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Exact upper tail probabilities of random series∗ Xiangfeng Yang† Abstract In this paper, we obtain new estimates on upper tail probabilities of suitable random series involving a distribution having an exponential tail. These estimates are exact, and the distribution is not heavy-tailed. Keywords and phrases: exponential tail, random series, upper tail probability, large deviation AMS 2010 subject classifications: 60F10, 60F15 1 Introduction In this paper we consider the random series P∞ j=1 aj ξj with deterministic coefficients aj ∈ R, and real-valued random variables {ξj }j=1,2,... . Such series appeared in literature under the name of linear processes (cf. [14] and [10]), and they are basic objects in time series analysis and in regression models (cf. [4]). Estimates on the upper tail probabilities of P∞ j=1 aj ξj have been an active research topic. In [15], [7] and [12], several estimates were obtained on the upper tail probabilities for suitable random variables, but those estimates are not exact. The first exact upper tail probability was derived in [19] with i.i.d. nonnegative {ξj } having regular variation at infinity, where the coefficients {aj } could be random. This result was later generalized in [8], [9] and [14]. Recently there are several fresh aspects in this research on searching exact upper tail probabilities for other families of random variables such as extended regular variation in [22], intermediate regular variation in [16], and so on. For specific applications of these upper tail probabilities and their related problems such as ruin theory (finite time ruin and ultimate ruin), we refer to [20] and [10]. ∗ † Date: December 25, 2014 [email protected]; Department of Mathematics, Linköping University, SE-581 83 Linköping, Sweden 1 All random variables mentioned above belong to a class which is known as the class of heavytailed distributions. A real-valued random variable ξ is said to be (right) heavy-tailed if (cf. [2]) lim ec·x · P {ξ > x} = ∞, for every c > 0. x→∞ This class of heavy-tailed distributions is somewhat insufficient in many applications. For instance, in the study of weak convergence of sums of products of two independent random variables in [3], distributions which are not heavy-tailed arise. In this paper, we aim to study the exact upper tail P probabilities for random series ∞ j=1 aj ξj with i.i.d. random variables {ξj } which are not heavytailed. More precisely, the distribution of ξj is assumed to have an exponential tail with rate γ ≥ 0 (cf. [3] and [5]): lim x→∞ P {ξj > x − y} = eγy , for every y ∈ R. P {ξj > x} The class of distributions having an exponential tail with rate γ ≥ 0 is denoted by Lγ . The special case L0 with γ = 0 corresponds to a very substantial class of long-tailed distributions which is in the class of heavy-tailed distributions. We thus only focus on Lγ with γ > 0 throughout this paper. The class Lγ with γ > 0 does not contain any heavy-tailed distributions. In fact, if ξ ∈ Lγ , then Z x P {ξ > x} = a(x) exp − b(v)dv , where a(x) → a > 0 and b(x) → γ, as x → ∞; (1.1) 0 see [3]. In Section 2, we formulate the main result (Theorem 2.1) of this paper in the most general possible form. Detailed remarks and examples regarding the assumptions are given as well. A proof is then included in Section 3 based on a convolution convergence result. 2 Statement of the main result Throughout this paper, we consider a random series P∞ j=1 aj ξj with deterministic coefficients aj ∈ R and i.i.d. real-valued random variables ξj ∈ Lγ (for a fixed γ > 0). Without loss of generality, it is always assumed that a1 > 0. The following notations will be used: G(x) := P {a1 ξ1 ≤ x} , Ḡ(x) := 1 − G(x), x ∈ R; k+1 X aj ξj ≤ x , H̄k (x) := 1 − Hk (x), x ∈ R and k ≥ 1; Hk (x) := P j=2 2 Z Mk (y) := ey·x dHk (x) = R k+1 Y E exp {y · aj ξj } , y ∈ R and k = 1, 2, . . . , ∞. j=2 Now we formulate our theorem as follows. Theorem 2.1. The limit lim x→∞ P nP ∞ o a ξ > x j=1 j j P {a1 ξ1 > x} = M∞ (γ/a1 ) (2.1) holds under the following assumptions: P (A0). The series ∞ j=1 aj ξj converges almost surely; (A1). limk→∞ Mk (γ/a1 ) = M∞ (γ/a1 ) in the sense of an infinite product; (A2). Mk (β) ≤ c < ∞ with some β > γ/a1 for all k ≥ 1; o o nP nP ∞ k+1 a ξ > x for each x ∈ R. a ξ > x = P (A3). limk→∞ P j=1 j j j=1 j j Remark 2.1. Upper tail probabilities are also referred to as large deviations (cf. [13] and [6]). Exact upper tail probabilities for random series with general normal (Gaussian) random variables were derived in [13]. We note that a normal random variable does not belong to any class mentioned in this paper. What is more, interesting comparisons on the ratio of two random series with normal random variables were studied in [6]. Exact upper tail probabilities of random series in a weaker sense (namely, after taking logarithm) were obtained for more general random variables other than normal in [1] by using large deviation techniques. Remark 2.2. For a random series X := P∞ j=1 aj ηj with deterministic coefficients {aj } and i.i.d. {ηj } having regular variation, it has been remarked in [14] (see page 1055 therein), as a part of the folklore, that X and η1 have similar behaving tails. Namely, P {X > x} /P {a1 η1 > x} ∼ c(a), as x → ∞ for some positive constant c depending on {aj }. Our result (2.1) totally coincides with this folklore. 2.1 Examples In this section we present examples satisfying all the assumptions in the theorem. Example 1: Let aj = 1/j 2 and {ξj } be i.i.d. exponential distributions with rate 1. In this Q 1 case, γ = 1 and M∞ (1) = ∞ 1 + . j=2 n2 −2 3 Example 2: Let a1 > a2 > a3 ≥ a4 ≥ . . . ≥ 0, and {ξj } be i.i.d. such that ξ1 ∈ Lγ (with P γ > 0), ξ1 ≥ 0 almost surely, and ξ1 has no atoms. If ∞ j=1 aj < ∞, then (2.1) holds. In this case the constant β can be chosen as β = γ2 a11 + a12 (see Section 2.2.3). The assumption that ξ1 has no atoms is for (A3); see Section 2.2.4. The detailed discussion of this example is included in Section 2.2. Example 3: As a comparison with Example 2, if {aj }j≥2 are not nonnegative, then we P γ 1 1 assume that a1 > |a2 | > |a3 | ≥ |a4 | ≥ . . . ≥ 0, with ∞ |a | < ∞ and β = + j=1 j 2 a1 |a2 | . If ξ1 is not nonnegative, then one additional assumption is imposed: E exp {α|ξ1 |} < ∞ for any α < γ. P∞ P∞ P If we rewrite the linear process ∞ j=1 |aj | < j=1 aj ξn−j , then the condition j=1 aj ξj as Xn = ∞ corresponds to a short memory process. It is also possible to associate a long memory process P 2 with ∞ j=1 aj < ∞. To produce such an example, we need one additional assumption Eξj = 0 on the random variables ξj . The reason is that we need to prove, in order to check the assumption (A1), the following convergence as explained in (2.2) below, ∞ X γaj j=2 a1 1 · Eξj + 2 γaj a1 2 E ξj2 exp γaj · θj · ξj a1 < ∞. P∞ γaj If Eξj = 0 is assumed, then the first term j=2 a1 · Eξj = 0. This suggests that there is no P∞ need to make the assumption j=1 |aj | < ∞, and it is enough to impose a weaker assumption P∞ 2 j=1 aj < ∞. We refer to [21] and [18] for some recent results on large (or moderate) deviations for linear processes with long range dependence. 2.2 2.2.1 Analysis of the assumptions (A0). The series P∞ j=1 aj ξj converges almost surely From Kolmogorov’s three-series theorem, this assumption is equivalent to (A0.1)+(A0.2)+(A0.3): P (A0.1) ∞ j=1 P {|aj ξj | > 1} < ∞; P (A0.2) The series ∞ j=1 E aj ξj 1{|aj ξj |≤1} is convergent; P (A0.3) ∞ V ar a ξ 1 < ∞. j j {|a ξ |≤1} j=1 j j P∞ min(2,p) < ∞ if For the convergence of these three series, we may simply assume that j=1 |aj | ξj ∈ Lp with p > 0. Note that ξj ∈ Lp is not an assumption in this paper. If ξj ∈ Lγ and ξj ≥ 0 almost surely, then all its moments are finite. Actually in this case E exp {αξj } < ∞ for any α < γ, P 2 but E exp {γξj } may or may not be finite; see [3]. It then follows that the condition ∞ j=1 aj < ∞ 4 is good enough to guarantee the convergence of the series since ξj has all the moments. 2.2.2 (A1). limk→∞ Mk (γ/a1 ) = M∞ (γ/a1 ) in the sense of an infinite product An infinite product Q∞ n=1 cn [11]) if the limit limk→∞ with positive elements cn is said to be convergent (see page 1413 in Qk n=1 cn = c∞ exists and 0 < c∞ < ∞. Since the essential part of a distribution with an exponential tail is the behavior near positive infinity, for simplicity we assume that ξj ≥ 0 almost surely. As mentioned before, in this case E exp {αξj } < ∞ for any α < γ. For the same reason we assume all coefficients aj ≥ 0. What is more, we arrange these coefficients as P follows: a1 > a2 ≥ a3 ≥ . . . ≥ 0. We now claim that if ∞ j=1 aj < ∞, then (A1) holds. To prove the claim, we recall the definition of Mk (γ/a1 ) : Z Mk (γ/a1 ) = e(γ/a1 )·x dHk (x) = R k+1 Y E exp {(γ/a1 ) · aj ξj } . j=2 It is clear that Mk (γ/a1 ) ≥ 1 for every k ≥ 1. From Taylor’s expansion, it follows γaj γaj 1 γaj 2 2 E exp {(γ/a1 ) · aj ξj } = 1 + · Eξj + E ξj exp · θj · ξj a1 2 a1 a1 2 h oi n γa γa γa for some random θj ∈ [0, 1]. Since a1j · Eξj + 21 a1j E ξj2 exp a1j · θj · ξj ≥ 0, the convergence Q∞ of j=2 E exp {(γ/a1 ) · aj ξj } is equivalent to ∞ X γaj j=2 a1 1 · Eξj + 2 γaj a1 2 E ξj2 exp γaj · θj · ξj a1 < ∞. (2.2) The convergence of (2.2) follows from the conditions at the beginning of this section by noticing the following facts: Eξj ≤ c1 E exp {c2 ξj } < ∞, for some c1 > 0 and c2 < γ; γaj 2 E ξj exp · θj · ξj ≤ c3 E exp {c4 ξj } < ∞, for some c3 > 0 and c4 < γ. a1 2.2.3 (A2). Mk (β) ≤ c < ∞, β > γ/a1 , k ≥ 1 Without loss of generality, let a1 > |a2 | > |a3 | ≥ . . . ≥ 0. If we impose that E exp {α|ξj |} < ∞ for any α < γ, and ∞ Y j=2 E exp γ 1 1 + aj ξj < ∞, 2 a1 |a2 | 5 (2.3) then (A2) is satisfied. Indeed, in this case β can be chosen as β = γ2 a11 + |a12 | . In order to prove P (2.3) by using the arguments in the previous section, we assume ∞ j=1 |aj | < ∞. 2.2.4 (A3). limk→∞ P o o nP ∞ a ξ > x , x∈R a ξ > x = P j=1 j j j=1 j j nP k+1 Based on (A0), the almost surely convergence implies the convergence in distribution. Thus (A3) o nP ∞ a ξ > x is continuous at x. If ξ1 has no atoms is satisfied for every point x such that P j j j=1 o nP ∞ a ξ > x is continuous at every x. This can (i.e. P {ξ1 = x} = 0 for every x ∈ R), then P j=1 j j be easily seen from the following well-known fact. Lemma 2.3. Suppose µ and ν are two probability distributions on R. If µ has no atoms, then the convolution µ ∗ ν has no atoms. P Here we can take µ as the distribution of a1 ξ1 , and ν as the distribution of ∞ j=2 aj ξj . It is not P∞ necessary that the random variable j=1 aj ξj has a density in this case. If we further assume that P ξ1 has a density, then ∞ j=1 aj ξj also has a density, which is from another well-known fact. Lemma 2.4. Suppose µ and ν are two probability distributions on R. If µ has a density, then the convolution µ ∗ ν has a density. 3 Proof of Theorem 2.1 Before we outline the proof, let us first establish two auxiliary results based on (A0)-(A3). The first result depends only on (A0). Lemma 3.1. Under the assumption (A0), we have limx→∞ P nP k+1 j=2 o aj ξj ≥ x = 0 uniformly in k ≥ 1. This lemma easily follows from the fact that the family { Pk+1 j=2 aj ξj }k≥1 is tight. The second result plays an important role in the proof of Theorem 2.1. Lemma 3.2. Under the assumption (A2), we have H̄k (x) = o(Ḡ(x)) as x → ∞ uniformly in k ≥ 1. Proof. It follows from ξ1 ∈ Lγ and (1.1) that ( Z Ḡ(x) = P {a1 ξ1 > x} = a(x/a1 ) exp − 0 6 ) x/a1 b(v)dv where a(x) → a > 0 and b(x) → γ as x → ∞. Thus, for large enough x, Ḡ(x) ≥ a · exp {−b · x} 2 for some 0 < b < β. Such b exists because β > γ/a1 in (A2). We then apply this inequality and Chebyshev’s inequality to get H̄k (x) = Ḡ(x) ≤ 3.1 P o a ξ > x j=2 j j nP k+1 Ḡ(x) ≤ c Mk (β) ≤ βx · Ḡ(x) e · Ḡ(x) eβx 2c · exp{−(β − b)x} → 0, as x → ∞. a Outline of the proof Because of the independence, the finite sums are associated with the corresponding convolutions. We first prove (2.1) for the finite sums by using an important lemma in [17], and then show that the convergence is uniform in k ≥ 1. The proof will be complete by switching the limits between x → ∞ and k → ∞, and taking into account our two auxiliary results Lemmas 3.1 and 3.2. Let us now recall another lemma regarding the convolution of two distributions. Lemma 3.3 (Lemma 2.1 in [17]). Let γ > 0 and G be defined in Section 2. Suppose H(x) is a distriR bution function with H̄(x) := 1−H(x) satisfying H̄(x) = o(Ḡ(x)) and MH (β) := R eβx dH(x) < ∞ for some β > γ/a1 . Then F̄ (x) := 1 − F (x) = 1 − G ∗ H(x) satisfies F̄ (x)/Ḡ(x) → MH (γ/a1 ). Lemmas 3.2 and 3.3 yield lim P x→∞ nP k+1 o a ξ > x j=1 j j P {a1 ξ1 > x} = Mk (γ/a1 ). By taking limit k → ∞ on both sides of (3.1), it follows from (A1) that nP o k+1 P a ξ > x j=1 j j lim lim = M∞ (γ/a1 ). k→∞ x→∞ P {a1 ξ1 > x} (3.1) (3.2) If we can switch the limits limk→∞ limx→∞ in (3.2) as limx→∞ limk→∞ , then Theorem 2.1 is proved with the help of (A3). The switch of the limits is feasible if the following uniform convergence is obtained lim x→∞ P nP k+1 j=1 o aj ξj > x P {a1 ξ1 > x} = Mk (γ/a1 ), uniformly in k ≥ 1. Lemma 3.1 plays an important part in proving (3.3). 7 (3.3) 3.2 Uniform convergence By definition, P o a ξ > x j=1 j j nP k+1 1 = Ḡ(x) P {a1 ξ1 > x} Z ∞ Ḡ(x − y)dHk (y). −∞ Since ξj ∈ Lγ , it is straightforward to have lim x→∞ Ḡ(x − y) = eγy/a1 Ḡ(x) for every y ∈ R. For a fixed y 0 > 0, this convergence is uniform for y ≤ y 0 since γ > 0. Taking into account this fact and Lemma 3.2, we thus break the integral into three parts as follows: 1 Ḡ(x) Z ∞ Ḡ(x − y)dHk (y) −∞ 1 = Ḡ(x) 3.2.1 Z y0 1 Ḡ(x − y)dHk (y) + Ḡ(x) −∞ Z x−y 0 y0 1 Ḡ(x − y)dHk (y) + Ḡ(x) Z ∞ x−y 0 Ḡ(x − y)dHk (y) The part on (−∞, y 0 ] On the first part, 1 Ḡ(x) Z y0 Z y0 eγy/a1 dHk (y) uniformly in k ≥ 1. Ḡ(x − y)dHk (y) → −∞ −∞ R y0 The uniform convergence on this part will be proved if we can show that −∞ eγy/a1 dHk (y) tends R∞ to −∞ eγy/a1 dHk (y) uniformly in k as y 0 → ∞. This is indeed the case since Z Z Z y0 ∞ ∞ γy/a1 γy/a1 e dHk (y) − e dHk (y) = eγy/a1 dHk (y) −∞ 0 −∞ y k+1 γ X aj ξj = E 1{Pk+1 aj ξj ≥y0 } exp j=2 a1 j=2 1/q k+1 X ≤ P aj ξj ≥ y 0 · (Mk (β))1/p j=2 where p = a1 β/γ > 1 and 1/p+1/q = 1. Lemma 3.1 together with (A2) for the uniform boundedness of Mk (β) imply the uniform convergence in k. 8 3.2.2 The part on [y 0 , x − y 0 ] Noticing that on this part, we have x − y ≥ y 0 and thus can choose a sufficiently large y 0 such that 0 Ḡ(x − y)/Ḡ(x) ≤ 2eβ y for some β 0 ∈ (γ/a1 , β). Therefore, 1 Ḡ(x) Z x−y 0 Z x−y 0 0 eβ y dHk (y) y0 y0 k+1 X = 2E 1{y0 ≤Pk+1 aj ξj ≤x−y0 } exp β 0 aj ξj j=2 Ḡ(x − y)dHk (y) ≤ 2 j=2 1/q k+1 X 0 0 ≤ P y ≤ a j ξj ≤ x − y · (Mk (β))1/p j=2 1/q k+1 X ≤ P y 0 ≤ aj ξj · (Mk (β))1/p j=2 with p = β/β 0 > 1 and 1/p + 1/q = 1, which approaches zero as y 0 → ∞ uniformly in k because of Lemma 3.1 and (A2). 3.2.3 The part on [x − y 0 , ∞) The function Ḡ ≤ 1, so 1 Ḡ(x) Z ∞ x−y 0 Ḡ(x − y)dHk (y) ≤ H̄k (x − y 0 ) H̄k (x − y 0 ) Ḡ(x − y 0 ) = · Ḡ(x) Ḡ(x − y 0 ) Ḡ(x) which tends to zero as x → ∞ uniformly in k by Lemma 3.2. Acknowledgment. I am grateful to the Editor and an anonymous referee for useful comments. References [1] Arcones, M., 2004. 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