The term moderate deviations is often used in the literature to mean a class of large deviation principles that, in some sense, fills the gap between a convergence in probability to zero (governed by a large deviation principle) and a weak convergence to a centered normal distribution. In this paper, some examples of classes of large deviation principles of this kind are presented, but the involved random variables converge weakly to Gumbel, exponential and Laplace distributions.
Sampled extremaoccupancy problemcoupon collector’s problemreplacement model for random lifetimes60F1060F0560G7060C05MIURCUP E83C18000100006University of Rome Tor VergataCUP E83C22001780005Indam-GNAMPAThis work has been partially supported by MIUR Excellence Department Project awarded to the Department of Mathematics, University of Rome Tor Vergata (CUP E83C18000100006), by University of Rome Tor Vergata (project “Asymptotic Methods in Probability” (CUP E89C20000680005) and project “Asymptotic Properties in Probability” (CUP E83C22001780005)) and by Indam-GNAMPA. Introduction
The theory of large deviations gives an asymptotic computation of small probabilities on exponential scale; see [4] as a reference of this topic. The basic definition of this theory is the concept of large deviation principle, which provides some asymptotic bounds for a family of probability measures on the same topological space; these bounds are expressed in terms of a speed function (that tends to infinity) and a lower semicontinuous rate function defined on the topological space.
The term moderate deviations is used for a class of large deviation principles which fills the gap between a convergence to a constant (governed by a large deviation principle with a suitable speed function) and an asymptotic normality result. In view of the examples studied in this paper we explain the concept of moderate deviations in the next Assertion 1.1, and our presentation will be restricted to sequences of real random variables defined on the same probability space (Ω,F,P); thus the speed function is a sequence {vn:n≥1} such that vn→∞ (as in the rest of the paper).
((Classical) moderate deviations).
Let{Cn:n≥1}be a sequence of real random variables such that the following asymptotic regimes hold.
R1:{Cn:n≥1}converges in probability to zero, and this convergence is governed by a large deviation principle with speedvnand rate functionILD(such thatILD(x)=0if and only ifx=0);
R2:vnCnconverges weakly to a centered normal distribution with (positive) varianceσ2.
Then we talk about moderate deviations when, for every family of positive numbers{an:n≥1}such thatan→0andanvn→∞,the sequence of random variables{anvnCn:n≥1}satisfies the large deviation principle with speed1/anand rate functionIMDdefined byIMD(x)=x22σ2for allx∈R.Moreover, one typically hasILD″(0)=1σ2.
We have the following remarks.
We can recover the asymptotic regimes R1 and R2 in Assertion 1.1 by setting an=1vn and an=1 respectively; note that, in both cases, one of the conditions in (1) holds and the other one fails. So the class of large deviation principles in Assertion 1.1 is determined by a family of positive scaling factors {an:n≥1} that fills the gap between the asymptotic regimes R1 and R2. Moreover, by (1), the speed 1/an for the random variables {anvnCn:n≥1} has a lower intensity than the speed vn (see R1).
Concerning the random variables {Cn:n≥1} in Assertion 1.1, one often has
Cn=Zn−z∞for alln≥1,
where {Zn:n≥1} is a sequence which converges in probability to z∞ for some z∞∈R (see, for instance, (23) concerning Example 6.1 where z∞=t); moreover this convergence is governed by a large deviation principle with some speed vn, say, and rate function IZ, and one has IZ(z)=0 if and only if z=z∞. Then, for the rate function ILD in Assertion 1.1, one has
ILD(x)=IZ(x+z∞)for allx∈R.
Moreover, if we refer to the equality (3) at the end of Assertion 1.1, one typically has IZ″(z∞)=1σ2 because IZ″(z∞)=ILD″(0).
Now we present a prototype example of the framework in Assertion 1.1, for which one can refer to the law of large numbers and the central limit theorem. Without loss of generality we restrict our attention to the case of centered random variables {Xn:n≥1} in order to have a convergence to zero (as in the asymptotic regime R1 in Assertion 1.1).
We set
Cn:=X1+⋯+Xnnfor alln≥1
where {Xn:n≥1} is a sequence of i.i.d. centered real random variables. Moreover, we also assume that E[eθX1]<∞ in a neighborhood of the origin θ=0, and therefore σ2=Var[X1] is finite (we also assume that σ2>00$]]> to avoid trivialities). In such a case the sequence {Cn:n≥1} satisfies the large deviation principle with speed vn=n and rate function ILD defined by
ILD(x)=supθ∈Rθx−logEeθX1for allx∈R.
This is a consequence of the Cramér theorem (see, e.g., Theorem 2.2.3 in [4]), and one can check that ILD″(0)=1σ2. Moreover, for every family of positive numbers {an:n≥1} such that (1) holds, the sequence of random variables {annCn:n≥1} satisfies the large deviation principle with speed 1/an and rate function IMD defined by (2); this is a consequence of Theorem 3.7.1 in [4] with d=1. Actually this prototype example can be presented for an arbitrary d (i.e. for multivariate random variables) by taking into account the multivariate version of the Cramér theorem (see, e.g., Theorem 2.2.30 in [4]).
The aim of this paper is to present some examples of noncentral moderate deviations. We use this terminology to mean a class of large deviation principles which fills the gap between a convergence to a constant (governed by a large deviation principle with some speed vn and some rate function which uniquely attains the value zero at that constant) and a weak convergence to some non-Gaussian law. This should happen in the same spirit of Assertion 1.1 for some positive scalings {an:n≥1} such that (1) holds.
The terminology of noncentral moderate deviations appears in three recent results: Proposition 3.3 in [13] (for possibly m-variate random variables), Proposition 4.3 in [15] and Proposition 3.3 in [2]. In the first two cases the weak convergence is trivial because one has a family of identically distributed random variables. Another result is Proposition 2.2 in [12], where the convergence in distribution is not trivial; however the term noncentral moderate deviations does not appear in that paper.
The common line of the examples studied in this paper can be summarized as follows.
(Common line of several examples in this paper).
We consider a sequence of real random variables{Cn:n≥1}such that the following asymptotic regimes hold.
R1:{Cn:n≥1}converges in probability to zero, and this convergence is governed by a large deviation principle with speedvn(we always havevn→∞) and rate functionILD(such thatILD(x)=0if and only ifx=0);
R2:vnCnconverges weakly to a non-Gaussian law.
Moreover, for every family of positive numbers{an:n≥1}such that (1) holds, the sequence of random variables{anvnCn:n≥1}satisfies the large deviation principle with speed1/anand a suitable rate functionIMD.
Some interesting common features of the examples studied in this paper are given by the following equalities: ILD(0)=IMD(0)=0 (as in Assertion 1.1),
IMD(x)=ILD′(0+)xorIMD(x)=∞ifx>00\]]]>
and
IMD(x)=ILD′(0−)xorIMD(x)=∞ifx<0.
Not all the noncentral moderate deviation results have these common features; for instance they do not appear in Proposition 3.3 in [2]. This explains the interest of non-central moderate deviations that, in our opinion, deserve to be investigated. We also recall that the common features of the examples studied in this paper can be seen as the analogue of the equality IMD(x)=ILD″(0)2x2 stated in Assertion 1.1 for the classical moderate deviations (it is a consequence of (2) and (3)).
Now we present the outline of the paper with a very brief description of the examples studied in each section. We start with some preliminaries in Section 2. Section 3 is devoted to Example 3.1 which concerns minima of i.i.d. nonnegative random variables. In Section 4 we study Example 4.1, which is based on maxima of i.i.d. random variables in the maximum domain attraction (MDA) of the Gumbel distribution (see, e.g., the family of distributions in Theorem 8.13.4 in [3] together with the well-known Fisher–Tippett theorem, e.g., Theorem 8.13.1 in [3]). In Section 5 we study Example 5.1 which concerns the classical occupancy problem (or the coupon collector’s problem); in particular that is an example with a weak convergence to the Gumbel distribution (as Example 4.1). Finally, in Section 6, we consider Example 6.1 which is inspired by a recent replacement model for random lifetimes in the literature (see, e.g., [5]).
We conclude with some notation used throughout the paper. We write an∼bn to mean that anbn→1 as n→∞; moreover we use the symbol [x] for the integer part of x∈R, i.e.
[x]:=max{k∈Z:k≤x}.
Preliminaries
We start with the definition of large deviation principle (see, e.g., [4], pages 4–5). Let (X,τX) be a topological space and let {Zn:n≥1} be a sequence of X-valued random variables defined on the same probability space (Ω,F,P). A sequence {vn:n≥1} such that vn→∞ (as n→∞) is called a speed function, and a lower semicontinuous function I:X→[0,∞] is called a rate function. Then the sequence {Zn:n≥1} satisfies the large deviation principle (LDP from now on) with speed vn and rate function I if
lim supn→∞1vnlogP(Zn∈C)≤−infx∈CI(x)for all closed setsC,
and
lim infn→∞1vnlogP(Zn∈O)≥−infx∈OI(x)for all open setsO.
The rate function I is said to be good if every level set {x∈X:I(x)≤η} (for η≥0) is compact.
The following Lemma 2.1 is quite a standard result (anyway we give briefly some hints of the proof for the sake of completeness). All the moderate deviation results in this paper concern the situation presented in Assertion 1.2, and they will be proved by applying Lemma 2.1; more precisely, for every choice of the positive scalings {an:n≥1} such that (1) holds (with a further condition for Example 6.1, i.e. (24)), we shall have sn=1/an and I=IMD for some rate function IMD. We shall apply Lemma 2.1 also to prove Proposition 6.1 (with sn=vn=n and I=ILD for a suitable rate function ILD), which provides the LDP for the asymptotic regime R1 concerning Example 6.1.
Let{sn:n≥1}be a speed and let{Cn:n≥1}be a sequence of real random variables defined on the same probability space(Ω,F,P). Moreover letI:R→[0,∞]be a rate function decreasing on(−∞,0), increasing on(0,∞)and such thatI(x)=0if and only ifx=0. Moreover, assume that:lim supn→∞1snlogP(Cn≥x)≤−I(x)for allx>0;0;\]]]>lim supn→∞1snlogP(Cn≤x)≤−I(x)for allx<0;lim infn→∞1snlogP(Cn∈O)≥−I(x)for everyx∈{y∈R:I(y)<∞}, andfor all open setsOsuch thatx∈O.Then{Cn:n≥1}satisfies the LDP with speedsnand rate function I.
It is known that the final statement (8) yields the lower bound for open sets (5) (see, e.g., condition (b) with eq. (1.2.8) in [4]). Here we prove that (6) and (7) yield the upper bound for closed sets (4). Such an upper bound trivially holds if C is the empty set or if 0∈C, and therefore in what follows we assume that C is nonempty and 0∉C; then at least one of the sets C∩(0,∞) and C∩(−∞,0) is nonempty. For simplicity we assume that both sets C∩(0,∞) and C∩(−∞,0) are nonempty (in fact, if one of them is empty, the proof can be adapted readily). Then we can find x1∈C∩(0,∞) and x2∈C∩(−∞,0) such that C⊂(−∞,x2]∪[x1,∞), and we have
P(Zn∈C)≤P(Zn≥x1)+P(Zn≤x2);
thus, by Lemma 1.2.15 in [4] (together with (6) and (7)), we have
lim supn→∞1snlogP(Zn∈C)≤lim supn→∞1snlog{P(Zn≥x1)+P(Zn≤x2)}=max{−I(x1),−I(x2)}=−min{I(x1),I(x2)}.
We conclude the proof noting that min{I(x1),I(x2)}=infx∈CI(x) by the hypotheses. □
An example with minima of i.i.d. nonnegative random variables
In this section we consider the following example.
Let {Xn:n≥1} be a sequence of i.i.d. real random variables, with a common distribution function F. We assume that F is strictly increasing on (αF,ωF), where
αF:=inf{x∈R:F(x)>0}=0andωF:=sup{x∈R:F(x)<1};0\}=0\hspace{2.5pt}\text{and}\hspace{2.5pt}{\omega _{F}}:=\sup \{x\in \mathbb{R}:F(x)<1\};\]]]>
so {Xn:n≥1} are nonnegative random variables. We also assume that F(0)=0, and that there exists
F′(0+):=limx→0+F(x)−F(0)x=limx→0+F(x)x∈(0,∞).
In what follows we use the notation
F(x−):=limy↑xF(y).
Throughout this section we set
Cn:=min{X1,…,Xn}for alln≥1.
We can recover the asymptotic regimesR1andR2in Assertion1.2as follows.
R1:{Cn:n≥1}converges in probability to zero (actually it is an a.s. convergence); moreover, by Lemma 1 in [14],{Cn:n≥1}satisfies the LDP with speedvn=nand rate functionILDdefined byILD(x):=−log(1−F(x−))ifx∈[0,ωF),∞otherwise.
R2:vnCn=nmin{X1,…,Xn}converges weakly to the exponential distribution with mean1/F′(0+); in fact, for everyx>00$]]>, by (9), we getP(nmin{X1,…,Xn}≤x)=1−P(nmin{X1,…,Xn}>x)=1−PnX1>xn=1−1−Fxnn=1−1−nFxnnn→1−e−F′(0+)xasn→∞.x)\\ {} & =1-{P^{n}}\left({X_{1}}>\frac{x}{n}\right)=1-{\left(1-F\left(\frac{x}{n}\right)\right)^{n}}\\ {} & =1-{\left(1-\frac{nF\left(\frac{x}{n}\right)}{n}\right)^{n}}\to 1-{e^{-{F^{\prime }}(0+)x}}\hspace{1em}\textit{as}\hspace{2.5pt}n\to \infty .\end{aligned}\]]]>
Thus we are studying a sequence of random variables that has interest for any possible concrete model related to the maximum domain of attraction of the Weibull distribution (indeed, exponential distribution is a particular case of the Weibull distribution). For instance, the random variables {Xn:n≥1} could represent wind speed data (to determine the viability of windmills).
Now we prove the moderate deviation result.
For every family of positive numbers{an:n≥1}such that (1) holds (i.e.an→0andann→∞), the sequence of random variables{annmin{X1,…,Xn}:n≥1}satisfies the LDP with speed1/anand rate functionIMDdefined byIMD(x):=F′(0+)xifx∈[0,∞),∞otherwise.
We apply Lemma 2.1 for every choice of {an:n≥1} such that (1) holds, with sn=1/an, I=IMD and {Cn:n≥1} as in (10).
We have to show that, for every x>00$]]>,
lim supn→∞anlogP(annmin{X1,…,Xn}≥x)≤−F′(0+)x.
For every δ∈(0,x) we have
P(annmin{X1,…,Xn}≥x)=PnX1≥xann=1−Fxann−n≤1−Fx−δannn
and, by the relation limx→0+F(x)=0 and the limit in (9),
lim supn→∞anlogP(annmin{X1,…,Xn}≥x)≤lim supn→∞annlog1−Fx−δann=lim supn→∞ann−Fx−δann=−F′(0+)(x−δ);
so we obtain (6) by letting δ go to zero.
We have to show that, for every x<0,
lim supn→∞anlogP(annmin{X1,…,Xn}≤x)≤−∞.
This trivially holds as −∞≤−∞ because the random variables {Cn:n≥1} are nonnegative (and therefore P(annmin{X1,…,Xn}≤x)=0 for every n≥1).
We want to show that, for every x≥0 and for every open set O such that x∈O, we have
lim infn→∞anlogP(annmin{X1,…,Xn}∈O)≥−F′(0+)x.
The case x=0 is immediate; indeed, since annmin{X1,…,Xn} converges in probability to zero by the Slutsky theorem (by an→0 and the weak convergence in R2 in Assertion 3.1), we have trivially 0≥0. So, from now on, we suppose that x>00$]]> and we take δ>00$]]> small enough to have
(x−δ,x+δ)⊂O∩(0,∞).
Then
P(annmin{X1,…,Xn}∈O)≥P(annmin{X1,…,Xn}∈(x−δ,x+δ))=Pmin{X1,…,Xn}∈x−δann,x+δann=PnX1>x−δann−PnX1≥x+δann=1−Fx−δannn−1−Fx+δann−n=1−Fx−δannn1−1−Fx+δann−n1−Fx−δannn=1−Fx−δannn1−expnlog1−Fx+δann−1−Fx−δann.\frac{x-\delta }{{a_{n}}n}\right)-{P^{n}}\left({X_{1}}\ge \frac{x+\delta }{{a_{n}}n}\right)\\ {} & \hspace{1em}={\left(1-F\left(\frac{x-\delta }{{a_{n}}n}\right)\right)^{n}}-{\left(1-F\left(\frac{x+\delta }{{a_{n}}n}-\right)\right)^{n}}\\ {} & \hspace{1em}={\left(1-F\left(\frac{x-\delta }{{a_{n}}n}\right)\right)^{n}}\left(1-\frac{{\left(1-F\left(\frac{x+\delta }{{a_{n}}n}-\right)\right)^{n}}}{{\left(1-F\left(\frac{x-\delta }{{a_{n}}n}\right)\right)^{n}}}\right)\\ {} & \hspace{1em}={\left(1-F\left(\frac{x-\delta }{{a_{n}}n}\right)\right)^{n}}\left\{1-\exp \left(n\log \left(\frac{1-F\left(\frac{x+\delta }{{a_{n}}n}-\right)}{1-F\left(\frac{x-\delta }{{a_{n}}n}\right)}\right)\right)\right\}.\end{aligned}\]]]>
Now notice that
log1−Fx+δann−1−Fx−δann=log1+Fx−δann−Fx+δann−1−Fx−δann≤Fx−δann−Fx+δann−1−Fx−δann=−Fx+δann−−Fx−δann1−Fx−δann≤−Fxann−Fx−δann1−Fx−δann
and, again by the relation limx→0+F(x)=0 and the limit in (9),
nlog1−Fx+δann−1−Fx−δann≤−annFxann−annFx−δannan1−Fx−δann→−∞asn→∞,
because an1−Fx−δann→0 and
annFxann−annFx−δann→F′(0+)(x−(x−δ))=F′(0+)δ>0.0.\]]]>
In conclusion, we get
lim infn→∞anlogP(annmin{X1,…,Xn}∈O)≥lim infn→∞anlog1−Fx−δannn+lim infn→∞anlog1−expnlog1−Fx+δann−1−Fx−δann=lim infn→∞annlog1−Fx−δann=lim infn→∞ann−Fx−δann=−F′(0+)(x−δ);
so we obtain (8) by letting δ go to zero.
□
An example with maxima of i.i.d. random variables in the MDA of Gumbel distribution
In this section we consider the following example.
Let {Xn:n≥1} be a sequence of i.i.d. real random variables, with a common distribution function F. Let ωF be as in Example 3.1, and assume that ωF=∞. We also assume that, for x large enough, F is strictly increasing with positive density f. Moreover let w be the function defined by
w(x):=F¯(x)f(x),whereF¯(x):=1−F(x),
and assume that w is differentiable for x large enough, limx→∞w′(x)=0 and w is a regularly varying function with exponent 1−μ for some μ>00$]]>, i.e. limx→∞w(tx)w(x)=t1−μ for all t>00$]]>. So, it is well known that
w(x)=x1−μL(x)
for a suitable slowly varying function L, i.e. a function such that
limx→∞L(tx)L(x)=1for allt>00\]]]>
(this can be immediately checked by the definitions of regularly varying and slowly varying functions).
Here we list some particular cases in which w is a regularly varying function with exponent 1−μ for some μ>00$]]>.
Standard normal distribution (see, e.g., [18] (pp. 48, 50–51, 88–90) for the asymptotic behavior of w(x))
F¯(x)=12π∫x∞e−t2/2dt(x∈R),w(x)=∫x∞e−t2/2dte−x2/2∼1x,μ=2.
Gamma distribution for a>00$]]> (see, e.g., [1] (formula 6.5.32, p. 263), for the limit of w(x))
F¯(x)=1Γ(a)∫x∞ta−1e−tdt(x>0),0),\]]]>w(x)=∫x∞ta−1e−tdtxa−1e−x→1,μ=1.
Weibull distribution for a>00$]]>F¯(x)=e−xa(x>0),0),\]]]>w(x)=e−xaaxa−1e−xa=x1−aa,μ=a.
Logistic distribution
F¯(x)=11+ex(x∈R),w(x)=11+exex(1+ex)2=1+e−x→1,μ=1.
We need several consequences of the assumptions in Example 4.1. We start with some results concerning the function L and the value μ (see (11)). In particular we provide an estimate for
ℓ:=lim supx→∞−L(x)logF¯(x)xμ.
Under the assumptions in Example4.1we haveL(x)xμ→0(asx→∞).
It is known (see, e.g., [8], Chapter VIII, Section 8, Lemma 2, p. 277) that, for every ε>00$]]>, there exists xε>00$]]> such that
x−ε<L(x)<xεfor allx>xε. {x_{\varepsilon }}.\]]]>
Then, if we take ε<μ, we get
x−(ε+μ)<L(x)xμ<xε−μfor allx>xε,{x_{\varepsilon }},\]]]>
and we immediately get the desired limit by letting x go to infinity. □
Under the assumptions in Example4.1we haveℓ≤1μ(where ℓ is defined by (12)).
Recall from (11) that F¯(x)f(x)=x1−μL(x); thus L(x) does not vanish because we have F¯(x)≠0 for all x since ωF=∞. Therefore, for some x1, we get
f(x)F¯(x)=xμ−1L(x)for allx>x1,{x_{1}},\]]]>
which yields
∫x1xf(t)F¯(t)dt︸−logF¯(x)+logF¯(x1)=∫x1xtμ−1L(t)dt.
Recall also the so-called Potter bound (see, e.g., Theorem 1.5.6(i) in [3]): for every A>11$]]> and δ>00$]]> there exists x0=x0(A,δ) such that
L(y)L(z)≤Amaxzyδ,yzδfor ally,z>x0.{x_{0}}.\]]]>
Then we have
−L(x)logF¯(x)xμ+L(x)logF¯(x1)xμ=L(x)xμ∫x1x0tμ−1L(t)dt+L(x)xμ∫x0xtμ−1L(t)dt;
moreover, by the Potter bound and some computations, we can estimate the last term as
L(x)xμ∫x0xtμ−1L(t)dt≤Axμ∫x0xxtδtμ−1dt=Aμ−δ1−x0xμ−δ.
Then, by the definition of ℓ in (12) and by Lemma 4.1, we get ℓ≤Aμ−δ by letting x go to infinity; so we conclude by the arbitrariness of A and δ. □
(A discussion on the inequality in Lemma 4.2).
If we consider the four distributions listed above for which the function w is regularly varying with exponent 1−μ for some μ>00$]]>, we have
limx→∞L(x)=L(∞)for someL(∞)∈(0,∞);
indeed we have L(∞)=1 for standard normal, Gamma and logistic distributions, and L(∞)=1/a for Weibull distribution. Then, by the L’Hôpital rule we have
limx→∞logF¯(x)xμ=limx→∞−f(x)/F¯(x)μxμ−1=−limx→∞1/w(x)μxμ−1=−limx→∞1μL(x)=−1μL(∞),
and therefore
limx→∞−L(x)logF¯(x)xμ=−limx→∞L(x)limx→∞logF¯(x)xμ=1μ.
In conclusion, we have shown that the limit of −L(x)logF¯(x)xμ as x→∞ exists if (13) holds (as it happens for the four distributions listed above). We are not aware of cases in which the inequality in Lemma 4.2 is strict.
Some further preliminaries are needed. Firstly, under the assumptions in Example 4.1, the following quantities are well defined for n large enough:
mn:=F−11−1n
and (in one of the following equalities we take into account (11))
hn:=mnnf(mn)=mnf(mn)F¯(mn)=mnw(mn)=mnmn1−μL(mn)=mnμL(mn).
The following lemmas provide some properties of the function w and the sequence {hn:n≥1}.
Under the assumptions in Example4.1we havew(xn)∼w(yn)forxn,yn→∞such thatxn∼yn.
By the well-known Karamata’s representation of slowly varying functions (see, e.g., Theorem 1.3.1 in [3]), there exists b>00$]]> such that the function L introduced above can be written as
L(x)=c1(x)exp∫bxc2(t)tdtfor allx≥b,
where c1 and c2 are suitable functions such that c1(x) tends to some finite limit and c2(x)→0 (as x→∞). Then we have to prove that
w(xn)w(yn)=c1(xn)exp∫bxnc2(t)tdtxn1−μc1(yn)exp∫bync2(t)tdtyn1−μ→1asn→∞.
By the hypotheses we only need to prove that
exp∫bxnc2(t)tdtexp∫bync2(t)tdt=exp∫ynxnc2(t)tdt→1,
which amounts to
limn→∞∫ynxnc2(t)tdt=0.
Let δ∈(0,1) and ε∈(0,δ1+δ) be arbitarily fixed. Then there exists nε≥1 such that |c2(t)|<ε for t>nε{n_{\varepsilon }}$]]> and 1−ε<xnyn<1+ε for n>nε{n_{\varepsilon }}$]]>. So, for n large enough to have xn∧yn>nε{n_{\varepsilon }}$]]>, we get
∫ynxnc2(t)tdt≤∫xn∧ynxn∨yn|c2(t)|tdt≤ε∫xn∧ynxn∨yn1tdt=εlogxn∨ynxn∧yn
and
1≤xn∨ynxn∧yn=xnynifxn≥ynynxnifxn<yn≤1+εifxn≥yn11−εifxn<yn<1+δ.
The proof is complete. □
Under the assumptions in Example4.1we havehn∼μlogn(asn→∞).
Firstly we recall that hn=mnμL(mn) (see (14)); then hn→∞ by taking into account that mn→∞ and by Lemma 4.1. Moreover, we have
∫0mn1w(t)dt=∫0mnf(t)F¯(t)dt=−∫0mnddtlogF¯(t)dt=−logF¯(mn)+logF¯(0)=logn+logF¯(0);
then hn∼μlogn if and only if hn∼μ∫0mn1w(t)dt, which amounts to
limn→∞μhn∫0mn1w(t)dt=1.
So we prove the lemma showing that (15) holds. We take η>00$]]> and we write
μhn∫0mn1w(t)dt=μhn∫0ηmn1w(t)dt︸=:In(1)(η)+μhn∫ηmnmn1w(t)dt︸=:In(2)(η).
Now we estimate In(1)(η) and In(2)(η) separately.
We have (make the change of variable r=F¯(t))
In(1)(η)=μhn∫0ηmnf(t)F¯(t)dt=μhn∫F¯(ηmn)F¯(0)drr=μhn(logF¯(0)−logF¯(ηmn)),
where F¯(0),F¯(ηmn)<1 because ωF=∞. Thus
0≤lim infn→∞In(1)(η)≤lim supn→∞In(1)(η)≤ημμℓ,
because μhnlogF¯(0)→0 (since hn→∞), −μhnlogF¯(ηmn)≥0,
−μhnlogF¯(ηmn)=−μmnμ/L(mn)logF¯(ηmn)=−μημL(mn)(ηmn)μlogF¯(ηmn)=μημL(mn)L(ηmn)−L(ηmn)logF¯(ηmn)(ηmn)μ,
and by taking into account that mn→∞, L is a slowly varying function, and Lemma 4.2.
We have (make the change of variable u=tmn)
In(2)(η)=μmnμ/L(mn)∫ηmnmn1t1−μL(t)dt=μL(mn)mnμ∫η1mndu(mnu)1−μL(mnu)=∫η1L(mn)L(mnu)μuμ−1du,
and, since L(mn)L(mnu)→1 uniformly on compact subsets of (0,∞) (with respect to u) by Theorem 1.2.1 in [3] (here we again take into account that mn→∞), for all ρ>00$]]> there exists n0=n0(η,ρ) such that for all n>n0{n_{0}}$]]>1−ρ<L(mn)L(mnu)<1+ρfor allu∈[η,1],
and therefore
(1−ρ)(1−ημ)<In(2)(η)<(1+ρ)(1−ημ).
Finally we combine the estimates for In(1)(η) and In(2)(η), and we get (15) by the arbitrariness of η>00$]]> and ρ>00$]]>. This completes the proof. □
Throughout this section we set
Cn:=Mnmn−1for alln≥n0, for somen0, whereMn:=max{X1,…,Xn};
we have to consider n0 large enough in order to have a well-defined mn for n≥n0.
We can recover the asymptotic regimesR1andR2in Assertion1.2as follows.
R1:{Cn:n≥n0}converges in probability to zero (actually the authors have checked the almost sure convergence with an argument based on Theorem 4.4.4 in [9], p. 268). Moreover, as an immediate consequence of Proposition 3.1 in [11],{Cn+1:n≥n0}satisfies the LDP with speedvn=μlogn(or equivalentlyvn=hnby Lemma4.4) and rate function J defined byJ(y):=yμ−1μify≥1,∞otherwise.Then, since we deal with a sequence of shifted random variables, we deduce that{Cn:n≥2}satisfies the LDP with speedvn=hnand rate functionILDdefined byILD(x):=J(x+1)=(x+1)μ−1μifx≥0,∞otherwise.
R2:vnCn=hnMnmn−1converges weakly to the Gumbel distribution by a well-known result by von Mises (see, e.g., Theorem 8.13.7 in [3]).
Thus we are studying a sequence of random variables that has interest for any possible concrete model related to the maximum domain of attraction of the Gumbel distribution. A possible example is when the random variables {Xn:n≥1} represent monthly and annual maximum values of daily rainfall and river discharge volumes.
Now we prove the moderate deviation result. We shall see that, for this example, the rate functions ILD and IMD coincide when μ=1.
For every family of positive numbers{an:n≥1}such that (1) holds (i.e.an→0andanhn→∞), the sequence of random variables{anhn(Mnmn−1):n≥1}satisfies the LDP with speed1/anand rate functionIMDdefined byIMD(x):=xifx∈[0,∞),∞otherwise.
We apply Lemma 2.1 for every choice of {an:n≥1} such that (1) holds, with sn=1/an, I=IMD and {Cn:n≥n0} as in (16).
We have to show that, for every x>0 0$]]>,
lim supn→∞anlogPanhnMnmn−1≥x≤−x.
For mn′:=(xanhn+1)mn we get
PanhnMnmn−1≥x=PMn≥mn′=1−PMn<mn′=1−Fnmn′=1−1−F¯mn′n=1−expnlog1−F¯mn′≤−nlog1−F¯mn′.
Thus
anlogPanhnMnmn−1≥x≤anlog−nlog1−F¯mn′=anlogn+anloglog1−F¯mn′−F¯mn′+anlogF¯mn′
and, noting that
limn→∞anloglog1−F¯mn′−F¯mn′=0,
because mn′→∞ (since mn→∞), we obtain
lim supn→∞anlogPanhnMnmn−1≥x≤lim supn→∞anlogn+logF¯mn′.
Now we apply the Lagrange theorem (also known as the mean value theorem) to the function g(z)=logF¯(z); so there exists ξn∈mn,mn′ such that
logF¯mn′=logF¯(mn)−f(ξn)F¯(ξn)mn′−mn.
Then, by the definitions of mn, w and hn, we get
logF¯mn′=−logn−mnxw(ξn)anhn=−logn−xw(ξn)annf(mn)=−logn−w(mn)xw(ξn)annF¯(mn)=−logn−w(mn)xw(ξn)an,
and therefore
anlogn+logF¯mn′=−xw(mn)w(ξn).
So we get
lim supn→∞anlogPanhnMnmn−1≥x≤lim supn→∞anlogn+logF¯mn′≤−x
by Lemma 4.3 with xn=mn and yn=ξn (indeed ξn∼mn because
1≤ξnmn≤1+xanhn
by construction, and 1+xanhn→1). Thus (6) holds.
We have to show that, for every x<0,
lim supn→∞anlogPanhnMnmn−1≤x≤−∞.
For mn′:=(xanhn+1)mn we get
PanhnMnmn−1≤x=PMn≤mn′=Fnmn′=1−F¯mn′n.
Thus
anlogPanhnMnmn−1≥x≤annlog1−F¯mn′∼−annF¯mn′=−explogannF¯mn′=−explogan+logn+logF¯mn′=−explogan+anlogn+logF¯mn′an.
Now we can repeat the computations above in the proof of (6) with some slight changes (we mean the part with the application of the Lagrange theorem; the details are omitted), and we find
limn→∞anlogn+logF¯mn′=−x>0,0,\]]]>
and therefore
limn→∞−explogan+anlogn+logF¯mn′an=−∞.
Thus (7) holds.
We want to show that, for every x≥0 and for every open set O such that x∈O, we have
lim infn→∞anlogPanhnMnmn−1∈O≥−x.
The case x=0 is immediate; indeed, since anhnMnmn−1 converges in probability to zero by the Slutsky theorem (by an→0 and the weak convergence in R2 in Assertion 4.1), we have trivially 0≥0. So, from now on, we suppose that x>00$]]> and we take δ>0 0$]]> small enough to have
(x−δ,x+δ]⊂O∩(0,∞).
Then, for mn(±):=1+x±δanhnmn, we get
PanhnMnmn−1∈O≥Px−δ<anhnMnmn−1≤x+δ=Pmn(−)<Mn≤mn(+)=Fnmn(+)−Fnmn(−)=1−F¯mn(+)n−1−F¯mn(−)n.
Now we apply the Lagrange theorem to the function g(z)=(1−F¯(z))n; so there exists ξn∈mn(−),mn(+) such that
1−F¯mn(+)n−1−F¯mn(−)n=n1−F¯ξnn−1f(ξn)mn(+)−mn(−)︸=2δmnanhn.
Then, by the definition of hn, we obtain
anlogPanhnMnmn−1∈O≥anlogn+(n−1)log1−F¯ξn+logf(ξn)+log2δmnanhn=an(n−1)log1−F¯ξn+logf(ξn)+log(2δ)−logan−logf(mn)
and therefore, by the definition of the function w,
lim infn→∞anlogPanhnMnmn−1∈O≥lim infn→∞an(n−1)log1−F¯ξn+logf(ξn)f(mn)=lim infn→∞an(n−1)log1−F¯ξn+logw(mn)F¯(ξn)F¯(mn)w(ξn)=lim infn→∞an(n−1)log(1−F¯(ξn))+logw(mn)w(ξn)+logF¯(ξn)F¯(mn).
Then we get (8) if we show the relations
limn→∞anlogw(mn)w(ξn)=0,lim infn→∞anlogF¯(ξn)F¯(mn)≥−(x+δ),limn→∞an(n−1)log(1−F¯(ξn))=0,
and by letting δ go to zero. The first limit in (17) holds by Lemma 4.3 with xn=mn and yn=ξn (indeed ξn∼mn because
1+x−δanhn≤ξnmn≤1+x+δanhn
by construction and 1+x±δanhn→1). The inequality in (17) holds since
anlogF¯(ξn)F¯(mn)≥anlogF¯(mn(+))F¯(mn),
by a new application of the Lagrange theorem to the function g(z)=−logF¯(z), and by applying again Lemma 4.3 (we omit the details to avoid repetitions). Finally we prove the last limit in (17) if we show that
limn→∞annF¯(mn(±))=0;
in fact
an(n−1)log(1−F¯(ξn))∼−annF¯(ξn)
and
annF¯(mn(+))≤annF¯(ξn)≤annF¯(mn(−)).
We have
annF¯(mn(±))=anF¯(mn(±))F¯(mn)=expan(logF¯(mn(±))−logF¯(mn))an+logan;
moreover,
limn→∞an(logF¯(mn(±))−logF¯(mn))=−(x±δ)<0
(this follows once more from an application of the Lagrange theorem to the function g(z)=−logF¯(z), and by applying again Lemma 4.3; the details are omitted). Thus
limn→∞an(logF¯(mn(±))−logF¯(mn))an+logan=−∞,
which yields the desired last limit in (17).
□
An example inspired by the classical occupancy problem
In this section we consider the following example.
Let {Tn:n≥1} be a family of random variables such that
Tn:=∑k=1nXn,k,
where {Xn,k:n≥1,1≤k≤n} is a triangular array of random variables (i.e. {Xn,k:1≤k≤n} are independent random variables for each fixed n), and each random variable Xn,k is geometric with parameter pn,k:=1−k−1n, i.e.
P(Xn,k=h)=(1−pn,k)h−1pn,kfor allh≥1.
It is well known that the random variable Tn can be seen as the number of balls required to fill n boxes with at least one ball when one puts balls in n boxes at random, and each ball is independently assigned to any fixed box with probability 1n; this is known in the literature as the classical occupancy problem. From a different point of view Tn can also be related to the coupon collector’s problem: a coupon collector chooses at random and independently among n coupon types, and Tn represents the number of coupons required to collect all the n coupon types. Throughout this section we set
Cn:=Tnnlogn−1for alln≥2.
We can recover the asymptotic regimesR1andR2in Assertion1.2as follows.
R1:{Cn:n≥2}converges in probability to zero (see Example 2.2.7 in [7] presented for the coupon collector’s problem). Moreover, by Proposition 2.1 in [10],{Cn+1:n≥2}satisfies the LDP with speedvn=lognand rate function J defined byJ(y):=y−1ify≥1,∞otherwise.Then, since we deal with a sequence of shifted random variables, we deduce that{Cn:n≥2}satisfies the LDP with speedvn=lognand rate functionILDdefined byILD(x):=J(x+1)=xifx≥0,∞otherwise.
R2:vnCn=lognTnnlogn−1converges weakly to the Gumbel distribution (see, e.g., Example 3.6.11 in [7]).
Now we prove the moderate deviation result. We shall see that, for this example, the rate functions ILD and IMD coincide. Several parts of the proof of the next Proposition 5.1 have some analogies with the one presented for Proposition 2.1 in [10]; so we shall omit some details to avoid repetitions.
For every family of positive numbers{an:n≥1}such that (1) holds (i.e.an→0andanlogn→∞), the sequence of random variables{anlogn(Tnnlogn−1):n≥1}satisfies the LDP with speed1/anand rate functionIMDdefined byIMD(x):=xifx∈[0,∞),∞otherwise.
We apply Lemma 2.1 for every choice of {an:n≥1} such that (1) holds, with sn=1/an, I=IMD and {Cn:n≥2} as in (18).
We have to show that, for every x>0 0$]]>,
lim supn→∞anlogPanlognTnnlogn−1≥x≤−x.
For every δ∈(0,x) we have (here the last inequality holds by a well-known estimate; see, e.g., Exercise 3.10 in [17], page 58)
PanlognTnnlogn−1≥x=PTn≥xanlogn+1nlogn≤PTn>x−δanlogn+1nlogn≤n1−(x−δanlogn+1)=n−x−δanlogn.\left(\frac{x-\delta }{{a_{n}}\log n}+1\right)n\log n\right)\le {n^{1-(\frac{x-\delta }{{a_{n}}\log n}+1)}}={n^{-\frac{x-\delta }{{a_{n}}\log n}}}.\end{aligned}\]]]>
Thus
lim supn→∞anlogPanlognTnnlogn−1≥x≤lim supn→∞anlogn−x−δanlogn=−x+δ,
and we obtain (6) by letting δ go to zero.
We have to show that, for every x<0,
lim supn→∞anlogPanlognTnnlogn−1≤x≤−∞.
We have xanlogn+1∈(0,1) eventually, and therefore (here the last inequality holds by a well-known estimate; for instance, it is a consequence of Theorem 5.10 and Corollary 5.11 in [16])
PanlognTnnlogn−1≤x=PTn≤xanlogn+1nlogn≤21−exp−xanlogn+1nlognnn=21−n−xanlogn+1n.
Thus
lim supn→∞anlogPanlognTnnlogn−1≤x≤lim supn→∞anlog21−n−xanlogn+1n=lim supn→∞annlog1−n−xanlogn+1=lim supn→∞−annn−xanlogn+1=lim supn→∞−ann−xanlogn=lim supn→∞−ane−xan=−∞,
and therefore (7) holds.
We want to show that, for every x≥0 and for every open set O such that x∈O, we have
lim infn→∞anlogPanlognTnnlogn−1∈O≥−x.
The case x=0 is immediate; indeed, since anlognTnnlogn−1 converges in probability to zero by the Slutsky theorem (by an→0 and the weak convergence in R2 in Assertion 5.1), we have trivially 0≥0. So, from now on, we suppose that x>00$]]> and we take δ>00$]]> small enough to have
(x−δ,x+δ]⊂O∩(0,∞).
Moreover, we also introduce the notation FTn(·)=P(Tn≤·) for the distribution function of Tn. Then
PanlognTnnlogn−1∈O≥Px−δ<anlognTnnlogn−1≤x+δ=P1+x−δanlognnlogn<Tn≤1+x+δanlognnlogn≥FTn1+x+δanlognnlogn−FTn1+x−δanlognnlogn+1.
Thus, by adapting some computations in the proof of Proposition 2.1 in [10], we have
FTn1+x+δanlognnlogn−FTn1+x−δanlognnlogn+1≥An−(An(+)+An(−))=An1−An(+)+An(−)An
where An, An(+), An(−) are the following nonnegative quantities:
An:=1−e−1n[(1+x+δanlogn)nlogn]n−1−e−1n([(1+x−δanlogn)nlogn]+1)n,An(+):=∑evenknke−kn[(1+x+δanlogn)nlogn]1−1−knkn[(1+x+δanlogn)nlogn]
and
An(−):=∑oddknke−kn([(1+x−δanlogn)nlogn]+1)1−1−knkn([(1+x−δanlogn)nlogn]+1).
Thus, we can say that
anlogPanlognTnnlogn−1∈O≥anlogAn+anlog1−An(+)+An(−)An,
and therefore
lim infn→∞anlogPanlognTnnlogn−1∈O≥lim infn→∞anlogAn+lim infn→∞anlog1−An(+)+An(−)An.
We conclude the proof of (8) (and therefore the proof of the proposition) if we show that
lim infn→∞anlogAn≥−(x+δ)
and
limn→∞An(+)An=0,limn→∞An(−)An=0;
in fact we obtain (8) by letting δ go to zero.
Concerning An in (19) we apply the Lagrange theorem to the function g(z)=(1−e−z)n; so there exists
ξn∈([(1+x−δanlogn)nlogn]+1)n,[(1+x+δanlogn)nlogn]n
such that
An=n(1−e−ξn)n−1e−ξn×[(1+x+δanlogn)nlogn]n−([(1+x−δanlogn)nlogn]+1)n.
Moreover,
(1−e−ξn)n−1e−ξn≥1−e−([(1+x−δanlogn)nlogn]+1)nn−1e−[(1+x+δanlogn)nlogn]n≥1−e−(1+x−δanlogn)lognn−1e−(1+x+δanlogn)logn=1−e−x−δannn−1e−x+δann
and
[(1+x+δanlogn)nlogn]n−([(1+x−δanlogn)nlogn]+1)n=[(1+x+δanlogn)nlogn]−1−[(1+x−δanlogn)nlogn]n≥(1+x+δanlogn)nlogn−1−1−(1+x−δanlogn)nlognn=−2n+2δan.
Then
anlogAn≥anlog1−e−x−δannn−1e−x+δan−2n+2δan=an(n−1)log1−e−x−δann−(x+δ)+anlog−2n+2δan,
whence we obtain (20) from
limn→∞anlog−2n+2δan=0
and
an(n−1)log1−e−x−δann∼−an(n−1)e−x−δann∼−ane−x−δan→0.
We remark that
limn→∞1−e−1n[(1+x+δanlogn)nlogn]n=1
and
limn→∞1−e−1n([(1+x−δanlogn)nlogn]+1)n=1,
because
1=lim infn→∞1−e−x+δan+1nnn≤lim infn→∞1−e−1n[(1+x+δanlogn)nlogn]n≤lim supn→∞1−e−1n[(1+x+δanlogn)nlogn]n≤lim supn→∞1−e−x+δannn=1,
and the second limit can be proved with a similar argument. Therefore
An=1−e−1n[(1+x+δanlogn)nlogn]n−1−e−1n([(1+x−δanlogn)nlogn]+1)n→1−1=0.
Moreover,
An=1−e−1n([(1+x−δanlogn)nlogn]+1)n×1−e−1n[(1+x+δanlogn)nlogn]n1−e−1n([(1+x−δanlogn)nlogn]+1)n−1∼1−e−1n[(1+x+δanlogn)nlogn]n1−e−1n([(1+x−δanlogn)nlogn]+1)n−1=expnlog1−e−1n[(1+x+δanlogn)nlogn]1−e−1n([(1+x−δanlogn)nlogn]+1)−1,
where
limn→∞nlog1−e−1n[(1+x+δanlogn)nlogn]1−e−1n([(1+x−δanlogn)nlogn]+1)=0.
Here we note that the limit (22) can be checked with some tedious computations:
nlog1−e−1n[(1+x+δanlogn)nlogn]1−e−1n([(1+x−δanlogn)nlogn]+1)=nlog1+e−1n([(1+x−δanlogn)nlogn]+1)−e−1n[(1+x+δanlogn)nlogn]1−e−1n([(1+x−δanlogn)nlogn]+1)
where
limn→∞e−1n([(1+x−δanlogn)nlogn]+1)=0
and
limn→∞e−1n[(1+x+δanlogn)nlogn]=0,
and therefore
nlog1−e−1n[(1+x+δanlogn)nlogn]1−e−1n([(1+x−δanlogn)nlogn]+1)∼ne−1n([(1+x−δanlogn)nlogn]+1)−e−1n[(1+x+δanlogn)nlogn]=elogn−1n([(1+x−δanlogn)nlogn]+1)×1−e−1n[(1+x+δanlogn)nlogn]+1n([(1+x−δanlogn)nlogn]+1).
Moreover,
limn→∞elogn−1n([(1+x−δanlogn)nlogn]+1)=0
and
limn→∞e−1n[(1+x+δanlogn)nlogn]+1n([(1+x−δanlogn)nlogn]+1)=0,
whence we obtain (22). Now, coming back to An, by (22) we can say that
An∼nlog1−e−1n[(1+x+δanlogn)nlogn]1−e−1n([(1+x−δanlogn)nlogn]+1);
moreover, by using some manipulations as above for
1−e−1n[(1+x+δanlogn)nlogn]1−e−1n([(1+x−δanlogn)nlogn]+1),
we get
An∼ne−1n([(1+x−δanlogn)nlogn]+1)−e−1n[(1+x+δanlogn)nlogn]=ne−1n[(1+x+δanlogn)nlogn]e−1n([(1+x−δanlogn)nlogn]+1−[(1+x+δanlogn)nlogn])−1.
Then
An∼ne−1n[(1+x+δanlogn)nlogn]×e−1n([(1+x−δanlogn)nlogn]+1−[(1+x+δanlogn)nlogn])−1≥ne−1n(1+x+δanlogn)nlogne−1n((1+x−δanlogn)nlogn+1−((1+x+δanlogn)nlogn−1))−1=n1ne−x+δane−2n+2δan−1=e−x+δane−2n+2δan−1=:A˜n.
Now, by using the same computations as in the proof of Proposition 2.1 in [10], there exists a constant C>00$]]> such that
0≤An(+)≤Cn1+x+δanlognnlogn×e−1n[(1+x+δanlogn)nlogn](1+e−1n[(1+x+δanlogn)nlogn])n−2
and
0≤An(−)≤Cn1+x−δanlognnlogn+1×e−1n([(1+x−δanlogn)nlogn]+1)(1+e−1n([(1+x−δanlogn)nlogn]+1))n−2.
Then, by observing that ann=anlognnlogn→∞,
limn→∞(1+e−1n[(1+x+δanlogn)nlogn])n−2=1
and
limn→∞(1+e−1n([(1+x−δanlogn)nlogn]+1))n−2=1,
we get
0≤lim supn→∞An(+)An≤lim supn→∞C(logn+x+δan)1ne−x+δanA˜n=lim supn→∞C(logn+x+δan)1ne−2n+2δan−1=0
and
0≤lim supn→∞An(−)An≤lim supn→∞C(logn+x−δan+1n)1ne−x−δanA˜n=lim supn→∞C(logn+x−δan+1n)1ne−2δan(e−2n+2δan−1)=0.
Thus the limits in (21) are checked.
□
An example inspired by the replacement model in [5]
In this section we consider the following example.
Let F and G be two continuous distribution functions on R such that F(0)=G(0)=0, and assume that they are strictly increasing on [0,∞). Moreover we assume that, for some t>00$]]>, there exist F′(t−) and G′(t+), i.e. the left derivative of F(x) at x=t and the right derivative of G(x) at x=t, and that F′(t−),G′(t+)>00$]]>. Then let {Zn:n≥1} be a family of random variables defined on the same probability space (Ω,F,P) such that, for some t>00$]]> and β∈(0,1), their distribution functions are defined by
P(Zn≤x):=βF(x)F(t)nifx∈[0,t],1−(1−β)1−G(x)1−G(t)nifx∈(t,∞).
Note that, for every n≥1, the distribution function P(Zn≤·) is continuous. Moreover, if β=F(t), after some easy computations one can check that
P(Z1≤x):=F(x)ifx∈[0,t],F(t)+1−F(t)1−G(t)(G(x)−G(t))ifx∈(t,∞).
Then Z1 is the random lifetime that appears in a replacement model recently studied in the literature (see eq. (5) in [5]) where, at a fixed time t, an item with lifetime distribution function F is replaced by another item having the same age and a lifetime distribution function G.
In general, the distribution functions of the random variables {Zn:n≥1} are suitable modifications of the distribution function of Z1, P(Zn≤t)=β for every n, and the distribution of Zn becomes more and more concentrated around t as n increases.
This example does not seem to have interesting applications to concrete models. However it could give some interesting ideas for the construction of more advanced examples based on other replacement models (see, e.g., the replacement model in [6]). The study of some possible new examples could be the subject of a future work.
Throughout this section we set
Cn:=Zn−tfor alln≥1.
We can recover the asymptotic regimesR1andR2in Assertion1.2as follows.
R1:{Cn:n≥1}converges in probability to zero because{Zn:n≥1}converges in probability to t; in fact, we havelimn→∞P(Zn≤x)=0ifx∈[0,t),βifx=t,1ifx∈(t,∞).Moreover,{Cn:n≥1}satisfies a suitable LDP with speedvn=n, that will be presented in Proposition6.1below.
R2:vnCn=n(Zn−t)converges weakly to a suitable asymmetric Laplace distribution with distribution function H (see below). In fact, we haveP(n(Zn−t)≤x)=PZn≤t+xn=βF(t+xn)F(t)nifx≤0,1−(1−β)1−G(t+xn)1−G(t)nifx>0,=βF(t)+F′(t−)xn+o(1/n)F(t)nifx<0,βifx=0,1−(1−β)1−G(t)−G′(t+)xn+o(1/n)1−G(t)nifx>0,=β1+F′(t−)F(t)xn+o(1/n)nifx<0,βifx=0,1−(1−β)1−G′(t+)1−G(t)xn+o(1/n)nifx>0,0,\end{array}\right.\\ {} & =\left\{\begin{array}{l@{\hskip10.0pt}l}\beta {\left(\frac{F(t)+{F^{\prime }}(t-)\frac{x}{n}+o(1/n)}{F(t)}\right)^{n}}& \hspace{2.5pt}\textit{if}\hspace{2.5pt}x<0,\\ {} \beta & \hspace{2.5pt}\textit{if}\hspace{2.5pt}x=0,\\ {} 1-(1-\beta ){\left(\frac{1-G(t)-{G^{\prime }}(t+)\frac{x}{n}+o(1/n)}{1-G(t)}\right)^{n}}& \hspace{2.5pt}\textit{if}\hspace{2.5pt}x>0,\end{array}\right.\\ {} & =\left\{\begin{array}{l@{\hskip10.0pt}l}\beta {\left(1+\frac{{F^{\prime }}(t-)}{F(t)}\frac{x}{n}+o(1/n)\right)^{n}}& \hspace{2.5pt}\textit{if}\hspace{2.5pt}x<0,\\ {} \beta & \hspace{2.5pt}\textit{if}\hspace{2.5pt}x=0,\\ {} 1-(1-\beta ){\left(1-\frac{{G^{\prime }}(t+)}{1-G(t)}\frac{x}{n}+o(1/n)\right)^{n}}& \hspace{2.5pt}\textit{if}\hspace{2.5pt}x>0,\end{array}\right.\end{aligned}\]]]>and thereforelimn→∞P(n(Zn−t)≤x)=H(x)for everyx∈R,where H is defined byH(x):=βexpF′(t−)F(t)xifx≤0,1−(1−β)exp−G′(t+)1−G(t)xifx>0.0.\end{array}\right.\]]]>So we have: an exponential distribution with mean1−G(t)G′(t+)on(0,∞)with weight1−β; an exponential distribution with meanF(t)F′(t−)on(−∞,0)with weight β.
Now we prove the LDP concerning the convergence to zero in R1 of Assertion 6.1.
Let{Cn:n≥1}be the sequence defined by (23). Then{Cn:n≥1}satisfies the LDP with speed n and rate functionILDdefined byILD(x):=∞ify∈(−∞,−t],−logF(x+t)F(t)ifx∈(−t,0],−log1−G(x+t)1−G(t)ifx∈(0,∞).
We apply Lemma 2.1 with sn=n, I=ILD and {Cn:n≥1} as in (23).
For every x>0 0$]]> we have
lim supn→∞1nlogP(Cn≥x)=lim supn→∞1nlog(1−P(Zn<x+t))=lim supn→∞1nlog(1−β)1−G(x+t)1−G(t)n=log1−G(x+t)1−G(t)=−ILD(x).
For every x<0 we have (when x≤−t we use the convention log0=−∞)
lim supn→∞1nlogP(Cn≤x)=lim supn→∞1nlogP(Zn≤x+t)=lim supn→∞1nlogβF(x+t)F(t)n=logF(x+t)F(t)=−ILD(x).
We want to show that, for every x>−t-t$]]> and for every open set O such that x∈O, we have
lim infn→∞1nlogP(Cn∈O)≥logF(x+t)F(t)ifx∈(−t,0],log1−G(x+t)1−G(t)ifx∈(0,∞).
The case x=0 is immediate; indeed, since Cn converges in probability to zero, we have trivially 0≥0. So, from now on, we suppose that x>−t-t$]]> with x≠0 and we have two cases: x∈(−t,0) and x∈(0,∞).
If x∈(−t,0) we take δ>00$]]> small enough to have
(x−δ,x+δ)⊂O∩(−t,0).
Then
P(Cn∈O)≥P(Cn∈(x−δ,x+δ))=P(Zn<x+δ+t)−P(Zn≤x−δ+t)=βF(x+δ+t)F(t)n−F(x−δ+t)F(t)n=βF(t)(F(x+δ+t)−F(x−δ+t))︸>0×∑k=0n−1F(x+δ+t)F(t)kF(x−δ+t)F(t)n−1−k︸≥F(x+δ+t)F(t)n−1>0, 0}{\underbrace{(F(x+\delta +t)-F(x-\delta +t))}}\\ {} & \hspace{1em}\times \underset{\ge {\left(\frac{F(x+\delta +t)}{F(t)}\right)^{n-1}} > 0}{\underbrace{{\sum \limits_{k=0}^{n-1}}{\left(\frac{F(x+\delta +t)}{F(t)}\right)^{k}}{\left(\frac{F(x-\delta +t)}{F(t)}\right)^{n-1-k}}}},\end{aligned}\]]]>
and therefore
lim infn→∞1nlogP(Cn∈O)≥logF(x+δ+t)F(t).
Then we get (8) for x∈(−t,0) by letting δ go to zero (by the continuity of F).
If x∈(0,∞) we take δ>00$]]> small enough to have
(x−δ,x+δ)⊂O∩(0,∞).
Then
P(Cn∈O)≥P(Cn∈(x−δ,x+δ))=P(Zn<x+δ+t)−P(Zn≤x−δ+t)=1−(1−β)1−G(x+δ+t)1−G(t)n−1−(1−β)1−G(x−δ+t)1−G(t)n=(1−β)1−G(x−δ+t)1−G(t)n−1−G(x+δ+t)1−G(t)n=1−β1−G(t)(G(x+δ+t)−G(x−δ+t))︸>0×∑k=0n−11−G(x−δ+t)1−G(t)k1−G(x+δ+t)1−G(t)n−1−k︸≥1−G(x−δ+t)1−G(t)n−1>0 0}{\underbrace{\frac{1-\beta }{1-G(t)}(G(x+\delta +t)-G(x-\delta +t))}}\\ {} & \hspace{2em}\times \underset{\ge {\left(\frac{1-G(x-\delta +t)}{1-G(t)}\right)^{n-1}}>0}{\underbrace{{\sum \limits_{k=0}^{n-1}}{\left(\frac{1-G(x-\delta +t)}{1-G(t)}\right)^{k}}{\left(\frac{1-G(x+\delta +t)}{1-G(t)}\right)^{n-1-k}}}}\end{aligned}\]]]>
and therefore
lim infn→∞1nlogP(Cn∈O)≥log1−G(x−δ+t)1−G(t).
Then we get (8) for x∈(0,∞) by letting δ go to zero (by the continuity of G).
□
Now we prove the moderate deviation result. We shall see that, for this example, the family of positive numbers {an:n≥1} such that (1) holds (i.e. an→0 and ann→∞) has to verify also the stricter condition
anlogn→0.
Obviously (24) yields an→0.
For every family of positive numbers{an:n≥1}such that (1) and (24) hold (i.e.anlogn→0andann→∞), the sequence of random variables{ann(Zn−t):n≥1}satisfies the LDP with speed1/anand rate functionIMDdefined byIMD(x):=−F′(t−)F(t)xifx≤0,G′(t+)1−G(t)xifx>0.0.\end{array}\right.\]]]>
We apply Lemma 2.1 for every choice of {an:n≥1} such that (1) and (24) hold, with sn=1/an, I=IMD and {Cn:n≥1} as in (23).
For every x>0 0$]]> we have
lim supn→∞anlogP(ann(Zn−t)≥x)=lim supn→∞anlog1−PZn<t+xann=lim supn→∞anlog(1−β)1−Gt+xann1−G(t)n=lim supn→∞annlog1−Gt+xann1−G(t)=lim supn→∞annlog1−G(t)−G′(t+)xann+o(1ann)1−G(t)=lim supn→∞annlog1−G′(t+)1−G(t)xann+o1ann=−G′(t+)1−G(t)x=−IMD(x).
For every x<0 we have (note that t+xann>00$]]> for n large enough)
lim supn→∞anlogP(ann(Zn−t)≤x)=lim supn→∞anlogPZn≤t+xann=lim supn→∞anlogβFt+xannF(t)n=lim supn→∞annlogFt+xannF(t)=lim supn→∞annlogF(t)+F′(t−)xann+o(1/(ann))F(t)=lim supn→∞annlog1+F′(t−)F(t)xann+o1ann=F′(t−)F(t)x=−IMD(x).
We show that, for every x∈R and for every open set O such that x∈O, we have
lim infn→∞anlogP(ann(Zn−t)∈O)≥F′(t−)F(t)xifx≤0,−G′(t+)1−G(t)xifx>0.0.\end{array}\right.\]]]>
The case x=0 is immediate; indeed, since ann(Zn−t) converges in probability to zero by the Slutsky theorem (by an→0 and the weak convergence in R2 in Assertion 6.1), we have trivially 0≥0. So, from now on, we suppose that x≠0 and we have two cases: x∈(−∞,0) and x∈(0,∞).
If x∈(−∞,0) we take δ>00$]]> small enough to have
(x−δ,x+δ)⊂O∩(−∞,0).
Then (note that t+x±δann>00$]]> for n large enough)
P(ann(Zn−t)∈O)≥P(ann(Zn−t)∈(x−δ,x+δ))=PZn<t+x+δann−PZn≤t+x−δann=β(F(t))nFt+x+δannn−Ft+x−δannn=β(F(t))nFt+x+δann−Ft+x−δann︸=F′(t−)2δann+o1ann×∑k=0n−1Ft+x+δannkFt+x−δannn−1−k︸≥Ft+x+δannn−1,
and therefore (in the final step we use the condition anlogn→0 in (24))
lim infn→∞anlogP(ann(Zn−t)∈O)≥lim infn→∞anlogF′(t−)2δann+o1annFt+x+δannF(t)n−1=lim infn→∞anlogF′(t−)2δann+o1ann+an(n−1)logFt+x+δannF(t)=lim infn→∞−anlogn+an(n−1)logF(t)+F′(t−)x+δann+o(1/(ann))F(t)=lim infn→∞−anlogn+an(n−1)log1+F′(t−)F(t)x+δann+o1ann=F′(t−)F(t)(x+δ).
So we obtain (8) for x∈(−∞,0) by letting δ go to zero.
If x∈(0,∞) we take δ>00$]]> small enough to have
(x−δ,x+δ)⊂O∩(0,∞).
Then
P(ann(Zn−t)∈O)≥P(ann(Zn−t)∈(x−δ,x+δ))=PZn<t+x+δann−PZn≤t+x−δann=1−β(1−G(t))n1−Gt+x−δannn−1−Gt+x+δannn=1−β(1−G(t))nGt+x+δann−Gt+x−δann︸=G′(t+)2δann+o1ann×∑k=0n−11−Gt+x−δannk1−Gt+x+δannn−1−k︸≥1−Gt+x−δannn−1,
and therefore (in the final step we use the condition anlogn→0 in (24))
lim infn→∞anlogP(ann(Zn−t)∈O)≥lim infn→∞anlogG′(t+)2δann+o1ann1−Gt+x−δann1−G(t)n−1=lim infn→∞anlogG′(t+)2δann+o1ann+an(n−1)log1−Gt+x−δann1−G(t)=lim infn→∞−anlogn+an(n−1)log1−G(t)−G′(t+)x−δann+o(1ann)1−G(t)=lim infn→∞−anlogn+an(n−1)log1−G′(t+)1−G(t)x−δann+o1ann=−G′(t+)1−G(t)(x−δ).
So we obtain (8) for x∈(0,∞) by letting δ go to zero.
□
Acknowledgement
We thank the referees for their comments.
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