Estimating the logarithm of characteristic function and stability parameter for symmetric stable laws
EEstimating the logarithm of characteristic functionand stability parameter for symmetric stable laws
J¨uri Lember ∗ Institute of Mathematics and Statistics, University of Tartu, Estonia,[email protected]
Annika Krutto † Institute of Mathematics and Statistics, University of Tartu, EstoniaDepartment of Biostatistics, University of Oslo, Norway,[email protected]
Abstract
Let X , . . . , X n be an i.i.d. sample from symmetric stable distribution withstability parameter α and scale parameter γ . Let ϕ n be the empirical character-istic function. We prove an uniform large deviation inequality: given preciseness (cid:15) > p ∈ (0 , (cid:15) and p butnot depending on α and γ ) constant ¯ r > P (cid:0) sup u> r ( u ) ≤ ¯ r | r ( u ) − ˆ r ( u ) | ≥ (cid:15) (cid:1) ≤ p, where r ( u ) = ( uγ ) α and ˆ r ( u ) = − ln | ϕ n ( u ) | . As an applications of the result, weshow how it can be used in estimation unknown stability parameter α . Keywords : Stable laws, large deviation inequalities, parameter estimation;
MSC codes : Stable laws 60E07; Confidence regions 62F25 ∗ Estonian institutional research funding IUT34-5, Estonian Research Council grant PRG865 † Estonian institutional research funding IUT34-5 a r X i v : . [ m a t h . S T ] A ug Introduction and preliminaries
Let X , . . . , X n be an i.i.d. sample from stable law with characteristic function ϕ ( u ) =exp [ − ( γ | u | ) α + iω ( u )], where ω ( u ) = u [ βγ tan πα ( | γu | α − −
1) + δ ] for α (cid:54) = 1 and ω ( u ) = u [ − βγ π ln( γ | u | ) + δ ] for α = 1, and α ∈ (0 , β ∈ [ − , γ > δ ∈ R are (unknown) stability, skewness, scale and shift parameters, respectively. Basicproperties of stable distributions can be found in [15, 13, 9]. Let F n be the empiricaldistribution function, and ϕ n the empirical characteristic function, i.e. ϕ n ( u ) = (cid:90) R exp { iux } d F n ( x ) , u ∈ R . (1.1)Let r ( u ) = − ln | ϕ ( u ) | = ( γ | u | ) α and ˆ r ( u ) = − ln | ϕ n ( u ) | . Estimating the parametersof stable law is a notoriously hard problem (see, e.g., [8, Section 2], [15, Chapter 4]).Simple empirical characteristic function based closed form estimates were proposed in[11]. In particular, the stability parameter estimator isˆ α = ln ˆ r ( u ) − ln ˆ r ( u )ln u − ln u , (1.2)where 0 < u < u are fixed arguments and ˆ r ( u ) = − ln | ϕ n ( u ) | . Since for any u ,ˆ r ( u ) → ( γ | u | ) α , a.s., we see that any choice of 0 < u < u gives consistent estimator.The same points 0 < u < u can be used to give consistent closed form estimatorsalso to other parameters γ , β , δ of stable law (see [5, Theorem 1]). Despite theasymptotic consistency holds for any pair u , u , in practice the right choice of u and u is crucial and the universal selection of these values has remained unsolved (e.g.,[10, 1, 3]). Recently, [6] suggests that the choice of u and u should be fixed based onthe preciseness of ˆ α , that is, on the preciseness of ˆ r ( u ). Clearly, the preciseness of ˆ α depends on how well ˆ r ( u i ) and estimates r ( u i ) for i = 1 , . . . . . . u (a) a =0.2 0.0 0.1 0.2 0.3 0.4 0.5 . . . . . . u (b) a =1 0.0 0.1 0.2 0.3 0.4 0.5 . . . . . . u (c) a =1.8r ( u ) r^ ( u ) for n=10 r^ ( u ) for n=100 r^ ( u ) for n=1000 Figure 1: The values of r ( u ) = u α vs ˆ r ( u ) of single replicates (simulated with [12]) ofstable law with γ = 1, δ = 0, β = 0 for α = 0 . α = 1 in (b) and α = 1 . u (a) a =0.2 0 1 2 3 4 5 6 7 u (b) a =1 0 1 2 3 4 5 6 7 u (c) a =1.8r ( u ) r^ ( u ) for n=10 r^ ( u ) for n=100 r^ ( u ) for n=1000 Figure 2: The values of r ( u ) = u α vs ˆ r ( u ) of single replicates (simulated with [12]) ofstable law with γ = 1, δ = 0, β = 0 for α = 0 . α = 1 in (b) and α = 1 . r ( u ) is relatively accurate estimate of r ( u ) only in asmall interval (0 , ¯ u ], where ¯ u obviously depends on sample size n , but unfortunatelyalso on α – the smaller α , the smaller also ¯ u . So, there seems not to exist an universal(that applies for any α ∈ (0 , u so that sup u ∈ (0 , ¯ u ] | ˆ r ( u ) − r ( u ) | were smallfor any α even when γ > u makes the estimate of α very imprecise. Observe that 0 < u < u cannot also be very small, because then ln u − ln u is also very small and that affectsthe preciseness of ˆ α even when ˆ r ( u i ) ≈ r ( u i ), i = 1 ,
2. On the other hand, Figure 1 andFigure 2 as well as simulations in [6] suggest that despite the possible non-existenceof universal ¯ u , there might exists an universal (not depending on α and γ ) ¯ r so thatsup
Let X , . . . , X n be an i.i.d. sample from symmetric stable law. Fix (cid:15) > , p ∈ (0 , . Then there exists n o ( (cid:15), p ) < ∞ such that for every α ∈ (0 , and n > n o P (cid:16) sup (cid:15) (cid:17) ≤ p, (1.3) where ¯ r ( (cid:15), p, n ) > is independent of α and γ . Theorem is proved in Subsection 2. From Theorem 1.1 it follows that with probability1 − p , | r ( u i ) − ˆ r ( u i ) | ≤ (cid:15) , given n is big enough and u i is cosen such that r ( u i ) ≤ ¯ r , i = 1 ,
2. Therefore, in order to apply (1.2), it makes sense not to fix arguments u and For mathematical tractability in formulas, in particular in tail estimation in (1.7) and (2.3), weprovide our results for symmetric stable laws. Similar construction of proof can be applied for generalstable laws. , but the values ˆ r ( u ) and ˆ r ( u ) instead. The obtained estimate is thenˆ α = ln(¯ r − (cid:15) ) − ln( r + (cid:15) )ln u − ln u , (1.4)where, (cid:15) , ¯ r and r are carefully chosen constants and u = inf { u : ˆ r ( u ) = ¯ r − (cid:15) } , u = sup { u : ˆ r ( u ) = r + (cid:15) } . (1.5)The estimates of parameters α, β, γ, δ at u and u based on ¯ r − (cid:15) = 0 . r + (cid:15) =0 . u and u as in (1.2) deserves moreattention in the recent literature and specifying u and u via ˆ r function as in (1.5)is not the only option. In [1], u = 1 and u is taken such that the distance betweenCauchy ( α = 1) and Gaussian ( α = 2) characteristic functions at u were maximal. Theidea of maximizing the distance between characteristic functions is further developed in[3], where an iterative 8-step algorithm for specifying u and u is proposed. Althoughthe idea of maximizing discrepancy between characteristic functions (or rather between r -functions) is quite natural, and the simulations in [3] show good behaviour of thatchoice in practice, the proposed algorithm in [3] is still ad hoc in nature and lackstheoretical justification. In particular, it is not clear that it allows to choose u and u so that the inequality (1.6) below holds.The proof of the Theorem 1.1 is constructive, but the goal of it is to show that universal¯ r exists, not to optimize the constant, i.e. to find the biggest possible ¯ r and smallestpossible n o . It means that the ¯ r constructed in the proof is probably too small forpractical use. Although, for every u , γ and α , ˆ r ( u ) → r ( u ), a.s. as n grows, our upperbound satisfies ¯ r ( (cid:15), p, n ) ≤ w o <
1, where w o is a constant depending on (cid:15) . Thus ¯ r isalways bounded away from 1 and does not increase to the infinity as n grows. Thisneed not necessarily be the deficiency of the proof, rather than necessary property. Tosee that, assume the inequality (1.3) holds with some ¯ r >
1. Then it follows that P (cid:16) sup u ∈ (0 , ¯ u ] | r ( u ) − ˆ r ( u ) | > (cid:15) (cid:17) ≤ p, where ¯ u = (cid:113) ¯ rγ , so that the universal upper bound (that applies for any α ) ¯ u wouldexists. However, there is no evidence at all that such an ¯ u exists, suggesting that thea.s. convergence ˆ r ( u ) → r ( u ) is not uniform over small α – for every u and n thereexists α small enough so that the difference | r ( u ) − ˆ r ( u ) | is still big. If so, then ¯ r mustalways remain smaller than 1. The situation is different, when we bound the unknownstability parameters α below from zero, i.e. we assume the existence of α > α belongs to [ α, r ( (cid:15), p, n, α ) satisfies lim n ¯ r ( (cid:15), p, n, α ) = ∞ (Corollary 2.1), andin this case also the upper bound ¯ u exists. The existence of α is common assumption4n practice (e.g., [7, 4, 8] suggest α = 0 .
5) and we keep this additional assumptionin Theorem 2.1, that provides a uniform bound similar to (1.3) to the difference oflogarithms | ln ˆ r ( u ) − ln r ( u ) | . Theorem 2.1 is actually a simple corollary of Theorem1.1, but the additional assumption about the existence of α is necessary, because if r is very small, then | ln ˆ r − ln r | can be rather big even when r is very close to ˆ r .Section 3 is devoted to the applications of Theorem 2.1 in the light of large devia-tion inequalities for ˆ α . We show how the upper bound from Theorem 2.1 can be usedto solve the two basic questions related with estimate ˆ α in (1.4): • Given precision (cid:15) >
0, probability p and lower bound α , find ¯ r, r, (cid:15) (needed toconstruct ˆ α in (1.4)) and possibly small sample size n so that P (cid:0) | ˆ α − α | > (cid:15) (cid:1) ≤ p. (1.6) • Given sample size n and α , find ¯ r, r, (cid:15) and possibly small (cid:15) so that (1.6) holds.In other words, find exact i.e. non-asymptotic confidence interval to α .The solutions of these questions are formulated as Theorem 3.1 and Theorem 3.2.The bound in Theorem 2.1 is constructed using the basic bound ¯ r ( (cid:15), p, n ) providedby Theorem 1.1. Thus the (cid:15) and required sample size n in the inequality (1.6) dependheavily on the function ¯ r . Unfortunately, the function ¯ r constructed in Section 2 isnot explicitly given and, hence, difficult to work with. Although the main goal of thepresent paper is just to show that the function ¯ r exists, in Section 4, we discuss an-other possibility to construct ¯ r . The new construction gives analytically more tractablebound, the price for it is bigger minimal required sample size n o and lower bound. Alsothe upper bound ¯ r constructed in Section 4 is also strictly smaller than 1 for every n .So we have two different constructions with the same property, and this allows us toconjecture that even the best bound ¯ r is always smaller than 1 and we also conjecturethat the universal upper bound ¯ u does not exist. Preliminaries.
For 0 < α < F ( t ) ∼ c α γ α (1 − β ) | t | − α , when t → −∞ , − F ( t ) ∼ c α γ α (1 + β ) t − α , when t → ∞ , where c α = Γ( α ) π sin πα ≤ , lim α → c α = , and lim α → c a = 0 . For symmetric stablelaws ( β = 0) these results imply the existence of constants L ( α ) so that F ( t ) ≤ L ( α ) (cid:18) | t | γ (cid:19) − α , ∀ t < , (1 − F ( t )) ≤ L ( α ) (cid:18) tγ (cid:19) − α , ∀ t > , (1.7) For general stable laws it implies for the exitsence of constants L ( α, β ) = L ( a )(1 − β ) and L ( α, β ) = L ( a )(1 + β ) with(1 − β ) ∈ [ − ,
0] and (1 + β ) ∈ [0 , t (cid:55)→ F ( t ) | t | α for different α -values and γ = 1 in the range[ − ,
0] and [ − , L < ∞ so that sup α ∈ (0 , L ( α ) ≤ L . Throughout the paper, we keep L undetermined, although one can take it as . It is also obvious that L is independentof γ . −5 −4 −3 −2 −1 0 . . . . . t(a) a . . . . . . t(b) a Figure 3: Plotting t (cid:55)→ F ( t ) | t | α (calculated with [12])for different α -values with γ =1 , delta = 0 , β = 0 in the range [ − ,
0] and [ − ,
0] .In what follows, we shall use the following elementary inequalities: for any x, y > | ln x − ln y | ≤ | x − y | x ∧ y = | x − y | x ∨ | x − y | x − | x − y | , we obtain that | ln x − ln y | > (cid:15) implies | x − y | x > (cid:15) (cid:15) . Thus, for any ¯ u > (cid:15) > P (cid:0) sup u ∈ (0 , ¯ u ] | r ( u ) − ˆ r ( u ) | > (cid:15) (cid:1) ≤ P (cid:16) sup u ∈ (0 , ¯ u ] e r (¯ u ) | ϕ ( u ) − ϕ n ( u ) | > (cid:15) (cid:15) (cid:17) . (1.8)Observe that (1.8) holds for any estimate ϕ n . Bounding the difference of characteristic functions.
Recall ϕ n is the standardempirical estimate given by (1.1). To bound | ϕ ( u ) − ϕ n ( u ) | we use the approach in [2]6s follows. For every 0 < K < ∞ , | ϕ n ( u ) − ϕ ( u ) | = | (cid:90) e iux d ( F n ( x ) − F ( x )) | ≤ | (cid:90) − K −∞ e iux d ( F n ( x ) − F ( x )) | + | (cid:90) K − K e iux d ( F n ( x ) − F ( x )) | + | (cid:90) ∞ K e iux d ( F n ( x ) − F ( x )) | . Since | e iux | = 1 for every u and x , the first term can be bounded | (cid:90) − K −∞ e iux d ( F n ( x ) − F ( x )) | ≤ (cid:90) − K −∞ | e iux | dF n ( x ) + (cid:90) − K −∞ | e iux | dF ( x ) ≤ F n ( − K ) + F ( − K ) ≤ (cid:107) F n − F (cid:107) + 2 F ( − K ) , where (cid:107) F n − F (cid:107) := sup x | F n ( x ) − F ( x ) | . The last inequality holds, because F n ( − K ) ≤(cid:107) F n − F (cid:107) + F ( − K ). Similarly, the third term can be estimated above by (cid:107) F n − F (cid:107) +2(1 − F ( K )). To estimate the second term, we use the integration by parts | (cid:90) K − K e iux d ( F n ( x ) − F ( x )) | = (cid:12)(cid:12)(cid:12) e iux ( F n ( x ) − F ( x )) | K − K − iu (cid:90) K − K e iux ( F n ( x ) − F ( x )) dx (cid:12)(cid:12)(cid:12) ≤ | F n ( − K ) − F ( − K ) | + | F n ( K ) − F ( K ) | + | u | (cid:90) K − K | F n ( x ) − F ( x ) | dx ≤ (cid:107) F n − F (cid:107) + | u | · (cid:107) F n − F (cid:107) · K. Therefore, for any K | ϕ n ( u ) − ϕ ( u ) | ≤ (cid:107) F n − F (cid:107) + | u | · (cid:107) F n − F (cid:107) · K + 2 F ( − K ) + 2(1 − F ( K )) . (2.1)For any δ >
0, let K ( δ ) be so big that F ( − K ) ≤ δ . Then also 1 − F ( K ) ≤ δ and (2.1)implies | ϕ n ( u ) − ϕ ( u ) | ≤ δ (cid:107) F n − F (cid:107) + | u | · (cid:107) F n − F (cid:107) · K ( δ ) . (2.2)By (1.7), we can take K ( δ ) = (cid:16) Lδ (cid:17) α γ (2.3)and so with u >
0, by (2.2) and (2.3) | ϕ n ( u ) − ϕ ( u ) | ≤ δ (cid:107) F n − F (cid:107) (cid:16) (cid:0) Lr ( u ) δ (cid:1) α (cid:17) . (2.4) Bounding r . Recall ¯ r = r (¯ u ). Fix ¯ u > (cid:15) = (cid:15)e ¯ r . We now use Dworetzky-Kiefer-Wolfowitz inequality [14, p. 268]): P ( (cid:107) F n − F (cid:107) > (cid:15) ) ≤ − n(cid:15) ]7o estimate P (cid:16) e r (¯ u ) sup u ≤ ¯ u | ϕ n ( u ) − ϕ ( u ) | > (cid:15) (cid:17) ≤ P (cid:16) (cid:107) F n − F (cid:107) ≥ (¯ (cid:15) − δ/ δ α (cid:0) δ α + (8 L ¯ r ) α (cid:1) (cid:17) = P (cid:16) (cid:107) F n − F (cid:107) ≥ α (¯ (cid:15) − δ/ δ ) α (cid:0) α ( δ ) α + (8 L ¯ r ) α (cid:1) (cid:17) = P (cid:16) (cid:107) F n − F (cid:107) ≥ (¯ (cid:15) − δ/ δ ) α (cid:0) δ ) α + (4 L ¯ r ) α (cid:1) (cid:17) ≤ (cid:104) − n · (cid:16) (¯ (cid:15) − δ/ δ ) α ( δ ) α + (4 L ¯ r ) α (cid:17) (cid:105) . Define k ( α, (cid:15), ¯ r ) = 18 (cid:104) max ≤ x ≤ ¯ (cid:15) (¯ (cid:15) − x )1 + (cid:0) L ¯ rx (cid:1) α (cid:105) . From (1.8), we obtain P (cid:0) sup u ∈ (0 , ¯ u ] | r ( u ) − ˆ r ( u ) | > (cid:15) (cid:1) ≤ − nk (cid:0) α, (cid:15) (cid:15) , ¯ r (cid:1) ] = p which is equivalent to k (cid:0) α, (cid:15) (cid:15) , ¯ r (cid:1) = ln(2 /p ) n , and so the desired upper bound for any α ,denoted by r n ( α ) is the solution of the following equality k (cid:0) α, (cid:15) (cid:15) , r n (cid:1) = ln(2 /p ) n . (2.5)Observe that lim r → k (cid:0) α, (cid:15) (cid:15) , r (cid:1) = 18 (cid:0) (cid:15) (cid:15) (cid:1) . Hence the following condition gives a lower bound for minimal sample size n so that r n ( α ) > n > /p ) (cid:0) (cid:15)(cid:15) (cid:1) . (2.6) The existence of ¯ r = inf α ∈ (0 , r n ( α ) . The following lemma shows that inf α ∈ (0 , r n ( α ) >
0, hence the universal (not depending on α and γ ) bound ¯ r exists. Since (cid:15)/ (1 + (cid:15) ) < (cid:15) ∈ (0 , Lemma 2.1.
Fix > (cid:15) > and n such that n > ln(2 /p ) (cid:15) . Let r n ( α ) be the solution ofthe equality k ( α, (cid:15), r ) = ln(2 /p ) n . Then r n ( α ) is continuous strictly positive function on (0 , and lim α → r n ( α ) > . In particular, ¯ r = inf α ∈ (0 , r n ( α ) > . roof. For every α > (cid:15) >
0, define function h ( r, x ) = ( (cid:15)e − r − x )1 + ( Lrx ) α , x ∈ (0 , (cid:15)e − r ] , r > . Let us fix r > c = 4 Lr . Let x (cid:48) ( (cid:15), r ) = arg max ≤ x ≤ (cid:15)e − r ( (cid:15)e − r − x )1 + (cid:0) cx (cid:1) α = arg max ≤ x ≤ (cid:15)e − r h ( x, r ) . Since x (cid:55)→ h ( x, r ) is continuous and strictly decreasing function, the maximizer x (cid:48) existsand is unique. It is not difficult to see that x (cid:48) must satisfy the following equalities:( (cid:15)e − r − x (cid:48) )1 + ( cx (cid:48) ) α = (cid:15)e − r − (1 + α ) x (cid:48) ⇔ αc α ( x (cid:48) ) α = (cid:15)e − r − (1 + α ) x (cid:48) . The inequality in the left implies that k ( α, (cid:15), r ) = 18 (cid:104) sup x ∈ (0 ,(cid:15)e − r ] h ( x, r ) (cid:105) = 18 (cid:0) (cid:15)e − r − (1 + α ) x (cid:48) (cid:1) . (2.7)The equality in the right implies x (cid:48) < (cid:15)e − r α and is equivalent to x (cid:48) (cid:0) αc α ( x (cid:48) ) α + (1 + α ) (cid:1) = (cid:15)e − r . Hence we obtain (cid:15)e − r (cid:16) αc α (cid:18) (cid:15)e − r α (cid:19) α + (1 + α ) (cid:17) − < x (cid:48) < (cid:15)e − r α . (2.8)Define function f ( x ) = x (cid:16) α (cid:16) xc (cid:17) α + (1 + α ) (cid:17) , x ∈ (0 , (cid:15)e − r ] . (2.9)Thus x (cid:48) is the solution of the equality f ( x ) = (cid:15)e − r .Suppose now α m → α o >
0. Let f m and f o be as f with α m and α o instead of α , respectively and let x m and x o be the solutions of the equalities f m ( x ) = (cid:15)e − r and f o ( x ) = (cid:15)e − r . Clearly, for any x ∈ (0 , (cid:15)e − r ], it holds f m ( x ) → f o ( x ). However,since α o >
0, by (2.8) we see that there exists y > x m ∈ ( y, (cid:15)e − r ] even-tually since obviously sup x ∈ [ y,(cid:15)e − r ] | f m ( x ) − f ( x ) | →
0, we obtain that x m → x o andby (2.7), thus k ( α m , (cid:15), r ) → k ( α , (cid:15), r ). Now observe that for any α > (cid:15) > r → k ( α, (cid:15), r ) = (cid:15) and lim r →∞ k ( α, (cid:15), r ) = 0. Moreover r (cid:55)→ k ( α, (cid:15), r ) is strictlydecreasing function. For such functions, pointwise convergence implies uniform con-vergence, so that as m growssup r ≥ | k ( α m , (cid:15), r ) − k ( α , (cid:15), r ) | → . (2.10)The uniform convergence implies that the solutions of the equalities k ( α m , (cid:15), r ) =ln(2 /p ) /n converge as well, i.e. r n ( α m ) → r n ( α o ), provided n > ln(2 /p ) (cid:15) so that the9olutions exists.We have proven that the function r n ( α ) is continuous on the set (0 , α m →
0. From the equalities f m ( x m ) = (cid:15)e − r , it follows thatlim m x m = (cid:26) c, when c ≤ (cid:15)e − r ; (cid:15)e − r , when c > (cid:15)e − r ,where c = 4 Lr . Therefore, for every fixed r > k ( α m , (cid:15), r ) = 18 (cid:0) (cid:15)e − r − (1 + α m ) x m ) → k (0 , (cid:15), r )with k (0 , (cid:15), r ) = (cid:26) ( c − (cid:15)e − r ) , if c ≤ (cid:15)e − r ⇔ r ≤ W ( (cid:15) L );0 , else,where W stands for Lambert W -function. Observe that r (cid:55)→ k (0 , (cid:15), r ) is strictly de-creasing function with limits lim r → k (0 , (cid:15), r ) = (cid:15) and lim r →∞ k (0 , (cid:15), r ) = 0. Hence(2.10) holds with α o = 0. This, in turn, implies that r n ( α m ) → r n (0), where r n (0)is the solution of the equality k (0 , (cid:15), r n (0)) = ln(2 /p ) n . Since lim r → k (0 , (cid:15), r ) = (cid:15) , andby assumption (cid:15) > ln(2 /p ) n , we see that r n (0) >
0. Hence r n ( α ) is continuous strictlypositive function on [0 , α ∈ (0 , r n ( α ) > ln r We are now interested in finding the probabilistic bounds on difference | ln r ( u ) − ln ˆ r ( u ) | , where, as previously, X , ..., X n is iid sample from symmetric stable law, r ( u ) = − ln | ϕ ( u ) | = ( γ | u | ) α and ˆ r ( u ) = − ln | ϕ n ( u ) | . For that an additional assumption has tobe made. In the present subsection, we assume that there exists a known lower bound0 < α such that the parameter space is [ α,
2] instead of (0 , α is the fact that the bound ¯ r from Theorem 1.1 increasesto infinity as n grows. Recall that the bound ¯ r in lacks this property: although ¯ r increases with n , it always satisfies ¯ r < W ( (cid:15) L (1+ (cid:15) ) ) < n increases. Let us formulate it as acorollary. Corollary 2.1.
Assume α > to be given. Fix (cid:15) > , p ∈ (0 , and let n o ( (cid:15), p ) < ∞ be as in Theorem 1.1. Then for every n > n o there exists ¯ r ( (cid:15), p, n, α ) independent of α and γ such that lim n ¯ r ( (cid:15), p, n, α ) = ∞ and P (cid:16) sup u> r ( u ) ≤ r | r ( u ) − ˆ r ( u ) | > (cid:15) (cid:17) ≤ p, for every α ∈ [ α, and γ > . roof. Fix 0 < α < r o so big that 4 Lr o > (cid:15)e − r o . Then for every r ≥ r o andfor every x ∈ [0 , (cid:15)e − r ], the function α (cid:55)→ ( (cid:15)e − r − x )1 + ( c Lrx ) α is increasing in α . Therefore k ( α, (cid:15), r ) ≥ k ( α, (cid:15), r ) , ∀ α ≥ α r ≥ r o . (2.11)Now observe that when α >
0, then for every r , it holds x (cid:48) < (cid:15)e − r α , thus k ( α, (cid:15), r ) = (cid:0) (cid:15)e − r − (1 + α ) x (cid:48) ) >
0. Since lim r →∞ k ( α, (cid:15), r ) = 0, it follows that lim n r n ( α ) = ∞ .Therefore, there exists n ( r o , α ) so big that r n ( α ) > r o . From (2.11), it follows that when n > n o , it holds r n ( α ) ≥ r n ( α ), ∀ α ≥ α . Hence lim n inf α ∈ [ α, r n ( α ) = lim n r n ( α ) = ∞ . The inequality in the statement now follows from Theorem 1.1.
Theorem 2.1.
Assume α > to be given. Fix (cid:15) > , p ∈ (0 , and r > . Take n ( (cid:15), p, α, r ) as minimal n such that ¯ r ( (cid:15) (cid:15) r, p, n, α ) > r, n ( (cid:15), p, α, r ) ≥ n o ( (cid:15) (cid:15) r, p ) , (2.12) where n o ( (cid:15), p ) is as in Theorem 1.1 and the function ¯ r is as in Corollary 2.1. Then forevery n ≥ n , α ∈ [ α, and γ > P (cid:16) sup u : r ( u ) ∈ [ r,r ] | ln r ( u ) − ln ˆ r ( u ) | > (cid:15) (cid:17) ≤ p, (2.13) where ¯ r = ¯ r ( (cid:15) (cid:15) r, p, n, α ) > r .Proof. By Corollary 2.1, such a finite n exists. Now, for any n > n , by Corollary 2.1again, P (cid:16) sup u : r ( u ) ∈ [ r, ¯ r ] | ln r ( u ) − ln ˆ r ( u ) | > (cid:15) (cid:17) ≤ P (cid:16) sup u : r ( u ) ∈ [ r, ¯ r ] | r ( u ) − ˆ r ( u ) | > (cid:15) (cid:15) r (cid:17) ≤ p. Recall the estimate ˆ α in (1.4). The construction of ˆ α requires fixing the constants r, ¯ r and (cid:15) . The following simple lemma shows how Theorem 2.1 can be used to prove alarge deviation inequality for ˆ α . 11 emma 3.1. Suppose r < ¯ r and (cid:15) are chosen such that ¯ r − (cid:15) > r + 2 (cid:15) and (2.13)holds for some p ∈ (0 , . Take (cid:15) > so big that (cid:15) (cid:18) ¯ r − (cid:15)r + 2 (cid:15) (cid:19) ≥ (cid:15). (3.1) Then P (cid:16) | ˆ α − α | ≥ (cid:15) (cid:17) ≤ p. (3.2) Proof.
Since u (cid:55)→ ˆ r ( u ) is continuous, we have by (1.4) that ˆ r ( u ) = ¯ r − (cid:15) and ˆ r ( u ) = r + (cid:15) . Let E = (cid:8) sup u : r ( u ) ∈ [ r, ¯ r ] | ln r ( u ) − ln ˆ r ( u ) | ≤ (cid:15) (cid:9) . (3.3)On the event E , it holds | ˆ r ( u i ) − r ( u i ) | ≤ (cid:15) for i = 1 , r − (cid:15) ≤ r ( u ) ≤ ¯ r and r ≤ r ( u ) ≤ r + 2 (cid:15) . Therefore, on the set E , for any (cid:15) > {| ˆ α − α | ≥ (cid:15) } = (cid:110)(cid:12)(cid:12)(cid:12) ln(¯ r − (cid:15) ) − ln( r + (cid:15) )ln u − ln u − ln r ( u ) − ln r ( u )ln u − ln u (cid:12)(cid:12)(cid:12) > (cid:15) (cid:111) = (cid:8) | (ln ˆ r ( u ) − ln r ( u )) + (ln r ( u ) − ln ˆ r ( u )) | > (cid:15) ln( u /u ) (cid:9) ⊆ (cid:8) sup u : r ( u ) ∈ [ r, ¯ r ] | ln r ( u ) − ln ˆ r ( u ) | > (cid:15) u /u ) (cid:9) = (cid:8) sup u : r ( u ) ∈ [ r, ¯ r ] | ln r ( u ) − ln ˆ r ( u ) | > (cid:15) α ln( r ( u ) /r ( u )) (cid:9) ⊆ (cid:110) sup u : r ( u ) ∈ [ r, ¯ r ] | ln r ( u ) − ln ˆ r ( u ) | > (cid:15) (cid:18) ¯ r − (cid:15)r + 2 (cid:15) (cid:19) (cid:111) . We have thus shown that E ∩ {| ˆ α − α | ≥ (cid:15) } ⊆ (cid:110) sup u : r ( u ) ∈ [ r, ¯ r ] | ln r ( u ) − ln ˆ r ( u ) | > (cid:15) (cid:0) ¯ r − (cid:15)r + 2 (cid:15) (cid:1)(cid:111) . Taking now (cid:15) so big that (3.1) holds, we obtain that E ∩ {| ˆ α − α | ≥ (cid:15) } ⊆ E c which obviously implies that {| ˆ α − α | ≥ (cid:15) } ⊆ E c and so (3.2) holds. Exact confidence intervals.
In what follows, we shall briefly discuss how to choose (cid:15) > , r, ¯ r such that (3.2) holds. In particular, we shall address the following classicalproblems of parameter estimation: 12 Given lower bound α >
0, precision (cid:15) and probability p >
0, find possibly smallsample size n and constants r, ¯ r, (cid:15) (for constructing ˆ α ) so that the estimate (1.4)satisfies inequality (3.2). Q2:
Given lower bound α >
0, sample size n and probability p >
0, find possiblysmall (cid:15) > r, ¯ r, (cid:15) (for constructing ˆ α ) so that the estimate (1.4) satisfiesinequality (3.2). In other words, find exact (non-asymptotic) confidence intervals:with probability 1 − p : ˆ α − (cid:15) ≤ α ≤ ˆ α + (cid:15) .To solve Q1 , define for any ρ > n and (cid:15) > F ( n, ρ, (cid:15) ) := ¯ r (cid:0) ρ, p, n, α (cid:1) − (cid:15)ρ (cid:15)(cid:15) + 2 (cid:15) − exp[ 4 (cid:15) (cid:15) ] , (3.4)where ¯ r (cid:0) ρ, p, n, α (cid:1) is as in Corollary 2.1. The function F depends also on α, (cid:15) and p ,but these parameters are fixed and left out from notation. Define F ( n, ρ ) = sup (cid:15)> F ( n, ρ, (cid:15) ) , F ( n ) = sup ρ> (cid:104) F ( n, ρ ) ∧ ( n − n o ( ρ, p )) (cid:105) , where n o ( ρ, p ) is as in Theorem 1.1. Now take n minimal n such that F ( n ) > F ( n ) is increasing, so that when F ( n ) >
0, then F ( n ) > n > n . The estimation procedure is now the following. EstimationProcedure1:
1. Find n such that F ( n ) > n ≥ n find ρ > F ( n, ρ ) > ρ to find (cid:15) > F ( n, ρ o , (cid:15) ) > ρ to determine ¯ r = ¯ r ( ρ, p, n, α ), where ¯ r ( ρ, p, n, α ) is as in Corollary 2.1.5. Use (cid:15) and ρ to find r = ρ (cid:15)(cid:15) .6. Use r and ¯ r to find u and u as in (1.5).7. Estimate ˆ r ( u ) based on the sample of size n and the estimate of characteristicfunction.8. Find ˆ α as in (1.4). Theorem 3.1.
Let α , (cid:15) > and p ∈ (0 , be given. Let n be the sample sizesatisfying F ( n ) > . Then the estimate ˆ α obtained via EstimationProcedure1 satisfiesthe inequality (3.2), provided the true parameter α satisfies the inequality α ≥ α . roof. According to definition of F ( n ), the parameters ρ > (cid:15) > EstimationProcedure1 are such that F ( n, ρ, (cid:15) ) > n > n o ( ρ, p ). With r = ρ (cid:15)(cid:15) ,we see that ¯ r − (cid:15)r + 2 (cid:15) > exp[ 4 (cid:15) (cid:15) ]so that the equation (3.1) holds. This equation also implies that ¯ r > r + 4 (cid:15) . Since n > n o ( ρ, p ), we have n > n o ( (cid:15) (cid:15) r, p ). Thus both inequalities in (2.12) hold andtherefore n ≥ n , where n ( (cid:15), p, α, r ) is as in Theorem 2.1. Hence all assumptions ofTheorem 2.1 are fulfilled and so (2.13) holds. Both assumptions of Lemma 3.1 arefulfilled and so the inequality (3.2) holds as well.To solve Q2 , we need to assume some minimal requirements about the given samplesize n . In what follows, we assume that there exists ρ o > n > n o ( ρ o , p ),where n o ( ρ, p ) is as in Theorem 1.1. Now we define F ( (cid:15) , ρ, (cid:15) ) := ¯ r (cid:0) ρ, p, n, α (cid:1) − (cid:15)ρ (cid:15)(cid:15) + 2 (cid:15) − exp[ 4 (cid:15) (cid:15) ] . The function F ( (cid:15) , ρ, (cid:15) ) also depends on n and p , but these are fixed. As previously,define F ( (cid:15) , ρ ) := sup (cid:15)> F ( (cid:15) , ρ, (cid:15) ) , F ( (cid:15) ) := sup ρ ≥ ρ o F ( (cid:15) , ρ ) . Now find (as small as possible) (cid:15) > F ( (cid:15) ) > EstimationProcedure2:
1. Given (cid:15) > F ( (cid:15) ) > ρ ≥ ρ o such that F ( (cid:15) , ρ ) > ρ to find (cid:15) > F ( (cid:15) , ρ, (cid:15) ) >
03. Use ρ to determine r = r ( ρ, p, n ), where ¯ r ( ρ, p, n, α ) is as in Corollary 2.1.4. Use (cid:15) and ρ to find r = ρ (cid:15)(cid:15) .5. Use r and ¯ r to find u and u as in (1.5).6. Estimate ˆ r ( u ) based on the sample of size n and the estimate of characteristicfunction.7. Find ˆ α as in (1.4). Theorem 3.2.
Let α > , the sample size n > n o ( ρ o , p ) , where ρ o > and p ∈ (0 , be given. Let (cid:15) satisfy F ( (cid:15) ) > . Then the estimate ˆ α obtained via Estimation-Procedure2 satisfies the inequality (3.2), provided the true parameter α satisfies theinequality α ≥ α . roof. According to definition of F ( (cid:15) ), the parameters ρ > (cid:15) > EstimationProcedure2 are such that F ( (cid:15) , ρ, (cid:15) ) >
0. Since ρ (cid:55)→ n o ( ρ, p ) is decreasing,it holds that n > n o ( ρ, p ). With r = ρ (cid:15)(cid:15) , we see that equation (3.1) holds. Thisequation also implies that ¯ r > r + 4 (cid:15) and hence both inequalities in (2.12) hold andtherefore n ≥ n , where n ( (cid:15), p, α, r ) is as in Theorem 2.1. Hence all assumptions ofTheorem 2.1 are fulfilled and so (2.13) holds. Then Lemma 3.1 implies that (3.2) holdsas well. Recall the construction of the ¯ r in the proof of Theorem 1.1: the key of the constructionis the large deviation inequality P (cid:16) e ¯ r sup u ∈ (0 , ¯ u ] | ϕ n ( u ) − ϕ ( u ) | > (cid:15) (cid:17) ≤ A exp[ − n · k ( α, (cid:15), ¯ r )] , (4.1)where A = 2 and k ( α, (cid:15), ¯ r ) >
0. Then r n ( α ) was defined as the solution of the equality A exp[ − n · k ( α, (cid:15) (cid:15) , r n )] = p (4.2)and so the the desired bound ¯ r = inf α ∈ (0 ,α ] r n ( α ) was obtained. In order (4.2) to havepositive solution, the sample size n must satisfy n > n o ( (cid:15), p ). The large deviationbound (4.2) constructed in Section 2.1 is not the one possible option. We now sketchanother possible construction yielding to a different inequality, and therefore, also tothe different function r n ( α ) and different bound ¯ r . Unlike the ¯ r obtained in Section 2.1,the new ¯ r has more explicit form. The price for it is much bigger required sample sice n o .Observe that by (2.2) and (2.3) the inequality (cid:107) F n − F (cid:107) ≤ δ implies that | ϕ n ( u ) − ϕ ( u ) | ≤ δ + 2 (cid:107) F n − F (cid:107) (cid:16) Lr ( u ) δ (cid:17) α . Hence for every ¯ u > { e r (¯ u ) sup u ∈ (0 , ¯ u ] | ϕ ( u ) − ϕ n ( u ) | > (cid:15) } ⊂ {(cid:107) F n − F (cid:107) > δ } ∪ (cid:110) (cid:107) F n − F (cid:107) (cid:16) Lr (¯ u ) δ (cid:17) α ≥ (cid:15)e − r (¯ u ) − δ (cid:111) . Recall ¯ (cid:15) = (cid:15)e − ¯ r and ¯ r = r (¯ u ). Hence, we obtain P (cid:16) e ¯ r sup u ∈ (0 , ¯ u ] | ϕ n ( u ) − ϕ ( u ) | > (cid:15) (cid:17) ≤ P (cid:16) (cid:107) F n − F (cid:107) > δ (cid:17) + P (cid:16) (cid:107) F n − F (cid:107) ≥ (¯ (cid:15) − δ ) δ α L ¯ r ) α (cid:17) . ≤ − n · δ
32 ] + 2 exp (cid:104) − n · (¯ (cid:15) − δ ) δ α L ¯ r ) α (cid:105) . (4.3)15hoose δ = ¯ (cid:15) α = arg max δ ∈ [0 , ¯ (cid:15) ] (¯ (cid:15) − δ ) δ α . Thus plugging δ into (4.3), we obtain with s = 2(1 + α ), P (cid:16) e ¯ r sup u ∈ (0 , ¯ u ] | ϕ n ( u ) − ϕ ( u ) | > (cid:15) (cid:17) ≤ − n · ¯ (cid:15) α ) ] + 2 exp (cid:104) − n · (cid:16)(cid:0) ¯ (cid:15) α (cid:1) s α L ¯ r ) α (cid:17)(cid:105) ≤ − n · k ( α, (cid:15), ¯ r )] , where k ( α, (cid:15), ¯ r ) = (cid:0) (cid:15) e − r α ) (cid:1) ∧ (cid:16) (cid:15) s e − s ¯ r α ) s α (¯ r L ) α (cid:17) . The inequality (4.2) holds with an equality, when¯ r = r ,n ( α ) ∧ r ,n ( α )and r ,n , r ,n are solutions of the following equalities:exp[ − r ,n ] = ln(4 /p ) n (cid:0) (cid:15)(cid:15) (cid:1) α ) (4.4) d ( α ) exp[ − sr ,n ]( r ,n ) α = ln(4 /p ) n (cid:0) (cid:15)(cid:15) (cid:1) s , where d ( α ) = α α ) s L ) α (4.5)Thus r ,n ( α ) = −
12 ln (cid:104) ln(4 /p ) n (cid:0) (cid:15)(cid:15) (cid:1) α ) (cid:105) and r ,n ( α ) > n > ln(4 /p ) (cid:0) (cid:15)(cid:15) (cid:1) α ) . (4.6)From (4.6), we obtain necessary sample size n o ( (cid:15), p ) = ln(4 /p ) (cid:0) (cid:15)(cid:15) (cid:1) · . (4.7)The solution of equality (4.5) is r ,n ( α ) = 1 α + 1 W (( α + 1) g n ( α )) , where W is Lambert’s W-function and g n is defined as follows g n ( α ) = α α α ( α + 1) α +1 L (cid:16) n ln(4 /p ) (cid:17) α (cid:16) (cid:15) (cid:15) (cid:17) α +1 . Observe that lim α → g n ( α ) = L (cid:15) (cid:15) =: g o and lim α → r ,n ( α ) = W ( g o ) ∈ (0 , . It canbe shown that¯ r ( (cid:15), p, n ) = min α { r ,n ( α ) , r ,n ( α ) } = ln κ − ln (cid:16) ∨ α α (cid:17) > , (4.8)16here κ = (cid:15)(cid:15) +1 (cid:113) n /p ) while r ,n > κ >
12 and n > n , where n o is given by(4.7), and α ∈ (0 ,
2] is the solution of the equality r ,n ( α ) = ln κ α α . Since lim α → r ,n ( α ) = W ( g o ), it holds that for any n ¯ r <
1, and so the alternativeconstruction satisfies our conjecture. However, it is possible to show the for any α > r ( (cid:15), p, n, α ) = inf α ∈ ( α, r n ( α ) → ∞ . So we have another construction that confirms our conjecture that the universal bound¯ r satisfies the inequality ¯ r < u does not exists. r n and ¯ r Fix (cid:15) = 0 . p = 0 . L = 1 /
2. We find the values of r n ( α ) obtained in Section2.1 as follows: first we numerically find x (cid:48) as the solution of the equality f ( x ) = e e − r ,where f ( x ) is given by (2.9), then we calculate k ( α, (cid:15), r ) given by (2.7) and then r n canbe found as solution of (2.5). We compare the obtained values of r n ( α ) with the onesobtained in Section 4: r n ( α ) = min { r ,n ( α ) , r ,n ( α ) } , where r ,n ( α ) is the solution ofequality (4.4) and r ,n ( α ) is the solution of equality (4.5). Note that by (4.7) we have r ,n ( α ) > n > n o ≈ r n versus α ∈ (0 . ,
2] for different sample sizes n . Obviously, in Figure 4 (a) the r n ( a ) obtained . . . . a n= 10000 (a) . . . . . a n= 128551 (b) . . . . a n= 3e+05 (c) r n by Section 2.1 (proof of Theorem 1.1) r n by Section 4 (the alternative construction) Figure 4: The values of r n vs α with (cid:15) = 0 . p = 0 . n = 1000 in (a), n =128551 > n o in (b) and n = 300000 in (c).by the alternative construction in Section 4 is 0 because n < n o . However, the valuesof r n obtained by Section 2.1 are small but positive. In Figure 4 (b) both constructionsgive positive results (because n = n o + 1) while alternative construction gives smaller(more conservative) values, with drop after α = 1 . α is at α = 2).In Figure 4 (c) the large sample size such as n = 3 · is used and both r n ( α ) behave17 e+00 4e+05 8e+05 . . . . . aa =0.01 (a) . . . . . aa =0.1 (b) . . . . . aa =0.5 (c) r by Section 2.1 (proof of Theorem 1.1) r by Section 4 (the alternative construction) Figure 5: The values of ¯ r vs n with (cid:15) = 0 . p = 0 . α = 0 .
01 in (a), α = 0 . α = 0 . α . Next we compare the values of ¯ r = min α ∈ ( α, r n ( α ) obtained by Section 2.1 andSection 4. Figure 5 plots ¯ r versus n for different lower limits α . It is clearly evidentfrom Figure 5 that ¯ r is increasing in n (while alternative construction requires n > n o ).Setting lower limit from α = 0 . α = 0 . r approximatelytwice. All in all, in our example the construction given by Section 2.1 yields muchbigger (less conservative) values of ¯ r than the more explicit form construction given bySection 4. References [1] Bibalan, M. H., H. Amindavar, and M. Amirmazlaghani (2017). Characteristicfunction based parameter estimation of skewed alpha-stable distribution: An ana-lytical approach.
Signal Process. 130 , 323–336.[2] Cs¨orgo, S. (1981). Limit behaviour of the empirical characteristic function.
TheAnnals of Probability 9 (1), 130–144.[3] Kakinaka, S. and K. Umeno (2020). Flexible two-point selection approach forcharacteristic function-based parameter estimation of stable laws. Online at https://arxiv.org/abs/2005.11499 .[4] Kogon, S. M. and D. B. Williams (1998). Characteristic function based estimationof stable distribution parameters. In R. J. Adler, R. E. Feldman, and M. S. Taqqu(Eds.),
A Practical Guide to Heavy Tails , pp. 311–335. Boston: Birkh¨auser.[5] Krutto, A. (2016). Parameter estimation in stable law.
Risks 4 (4), 43.186] Krutto, A. (2018). Empirical cumulant function based parameter estimation in sta-ble laws.
Acta et Commentationes Universitatis Tartuensis de Mathematica 22 (2),311–338.[7] McCulloch, J. H. (1996). Financial applications of stable distributions. In G. Mad-dala and C. Rao (Eds.),
Statistical Methods in Finance , Volume 14 of
Handbook ofStatistics , pp. 393 – 425. Elsevier.[8] Nolan, J. P. (2001). Maximum likelihood estimation and diagnostics for stabledistributions. In O. Barndorff-Nielsen, S. Resnick, and T. Mikosch (Eds.),
L´evyProcesses , pp. 379–400. Boston: Birkh¨auser.[9] Nolan, J. P. (2018).
Stable Distributions - Models for Heavy Tailed Data . Boston:Birkh¨auser. In progress, Chapter 1 online at .[10] Paulson, A. S., E. W. Holcomb, and R. A. Leitch (1975). The estimation of theparameters of the stable laws.
Biometrika 62 (1), 163–170.[11] Press, J. S. (1972). Estimation in univariate and multivariate stable distributions.
J. Amer. Statist. Assoc. 67 (340), 842–846.[12] Robust Analysis Inc. (2017).
STABLE 5.3 R Version for Windows . Washington,DC, USA: Robust Analysis Inc. .[13] Samorodnitsky, G. and M. S. Taqqu (1994).
Stable Non-Gaussian Random Pro-cesses: Stochastic Models with Infinite Variance . New York: Chapman & Hall.[14] van der Vaart, A. (1998).
Asymptotic Statistics . Cambridge: Cambridge UniversityPress.[15] Zolotarev, V. (1986).
One-dimensional Stable Distributions , Volume 65 of