaria
b
HUB-EHSAL, Stormstraat 2, 1000, Brussels, Belgium
article
info
Article history: Received 30 April 2013 Received in revised form 24 September 2013 Accepted 25 September 2013 Available online 29 September 2013
abstract ∗n It is hard to find explicit expressions for the renewal function U (x) = n=0 F (x). Many researchers have made attempts to find suitable approximations for U (x). In this paper we present simple approximations and show that they cover many of the known results. © 2013 Elsevier B.V. All rights reserved.
∞
MSC: 60K05 60E10 26A12 Keywords: Renewal function Approximations Regular variation The class gamma
1. Introduction Suppose that X , X1 , X2 , . . . are i.i.d. and nonnegative r.v. with distribution function F (x) = P (X ≤ x) and F (0+) = 0. If F a density, we denote it by f (x). The tail of F is denoted by F (x) = 1 − F (x). If X has a finite mean, we denote it by µ = EX = has ∞ ¯ F (x)dx. We assume that F is non lattice. Let F (s) = E (e−sX ) denote the Laplace transform of F . Let Sn denote the partial 0 sums Sn = X1 + X2 + · · · + Xn , n ≥ 1, and let S0 = 0. Clearly we have P (Sn ≤ x) = F ∗n (x), n ≥ 0. As F ∗ (.) denote the ∞usual ∗0 ∗1 ∗n ∗(n−1) ∗(n−1) n-fold convolution of F (.) with itself, that is F (x) = 1, F (x) = F (x), F (x) = F ∗ F ( x) = 0 F (x − y)dF (y), n = 2, 3, . . .. In what follows we will use the same notation in more general sense. If G(.) is a measure on the positive half ∞ line and g (.) is an arbitrary measurable function we will denote G ∗ g (x) = 0 g (x − y)dG(y). ∞ ∗n The renewal function U (x) is given by U (x) = n=0 F (x). It is well known that the renewal function satisfies the following renewal equation U (x) = 1+U ∗F (x). It is well known that U (x) < ∞, for all x ≥ 0. If µ < ∞, the renewal theorem states that U (x)/x → 1/µ as x → ∞ and Blackwell’s theorem shows that for any fixed y > 0, U (x + y) − U (x) → y/µ as x → ∞. For these results we refer to Blackwell (1948) or Feller (1971). There have been many successful attempts to obtain the rate of convergence in these results. Some authors such as Rogozin (1972) or Frenk (1983, 1987) use Banach algebra techniques. Alsmeyer (1991), Carlson (1983) or Stone (1965) used Fourier analysis. Alsmeyer (1991) and Ney (1981) used coupling methods. In Dohli et al. (2002), the authors give an overview of numerical approximations in the renewal theorems. In this paper we propose a simple and intuitive approach to approximate U (x). It turns out that our approximations cover all the results
∗
Corresponding author. E-mail addresses: [email protected], [email protected] (K.V. Mitov), [email protected] (E. Omey).
0167-7152/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.spl.2013.09.030
K.V. Mitov, E. Omey / Statistics and Probability Letters 84 (2014) 72–80
73
that we have found in the literature. To define our approximations, recall that the Laplace transform of U (x) is given by
U (s) =
∞
e−sx dU (x) = (1 − F (s))−1 ,
s > 0.
0
When µ = EX < ∞, we can define the equilibrium distribution function Fe (x) as Fe (x) = µ
−1
x
F (y)dy
for x ≥ 0.
0
If EX 2 < ∞ then the mean of the equilibrium distribution is finite. We will denote µe = Fe (x) has the density fe (x) = µ−1 F (x), x ≥ 0. Clearly we have Fe (s) = (1 − F (s))/(µs), s > 0 and it follows that 1
1
∞ 0
(1 − Fe (x))dx < ∞. Clearly
1
, s > 0. µs 1 − ( 1 − Fe (s)) ∞ ∞ n Using a Taylor expansion, we obtain that U (s) = µ1s n=0 Tn (s). Using Newton’s binomial theorem, we n=0 (1 − Fe (s)) = U (s) =
µs Fe (s)
=
obtain that
n 1 1 k (1 − Fe (s))n = C (−1)k Fek (s). Tn (s) = µs µs k=0 n
It follows that U (x) =
∞
Tn (x),
for almost all x ≥ 0,
(1)
n=0
x
k k where Tn (x) = µ1 F ∗k (y)dy. We denote by Fe∗k (x) the kfold convolution of Fe (x) with itself. It is tempting to k=0 Cn (−1) 0 e use formula (1) to obtain consecutive approximations for U (x). Based on (1) we propose the following approximations for the renewal function: for fixed k ≥ 0, we set
n
Uk (x) =
k
Tn (x).
n =0
In the paper we consider the cases 0 ≤ k ≤ 3 and show that our approximation corresponds to the approximations that have been published in many papers before. 2. The functions Tn (x) 2.1. Some alternative expressions Lemma 1. Let x ≥ 0. We have T0 (x) = x/µ and for n ≥ 1, we have Tn (x) = −
n 1
µ
Cnk (−1)k
x
(1 − Fe∗k (y))dy.
(2)
0
k=1
Moreover, if µe < ∞, for n ≥ 2 we have Tn (x) =
n 1
µ
Cnk (−1)k
∞
(1 − Fe∗k (y))dy.
x
k=1
Proof. The result for T0 (x) is clear. The result for Tn (x) follows because we have n
Cnk (−1)k = (1 − 1)n = 0.
k=0
µ
Now assume that µe < ∞. Clearly we have Tn (∞) = − µe then take the first derivative to find:
(x − 1)n =
n
Cnk xk (−1)n−k
and
n(x − 1)n−1 =
k=0
n
n
k=1
Cnk (−1)k k. Now we use Newton’s binomial formula and
Cnk kxk−1 (−1)n−k ,
x ∈ R.
k=1
k n−k Taking x = 1 this is 0 = = (−1)n k=1 Cnk k(−1)k and so we find that Tn (∞) = 0. The result follows. k=1 Cn k(−1) Now we consider into more details the terms Tk (x), k = 1, 2, 3.
n
n
74
K.V. Mitov, E. Omey / Statistics and Probability Letters 84 (2014) 72–80
Lemma 2. For n ≥ 2, let Re,n (x) = 1 − Fe∗n (x) − n(1 − Fe (x)). We have T1 (x) = µ−1 T3 (x) = µ−1
x
F e (y)dy,
0 x
T2 (x) = −µ−1
x
Re,2 (y)dy, 0
(Re,3 (y) − 3Re,2 (y))dy.
0
If µe < ∞, we have T2 (x) = µ−1
∞
Re,2 (y)dy,
T3 (x) = −µ−1
∞
(Re,3 (y) − 3Re,2 (y))dy.
x
x
Proof. Using (2) with n = 1 we have the result for T1 (x). Using (2) with n = 2, we have T2 (x) = −µ
−1
x
−2(1 − Fe (y)) + (1 − Fe∗2 (y)) dy = −µ−1
x
Re,2 (y)dy.
0
0
Using (2) with n = 3, we have T3 (x) = µ−1
x
3(1 − Fe (y)) − 3(1 − Fe∗2 (y)) + (1 − Fe∗3 (y))dy 0
= µ−1
x
(Re,3 (y) − 3Re,2 (y))dy. 0
If µe < ∞, we have
∞ 0
Re,k (y)dy = 0, k ≥ 2, and the second result follows.
2.2. Asymptotic behavior: regular variation In this section we study the asymptotic behavior of Tk (x), 1 ≤ k ≤ 3. More precisely we study the regularly varying behavior of the functions involved. Recall that a positive and measurable function g is regularly varying with index α, α ∈ R if it satisfies lim g (xt )/g (t ) = xα ,
∀x > 0.
t →∞
Notation: g ∈ RV (α). Typical examples are g (x) = xα or g (x) = xα log(x). For a standard book about regular variation and its applications, we refer to Bingham et al. (1987) or Geluk and de Haan (1981). In the first result we discuss the asymptotic behavior of Re,n (x). We need the following results of Omey and Willekens (1987). Lemma 3 (Omey and Willekens (1987, Corollary 2.4)). Suppose that G(x) = P (Y ≤ x) has a density g ∈ RV (−β) with β > 2. For n ≥ 2, let Rn (x) = 1 − G∗n (x)− n(1 − G(x)). Then λ = EY < ∞ and for all n ≥ 2, we have Rn (x) ∼ λn(n − 1)g (x), x → ∞. In our case, we have g (x) = (1 − F (x))/µ and G(x) = Fe (x). The lemma gives the following result. Proposition 1. Suppose that F (x) ∈ RV (−α) and α > 2. Then µe < ∞ and for all n ≥ 2 we have Re,n (x) ∼ n(n − 1)(µe /µ)F (x),
x → ∞.
Now we discuss the asymptotic behavior of Re,3 (x) − 3Re,2 (x). In the result we assume that F has a density f . Proposition 2. Suppose that f (x) ∈ RV (−β) and β > 3. Then µe < ∞ and Re,3 (x) − 3Re,2 (x) ∼ f (x)2µ2e /µ, x → ∞. Proof. First note that Re,3 (x) = Re,2 ∗ Fe (x) + 2Re,2 (x). It follows that Re,3 (x) − 3Re,2 (x) = Re,2 ∗ Fe (x) − Re,2 (x). Clearly we have Re,3 (x) − 3Re,2 (x) = µ−1
x
(Re,2 (x − y) − Re,2 (x))F (y)dy − Re,2 (x)F e (x) 0
x
(Re,2 (x − y) − Re,2 (x))fe (y)dy − Re,2 (x)F e (x),
= 0
where fe (x) = F (x)/µ is the density of Fe . Here and later we denote by fe⊗2 the usual convolution of the density fe with itself, that is fe⊗2 is the density of Fe∗2 . We have Re,2 (x − y) − Re,2 (x) = Fe∗2 (x) − Fe∗2 (x − y) − 2(Fe (x) − Fe (x − y))
x
(fe⊗2 (z ) − 2fe (z ))dz .
= x −y
K.V. Mitov, E. Omey / Statistics and Probability Letters 84 (2014) 72–80
75
It follows that Re,3 (x) − 3Re,2 (x) =
x
x
(fe⊗2 (z ) − 2fe (z ))fe (y)dzdy − Re,2 (x)F e (x).
(3)
x −y
y=0
First we analyze r2 (x) = fe⊗2 (x) − 2fe (x). We have r 2 ( x) = 2
x/ 2
fe (x − y)fe (y)dy − 2fe (x)
0 x/ 2
=2
(fe (x − y) − fe (x))fe (y)dy − 2fe (x)F e (x/2)
0
=
2
µ
x/ 2
x
f (z )dzfe (y)dy − 2fe (x)F e (x/2) = I − II . z =x −y
y=0
By assumption we have f ∈ RV (−β). Then for 0 ≤ y ≤ x/2, we have x/2 ≤ z ≤ x. For fixed y, it follows that 1
f (x)
x
f (z )dzfe (y) → yfe (y),
x → ∞.
z =x −y
On the other hand, since f ∈ RV (−β), one obtains 1
f (x)
f (z )
x
f (z )dz ≤ z =x −y
sup
x/2≤z ≤x
f ( x)
y = O(1)y.
Using Lebesgue’s theorem, we obtain that I /f (x) → (2/µ)
∞
yfe (y)dy = 2µe /µ,
x → ∞.
y=0
Because β > 3, we also have II = O(1)xf (x)x2 f (x) = o(1)f (x). We conclude that r2 (x) ∼ f (x)2µe /µ ∈ RV (−β). Now we consider Re,3 (x) − 3Re,2 (x). Using (3), we have Re,3 (x) − 3Re,2 (x) =
x y=0
x
r2 (z )dzfe (y)dy − Re,2 (x)F e (x).
z =x−y
Since r2 (x) ∈ RV (−β), we have
x y=0
x
r2 (z )dzfe (y)dy ∼ r2 (x)µe ∼ 2(µ2e /µ)f (x),
x → ∞.
z =x −y
Since Re,2 (x) = O(1)fe (x)F e (x) and f (x) ∈ RV (−β), β > 3, we obtain that Re,2 (x) = o(1)f (x), x → ∞. We conclude that Re,3 (x) − 3Re,2 (x) ∼ 2(µ2e /µ)f (x),
x → ∞.
In the case of Tk (x), 1 ≤ k ≤ 3, we have the following results. Theorem 1. (i) If F e (x) ∈ RV (−γ ), γ > 1 then µe /µ − T1 (x) ∼ (µ(γ − 1))−1 xF e (x), x → ∞. (ii) If F (x) ∈ RV (−α) with α > 2 then µe < ∞ and T2 (x) ∼ 2µe F e (x), x → ∞. (iii) If there exists the density f (x) of F (x) and f (x) ∈ RV (−β) with β > 3 then µe < ∞ and T3 (x) ∼ −2µ2e µ−2 F (x), x → ∞. 2.3. Asymptotic behavior: the class Γ The regularly varying case does not include examples such as the exponential distribution or the gamma distribution. In extreme value theory, the following class Γ (g ) appears in a natural way. A positive and measurable function h belongs to the class Γ (g ) with auxiliary function g if and only if for every fixed y ∈ R, lim h(x + yg (x))/h(x) = e−y .
x→∞
It is well known that the auxiliary function g should satisfy g (x) = o(x), x → ∞ and for every fixed y, g (x + yg (x)) ∼ g (x), x → ∞. For properties and representation theorems about the class Γ , we refer to Geluk and de Haan (1981). For distribution functions F , we consider the tail h(x) = F (x). It is well known that F ∈ Γ (g ) implies that µ < ∞ and that F e (x) = µ−1
x
∞
F (t )dt ∼ µ−1 g (x)F (x) ∈ Γ (g ),
x → ∞.
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K.V. Mitov, E. Omey / Statistics and Probability Letters 84 (2014) 72–80
As a consequence we have µe < ∞ and also that as x → ∞, ∞
F e (t )dt ∼ g (x)F e (x) ∼ µ−1 g 2 (x)F (x).
(4)
x
Recall that µe /µ − T1 (x) = µ−1
∞
F e (t )dt. The following result follows from (4).
x
Lemma 4. If F ∈ Γ (g ), then µe /µ − T1 (x) ∼ µ−2 g 2 (x)F (x), x → ∞. Now consider T2 (x). Recall that we have T2 (x) = µ−1 2fe (z ))dz. We have the following result.
∞ x
Re,2 (t )dt. Using densities, we can write Re,2 (x) =
∞ x
(fe⊗2 (z ) −
∞
Lemma 5. (i) If fe (x) = o(1)fe⊗2 (x), then T2 (x) ∼ µ−1 x (1 − Fe∗2 (z ))dz , x → ∞. (ii) If also there exists the density f of F such that f ∈ Γ (1), f (x) = o(1)f ⊗2 (x) and f 2 (x/2) = o(1)f ⊗2 (x), then T2 (x) ∼ µ−3 (1 − F ∗2 (x)), x → ∞. Proof. (i) If fe (x) = o(1)fe⊗2 (x) then r2 (x) = fe⊗2 (x) − 2fe (x) ∼ fe⊗2 (x), x → ∞ and the result follows from Re,2 (x) ∼
∞
fe⊗2 (z )dz = 1 − Fe∗2 (x),
x → ∞.
x
(ii) Clearly we have 1 − Fe∗2 (x) = 2
x/ 2
2
F e (x − y)dFe (y) + F e (x/2) = I + II ,
0
where I = (2/µ)
x/2 0
2
F e (x − y)F (y)dy, and II = F e (x/2).
Since f ∈ Γ (1) we have µF e (x) ∼ F (x) ∼ f (x), x → ∞, cf. (4). Now choose ε > 0 and then x◦ so that
(1 − ε)f (x) ≤ µF e (x) ≤ (1 + ε)f (x),
x ≥ x◦
and
(1 − ε)f (x) ≤ F (x) ≤ (1 + ε)f (x),
x ≥ x◦ .
Now take x ≥ 2x◦ and write I = (2/µ)
x◦
F e (x − y)F (y)dy +
x/2
x◦
0
F e (x − y)F (y)dy
= (2/µ) (I (A) + I (B)) .
For I (B), we have
(1 − ε)2 2µ−2
x/2
x◦
f (x − y)f (y)dy ≤ I (B) ≤ (1 + ε)2 2µ−2
x/2
x◦
f (x − y)f (y)dy.
Note that
x/ 2
2 x◦
f (x − y)f (y)dy = f ⊗2 (x) − 2
x◦
f (x − y)f (y)dy. 0
Since f ∈ Γ (1), we have 1 f (x)
x◦
f (x − y)f (y)dy → 0
x◦
ey f (y)dy,
x → ∞.
0
Using f (x) = o(1)f ⊗2 (x), it follows that
x◦ 0
f (x − y)f (y)dy = o(1)f ⊗2 (x). We conclude that
(1 − ε)2 /µ2 ≤ lim inf(I (B)/f ⊗2 (x)) ≤ lim sup(I (B)/f ⊗2 (x)) ≤ (1 + ε)2 /µ2 . x→∞
x→∞
Now consider I (A). Using F (x) ∼ f (x) ∈ Γ (1) and then f (x) = o(1)f ⊗2 (x), we obtain that I (A) = O(1)f (x)
x◦
ey F (y)dy = o(1)f ⊗2 (x). 0
The overall conclusion is that I ∼ f ⊗2 (x)/µ2 . We have II = o(1)f ⊗2 (x) by assumption. It follows that 1 − Fe∗2 (x) ∼ µ−2 f ⊗2 (x), x → ∞, and as a consequence also that T2 (x) ∼ µ−3 (1 − F ∗2 (x)),
x → ∞.
K.V. Mitov, E. Omey / Statistics and Probability Letters 84 (2014) 72–80
77
Some examples will show the use of the previous results. Note that in our case we have fe (x) = (1 − F (x))/µ so that fe⊗2 (x)/fe (x) = µ−1
x
F (x − y) F (x)
0
F (y)dy
holds.
• WEIBULL: If F (x) = C exp(−λxα ) with α > 0, λ > 0, x ≥ 0 then F ∈ Γ (g ) where g (x) = x1−α /λ. If α > 1, we have that as x → ∞, F (x − y)/F (x) ∼ exp(−λ((x − y)α − xα )) = exp(α y(1 + o(1))λxα−1 ) and it follows that F (x − y)/F (x) → ∞, ∀y > 0. Using the lemma of Fatou, we obtain that limx→∞ fe⊗2 (x)/fe (x) = ∞. • GAMMA-TYPE: If f (x) = Cxβ−1 e−x , β > 0, x ≥ 0 then f ∈ Γ (1) and F (x) ∼ f (x) ∈ Γ (1). In this case we have F (x − y)/ ∞ F (x) → ey , x → ∞ and again it follows that lim infx→∞ fe⊗2 (x)/fe (x) ≥ 0 ey F (y)dy = ∞. In the gamma-case, we have ⊗2 that f (x) is again gamma with parameter 2β . We clearly find that f (x) = o(1)f ⊗2 (x) and that f 2 (x/2) = o(1)f ⊗2 (x). Lemma 5(ii) can be applied.
• LOGISTIC: If F (x) = C /(1 + ex ), then F ∈ Γ (1). Now we have F (x − y)/F (x) ∼ (1 + ex )/(1 + ex−y ) ∼ e−x /(e−x + e−y ) → ey ,
∞
x → ∞,
and lim infx→∞ fe (x)/fe (x) ≥ 0 e F (y)dy = ∞. For the density f we have f (x) = Cex /(1 + ex )2 ∼ Ce−x . As in the gamma-case, Lemma 5 is applicable. • EXTREME VALUE DISTRIBUTION: If f (x) = e−x exp(−e−x ), then f ∈ Γ (1) and F (x) ∼ f (x) ∼ e−x . As in the gamma case, we find that f (x) = o(1)f ⊗2 (x) and as before we can apply Lemma 5. ⊗2
y
3. Consecutive approximations for U (x) 3.1. The regularly varying case Now we use the partial sums of Tn (x) as consecutive approximations for U (x). In each case we relate our result to known results. 3.1.1. Case 1: U0 (x) = T0 (x) = x/µ We have U0 (x) = x/µ and U0 (x + y)− U0 (x) = y/µ. The famous renewal theorem states that U (x) ∼ x/µ and Blackwell’s result states that U (x + y) − U (x) → y/µ, cf. Blackwell (1948) or Feller (1971). 3.1.2. Case 2: U1 (x) = T0 (x) + T1 (x) Clearly we have that U1 (x) − x/µ = T1 (x). If F ∈ RV (−α), α > 2, we have F e ∈ RV (1 − α), µe < ∞ and from Theorem 1 we have
µe /µ − U1 (x) + x/µ ∼ xF e (x)/(µ(α − 2)),
x → ∞.
To formulate a Blackwell type of result, note that U1 (x + y) − U1 (x) − y/µ = µ−1
y
F e (x + z )dz . 0
Since F e (x) → 0, as x → ∞, we have U1 (x + y) − U1 (x) → y/µ, x → ∞. Moreover, if F e ∈ RV (−γ ), γ > 1, we obtain that U1 (x + y) − U1 (x) − y/µ ∼ yF e (x)/µ,
x → ∞.
x • In the literature, the estimate U (x) − x/µ ∼ µ−1 0 F e (y)dy, x → ∞ appears in many places. • In his paper, Ney (1981) used a coupling argument to prove that |U (x + y) − U (x) − y/µ| = O(1)F e (x). • In his Ph.D. dissertation Frenk (1983, Theorem 4.1.9) states conditions under which we have U (x + y) − U (x) − y/µ ∼ (y/µ)F e (x),
x → ∞,
see also Frenk (1987, Theorem 3.1.11). 3.1.3. Case 3: U2 (x) = T0 (x) + T1 (x) + T2 (x) If F (x) ∈ RV (−α), α > 2, from Theorem 1 we have that T2 (x) ∼ 2µe F e (x), x → ∞, and it follows that U2 (x) − x/µ − µ−1
0
x
F e (y)dy ∼ 2µe F e (x),
x → ∞.
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K.V. Mitov, E. Omey / Statistics and Probability Letters 84 (2014) 72–80
Our approach suggests the following approximation for U (x): U (x) = x/µ + µ−1
x
F e (y)dy + 2µe F e (x).
0
As to a Blackwell type of result, note that U2 (x + y) − U2 (x) − y/µ −
x +y
F e (z )dz = −µ
x+y
−1
x
Re,2 (z )dz , x
or U2 (x + y) − U2 (x) − y/µ − (y/µ)F e (x) = µ−1
y
F e (x + z )dz − yF e (x) −
0
y
Re,2 (x + z )dz . 0
Using Proposition 1, we obtain that
µ
−1
y
Re,2 (x + z )dz ∼ 2yF (x)µe /µ2 ,
x → ∞.
0
For the other term, we have y
F e (x + z )dz − yF e (x) =
y
(Fe (x) − Fe (x + z ))dz y z y2 −1 = −µ F (x + u)dudz ∼ −µ−1 F (x) ,
0
0
0
2
0
x → ∞.
We conclude that U2 (x + y) − U2 (x) − y/µ − (y/µ)F e (x) ∼ −µ−2 F (x) 2µe y + y2 /2 ,
x → ∞.
• Baltrunas and Omey (2002) provided conditions under which x U (x) − x/µ − µ−1 F e (y)dy = O(1)F e (x) 0
holds. This result extends Frenk (1983, Lemma 4.1.12) and Embrechts and Omey (1983). • Frenk (1983, Lemma 4.1.13) showed that if EX β < ∞, β ≥ 2, then xβ−1
U (x) − x/µ − µ−1
x
F e (y)dy
→ 0, x → ∞.
0
This was also proved by Stone (1965) for strongly non lattice distributions. See also Carlson (1983).
• In Frenk (1983, Theorem 4.1.16), the author proves a result of the form (cf. also Rogozin (1972)), x +y 1 U (x + y) − U (x) − y/µ − µ−1 F e (z )dz = O(1) F e (x). x
x
• Also Frenk (1983, Theorem 4.1.21) and Frenk (1987, Theorem 3.1.23) obtains a result of the form x +y F e (z )dz ∼ cF (x), x → ∞, for some constant c . U (x + y) − U (x) − y/µ − µ−1 x
3.1.4. Case 4: U3 (x) = T0 (x) + T1 (x) + T2 (x) + T3 (x) If f ∈ RV (−β), β > 3, we have µe < ∞, Re,3 (x) − 3Re,2 (x) ∼ 2µ2e f (x)/µ, x → ∞, and T3 (x) ∼ −2µ2e µ−2 F (x), x → ∞. We obtain that U3 (x) − x/µ − µ−1
x
F e (y)dy − µ−1
∞
0
Re,2 (y)dy ∼ −2µ2e F (x)/µ2 ,
x
This suggests and approximation for U (x) of the form U (x) = x/µ + µ−1
x
F e (y)dy + µ−1
∞
0
Re,2 (y)dy − 2µ2e F (x)/µ2 . x
For a Blackwell type of result, note that we have U3 (x + y) − U3 (x) − y/µ − µ
−1
x +y
F e (z )dz + µ x
= T3 (x + y) − T3 (x) ∼ 2yµ2e f (x)/µ2 ,
−1
x +y
Re,2 (y)dy x
x → ∞.
We have found no corresponding result in the literature.
x → ∞.
K.V. Mitov, E. Omey / Statistics and Probability Letters 84 (2014) 72–80
79
3.2. The case of Γ (g ) First we consider U1 (x) = x/µ + T1 (x). If F ∈ Γ (g ), then Lemma 4 shows that
µe /µ − U1 (x) + x/µ ∼ µ−2 g 2 (x)F (x),
x → ∞.
Our approximation for the renewal function is given by U (x) = x/µ + µe /µ − µ−2 g 2 (x)F (x). Also, we have U1 (x + y) − U1 (x) − y/µ = µ−1
y
F e (x + z )dz . 0
If F ∈ Γ (1), we have F e (x) ∼ µ−1 F (x) ∈ Γ (1), x → ∞, and it follows that U1 (x + y) − U1 (x) − y/µ ∼ µ−1 F (x)
y
e−z dz = µ−1 F (x)(1 − e−y ),
x → ∞.
0
Now we consider U2 (x). Assume that the conditions of Lemma 5(ii) are satisfied. Then we find that U2 (x) − x/µ − T1 (x) ∼ µ−3 (1 − F ∗2 (x)),
x → ∞.
Our approximation for the renewal function is given by U (x) = x/µ + T1 (x) + µ−3 (1 − F ∗2 (x)). As an example, let f (x) = xe−x denote a gamma density of order 2. In this case we have µ = 2 and F (x) ∼ f (x) ∈ Γ (1). We have F (s) = 1/(1 + s)2 . For the renewal function, we find
U (s) =
1
=1+
1 − F (s)
1
=1+
s(2 + s)
1 2s
−
1 4(1 + s/2)
so that x 1 3 x 1 − (1 − e−2x ) = + + e−2x . 2 4 4 2 4 Now we calculate T1 (x). Clearly we have U (x) = 1 +
1 − F (s) Fe (s) = =
2+s 2(1 + s)2
2s
so that 1 2s
(1 − Fe (s)) =
3 + 2s 4(1 + s)2
=
1 2(1 + s)
+
1 4(1 + s)2
,
and hence 1 1 F1 (x) + F (x), 2 4 where F1 (x) = 1 − e−x . It follows that T1 (∞) = µe /µ = 3/4 and T 1 ( x) =
3
1
− T 1 ( x) =
4 For U1 (x) we find U1 (x) =
x 2
2
+
F 1 (x) +
3
1 4
F (x) ∼
1
1
2
4
1 4
F (x).
− e−x − F (x).
4
Since F (x) ∼ xe−x , x → ∞, we find for example that U (x) − U1 (x) =
1 −2x 1 1 1 xe−x e + e−x + F (x) ∼ F (x) ∼ , 4 2 4 4 4
x → ∞.
Remark 1. In Alsmeyer (1991, page 152) the author obtains the following formula in the case of X ∼ GAMMA(m, 1). He finds U (x) = 1 +
x m
−
m−1 2m
−
m−1 1
m j =1
α(m, j) (α(m,j)−1)x e , 1 − α(m, j)
where α(m, j) are the (complex) m-unit roots given by α(m, j) = exp(i2π j/m). In the case of m = 2, we have α(2, 1) = exp(iπ) = −1 and we find back U (x) = 1 +
x 2
−
1 4
1
1
4
4
+ e−2x =
3 + 2x + e−2x .
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K.V. Mitov, E. Omey / Statistics and Probability Letters 84 (2014) 72–80
Remark 2. In Frenk (1987, Chapter 3.4) the author also considers distributions for which Fˆ (s) can be written as P (s)/Q (s) where P (s) and Q (s) are polynomials without common factors. His Theorems 3.2.2 and 3.2.4 extend the formula of Remark 1. 4. Concluding remarks
• In this paper we only considered the case where µe < ∞. If µe = ∞, we can use the analogue of Lemma 3 that was proved in Omey and Willekens (1986). A sufficient condition for µe < ∞ is EX 2 < ∞. • For discrete random variables, it is not hard to obtain similar results for the renewal sequence un . • In this paper we restricted to regularly varying functions and obtained asymptotic equalities. It is possible to obtain O(.) types of results by using dominated varying functions. Acknowledgment The authors are very grateful to the referee for the very detailed comments and suggestions which improved the paper significantly. References Alsmeyer, G., 1991. Erneurerungstheorie, B.G. Teubner, Stuttgart. Baltrunas, A., Omey, E., 2002. Second order renewal theorem in the finite means case. Theory Probab. Appl. 47, 127–132. Bingham, N.H., Goldie, C.M., Teugels, J.L., 1987. Regular Variation. Encyclopedia of Math. and its Appl. Cambridge Universtiy Press, Cambridge. Blackwell, D., 1948. A renewal theorem. Duke Math. J. 15, 145–151. Carlson, H., 1983. Remainder term estimates of the renewal function. Ann. Probab. 11, 143–157. Dohli, T., Kaio, N., Osaki, S., 2002. Renewal processes and their computational aspects. In: Osaki, S. (Ed.), Stochastic Models in Reliability and Maintenance. Springer-Verlag, Berlin, pp. 1–24. Embrechts, P., Omey, E., 1983. A property of longtailed distributions. J. Appl. Probab. 21, 80–87. Feller, W., 1971. An Introduction to Probability Theory and its Applications, Vol. 2, second ed. Wiley, New York. Frenk, J.B.G., 1983. On renewal theory, Banach algebras and functions of bounded increase. Ph.D Thesis, Mathematisch Centrum, Amsterdam. Frenk, J.B.G., 1987. On Banach Algebras, Renewal Measures and Regenerative Processes. Centre for Mathematics and Computer Science, Amsterdam, CWI tract 38. Geluk, J.L., de Haan, L., 1981. Regular variation, extensions and Tauberian theorems. CWI Tract 40, Amsterdam. Ney, P., 1981. A refinement of the coupling method in renewal theory. Stochastic Process. Appl. 11, 11–26. Omey, E., Willekens, E., 1986. Second-order behaviour of the tail of a subordinated probability distribution. Stochastic Process. Appl. 21, 339–353. Omey, E., Willekens, E., 1987. Second-order behaviour of distributions subordinate to a distribution with finite mean. Comm. Statist. Stoch. Models 3 (3), 311–342. Rogozin, B.A., 1972. Asymptotics of renewal functions. Theory Probab. Appl. 21, 669–686. Stone, C., 1965. On characteristic functions and renewal theory. Trans. Amer. Math. Soc. 120, 327–342.