STABLE DISTRIBUTIONS

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1 STABLE DISTRIBUTIONS SVANTE JANSON Abstract. We give some explicit calculations for stable istributions an convergence to them, mainly base on less explicit results in Feller [3]. The main purpose is to provie ourselves with easy reference to explicit formulas an examples. There are probably no new results.) 1. Introuction We give some explicit calculations for stable istributions an convergence to them, mainly base on less explicit results in Feller [3]. The main purpose is to provie ourselves with easy reference to explicit formulas an examples, for example for use in [8]. There are probably no new results.) These notes may be extene later with more examples. 2. Infinitely ivisible istributions We begin with the more general concept of infinitely ivisible istributions. Definition 2.1. The istribution of a ranom variable X is infinitely ivisible if for each n 1 there exists i.i.. ranom variable Y n) 1,..., Y n n) such that X = Y n) Y n n). 2.1) The characteristic function of an infinitely ivisible istribution may be expresse in a canonical form, sometimes calle the Lévy Khinchin representation. We give several equivalent versions in the following theorem. Theorem 2.2. Let hx) be a fixe boune measurable real-value function on R such that hx) = x+ox 2 ) as x 0. Then the following are equivalent. i) ϕt) is the characteristic function of an infinitely ivisible istribution. ii) There exist a measure M on R such that 1 x 2 ) Mx) < 2.2) an a real constant b such that e itx 1 ithx) ϕt) = exp ibt + x 2 Date: 1 December, 2011; revise 20 December, Mathematics Subject Classification. 60E07. 1 ) Mx), 2.3)

2 2 SVANTE JANSON where the integran is interprete as t 2 /2 at x = 0. iii) There exist a measure Λ on R \ 0} such that x 2 1 ) Λx) < 2.4) an real constants a 0 an b such that ϕt) = exp ibt 1 2 at2 + e itx 1 ithx) ) Λx)). 2.5) iv) There exist a boune measure K on R an a real constant b such that ϕt) = exp ibt + e itx 1 itx ) 1 + x 2 ) 1 + x 2 x 2 Kx), 2.6) where the integran is interprete as t 2 /2 at x = 0. The measures an constants are etermine uniquely by ϕ. Feller [3, Chapter XVII] uses hx) = sin x. Kallenberg [9, Corollary 15.8] uses hx) = x1 x 1}. Feller [3, Chapter XVII.2] calls the measure M in ii) the canonical measure. The measure Λ in iii) is known as the Lévy measure. The parameters a, b an Λ are together calle the characteristics of the istribution. We enote the istribution with characteristic function 2.5) for a given h) by IDa, b, Λ). Remark 2.3. Different choices of hx) yiel the same measures M an Λ in ii) an iii) but ifferent constants b; changing h to h correspons to changing b to b := b + hx) hx) x 2 Mx) = b + hx) hx) ) Λx). 2.7) We see also that b is the same in ii) an iii) with the same h), an that see the proof below) b in iv) equals b in ii) an iii) when x = x/1 + x 2 ). Proof. i) ii): This is shown in Feller [3, Theorem XVII.2.1] for the choice hx) = sin x. As remarke above, 2.3) for some h is equivalent to 2.3) for any other h, changing b by 2.7). ii) iii): Given M in ii) we let a := M0} an Λx) := x 2 Mx), x 0. Conversely, given a an Λ as in iii) we efine Mx) = aδ 0 + x 2 Λx). 2.8) The equivalence between 2.3) an 2.5) then is obvious. ii) iv): Choose hx) = x/1 + x 2 ) an efine Kx) := 1 Mx); 2.9) 1 + x2 conversely, Mx) = 1 + x 2 ) Kx). Then 2.3) is equivalent to 2.6).

3 STABLE DISTRIBUTIONS 3 Remark 2.4. At least iii) extens irectly to infinitely ivisible ranom vectors in R. Moreover, there is a one-to-one corresponence with Lévy processes, i.e., stochastic processes X t on [0, ) with stationary inepenent increments an X 0 = 0, given by in the one-imensional case) E e iuxt = ϕu) t = exp t ibu 1 2 au2 + e iux 1 iuhx) ) Λx))) 2.10) for t 0 an u R. See Bertoin [1] an Kallenberg [9, Corollary 15.8]. Example 2.5. The normal istribution Nµ, σ 2 ) has Λ = 0 an a = σ 2 ; thus M = K = σ 2 δ 0 ; further, b = µ for any h. Thus, Nµ, σ 2 ) = IDσ 2, µ, 0). Example 2.6. The Poisson istribution Poλ) has M = Λ = λδ 1 an K = λ 2 δ 1; further b = λh1). Thus b = λ/2 in iv).) Example 2.7. The Gamma istribution Gammaα) with ensity function x α 1 e x /Γα), x > 0, has the characteristic function ϕt) = 1 it) α. It is infinitely ivisible with see Feller [3, Example XVII.3.]. Mx) = αxe x, x > 0, 2.11) Λx) = αx 1 e x, x > 0, 2.12) Remark 2.8. If X 1 an X 2 are inepenent infinitely ivisible ranom variables with parameters a 1, b 1, Λ 1 ) an a 2, b 2, Λ 2 ), then X 1 + X 2 is infinitely ivisible with parameters a 1 + a 2, b 1 + b 2, Λ 1 + Λ 2 ). In particular, if X IDa, b, Λ), then X = X 1 + Y with X 1 ID0, 0, Λ), Y IDa, b, 0) = Nb, a), 2.13) an X 1 an Y inepenent. Moreover, for any finite partition R = A i, we can split X as a sum of inepenent infinitely ivisible ranom variables X i with the Lévy measure of X i having supports in A i. Example 2.9 integral of Poisson process). Let Ξ be a Poisson process on R \ 0} with intensity Λ, where Λ is a measure with ) x 1 Λx) <. 2.14) Let X := x Ξx); if we regar Ξ as a finite or countable) set or possibly multiset) of points ξ i }, this means that X := i ξ i. The sum converges absolutely a.s., so X is well-efine a.s.; in fact, the sum ξ i >1 ξ i is a.s. finite, an the sum ξ i 1 ξ i has finite expectation 1 1 x Λx).) Then X has characteristic function ϕt) = exp e itx 1 ) Λx)). 2.15) See, for example, the corresponing formula for the Laplace transform in Kallenberg [9, Lemma 12.2], from which 2.15) easily follows.) Hence, 2.5)

4 4 SVANTE JANSON hols with Lévy measure Λ, a = 0 an b = hx) Λx). When 2.5) hols, we can take hx) = 0, a choice not allowe in general. Note that 2.15) is the same as 2.5) with h = 0, a = 0 an b = 0.) By aing an inepenent normal variable Nb, a), we can obtain any infinitely ivisible istribution with a Lévy measure satisfying 2.5); see Example 2.5 an Remark 2.8. Example 2.10 compensate integral of Poisson process). Let Ξ be a Poisson process on R \ 0} with intensity Λ, where Λ is a measure with x 2 x ) Λx) <. 2.16) Suppose first that x Λx) <. Let X be as in Example 2.9. Then X has finite expectation E X = x Λ. Define X := X E X = Then, by 2.15), X has characteristic function x Ξx) Λx) ). 2.17) ϕt) = exp e itx 1 itx ) Λx)). 2.18) Now suppose that Λ is any measure satisfying 2.16). Then the integral in 2.18) converges; moreover, by consiering the truncate measures Λ n := 1 x > n 1 }Λ an taking the limit as n, it follows that there exists a ranom variable X with characteristic function 2.18). Hence, 2.5) hols with Lévy measure Λ, a = 0 an b = hx) x) Λx). When 2.16) hols, we can take hx) = x, a choice not allowe in general. Note that 2.18) is the same as 2.5) with hx) = x, a = 0 an b = 0.) By aing an inepenent normal variable Nb, a), we can obtain any infinitely ivisible istribution with a Lévy measure satisfying 2.16); see Example 2.5 an Remark 2.8. Remark Any infinitely ivisible istribution can be obtaine by taking a sum X 1 + X 2 + Y of inepenent ranom variables with X 1 as in Example 2.9, X 2 as in Example 2.10 an Y normal. For example, we can take the Lévy measures of X 1 an X 2 as the restrictions of the Lévy measure to x : x > 1} an x : x 1}, respectively. Theorem If X is an infinitely ivisible ranom variable with characteristic function given by 2.5) an t R, then E e tx = exp bt at2 + e tx 1 thx) ) Λx)). 2.19)

5 In particular, E e tx < STABLE DISTRIBUTIONS 5 e tx 1 thx) ) Λx) < 1 e tx Λx) <, t > 0, 1 etx Λx) <, t < ) Proof. The choice of h satisfying the conitions of Theorem 2.2) oes not matter, because of 2.7); we may thus assume hx) = x1 x 1}. We further assume t > 0. The case t < 0 is similar an the case t = 0 is trivial.) Denote the right-han sie of 2.19) by F Λ t). We stuy several ifferent cases. i). If supp Λ is boune, then the integral in 2.19) converges for all complex t an efines an entire function. Thus F Λ t) is entire an 2.5) shows that E e itx = F Λ it). It follows that E e tx < an E e tx = F Λ t) for any complex t, see e.g. Marcinkiewicz [10]. ii). If supp Λ [1, ), let Λ n be the restriction Λ [1,n] of the measure Λ to [1, n]. By the construction in Example 2.9, we can construct ranom variables X n ID0, 0, Λ n ) such that X n X ID0, 0, Λ) as n. Case i) applies to each Λ n, an 2.19) follows for X, an t > 0, by monotone convergence. iii). If supp Λ, 1], let Λ n be the restriction Λ [ n, 1]. Similarly to ii) we can construct ranom variables X n ID0, 0, Λ n ) with X n 0 such that X n X ID0, 0, Λ) as n. Case i) applies to each Λ n, an 2.19) follows for X; this time by monotone convergence. iv). The general case follows by i) iii) an a ecomposition as in Remark Stable istributions Definition 3.1. The istribution of a non-egenerate) ranom variable X is stable if there exist constants a n > 0 an b n such that, for any n 1, if X 1, X 2,... are i.i.. copies of X an S n := n i=1 X i, then The istribution is strictly stable if b n = 0. S n = an X + b n. 3.1) Many authors, e.g. Kallenberg [9], say weakly stable for our stable.) We say that the ranom variable X is strictly) stable if its istribution is. The norming constants a n in 3.1) are necessarily of the form a n = n 1/α for some α 0, 2], see Feller [3, Theorem VI.1.1]; α is calle the inex [4], [9] or characteristic exponent [3] of the istribution. We also say that a istribution or ranom variable) is α-stable if it is stable with inex α.

6 6 SVANTE JANSON The case α = 2 is simple: X is 2-stable if an only if it is normal. For α < 2, there is a simple characterisation in terms of the Lévy Khinchin representation of infinitely ivisible istributions. Theorem 3.2. i) A istribution is 2-stable if an only if it is normal Nµ, σ 2 ). This is an infinitely ivisible istribution with M = σ 2 δ 0, see Example 2.5.) ii) Let 0 < α < 2. A istribution is α-stable if an only if it is infinitely ivisible with canonical measure Mx) c + x 1 α, x > 0, = x c x 1 α 3.2), x < 0; equivalently, the Lévy measure is given by Λx) x = c + x α 1, x > 0, c x α 1, x < 0, an a = 0. Here c, c + 0 an we assume that not both are 0. Proof. See Feller [3, Section XVII.5] or Kallenberg [9, Proposition 15.9]. Note that 3.2) is equivalent to for any interval with x 1 0 x 2, with Theorem 3.3. Let 0 < α ) M[x 1, x 2 ] = C + x 2 α 2 + C x 1 2 α 3.4) C ± = c ± 2 α. 3.5) i) A istribution is α-stable if an only if it has a characteristic function ) ) exp γ α t 1 α iβ tan πα 2 ϕt) = sgnt) + iδt, α 1, ) ) exp γ t 1 + iβ 2 π sgnt) log t 3.6) + iδt, α = 1, where 1 β 1, γ > 0 an < δ <. Furthermore, an α-stable istribution exists for any such α, β, γ, δ. If α = 2, then β is irrelevant an usually taken as 0.) ii) If X has the characteristic function 3.6), then, for any n 1, 3.1) takes the explicit form n 1/α X + n n 1/α )δ, α 1, S n = nx + 2 π βγn log n, α = ) In particular, X is strictly stable δ = 0, α 1, β = 0, α = )

7 STABLE DISTRIBUTIONS 7 iii) An α-stable istribution with canonical measure M satisfying 3.4) has γ α C+ + C ) Γ3 α) = α1 α) cos πα 2, α 1, C + + C ) π 2, α = 1, 3.9) β = C + C C + + C. 3.10) iv) If 0 < α < 2, then an α-stable istribution with Lévy measure Λ satisfying 3.3) has γ α c + + c ) Γ α) cos πα ) = 2, α 1, c + + c ) π 2, α = 1, 3.11) β = c + c c + + c. 3.12) We use the notation S α γ, β, δ) for the istribution with characteristic function 3.6). Proof. Feller [3, XVII.3.18) 3.19) an Theorem XVII.5.1ii)] gives, in our notation, for a stable istribution satisfying 3.4), the characteristic function Γ3 α) exp C + + C ) cos πα α1 α) 2 i sgnt)c + C sin πα ) ) t α + ibt C + + C 2 if α 1 an π exp C + + C ) 2 + i sgnt)c + C ) ) log t t + ibt C + + C if α = 1. This is 3.6) with 3.9) 3.10) an δ = b. This proves i) an iii), an iv) follows from iii) by 3.5). Finally, ii) follows irectly from 3.6). Remark 3.4. If 1 < α 2, then δ in 3.6) equals the mean E X. In particular, for α > 1, a stable istribution is strictly stable if an only if its expectation vanishes. Remark 3.5. If X S α 1, β, 0), then, for γ > 0 an δ R, S α γ, β, δ), α 1, γx + δ S α γ, β, δ 2 π βγ log γ), α = ) Thus, γ is a scale parameter an δ a location parameter; β is a skewness parameter, an α an β together etermine the shape of the istribution. Remark 3.6. More generally, if X S α γ, β, δ), then, for a > 0 an R, S α aγ, β, aδ + ), α 1, ax + S α aγ, β, aδ + 2 π βγa log a), α = )

8 8 SVANTE JANSON Remark 3.7. If X S α γ, β, δ), then X S α γ, β, δ). In particular, X has a symmetric stable istribution if an only if X S α γ, 0, 0) for some α 0, 2] an γ > 0. We may simplify expressions like 3.6) by consiering only t 0 or t > 0); this is sufficient because of the general formula ϕ t) = ϕt) 3.15) for any characteristic function. We use this in our next statement, which is an immeiate consequence of Theorem 3.3. Corollary 3.8. Let 0 < α 2. A istribution is strictly stable if an only if it has a characteristic function ϕt) = exp κ iλ)t α), t 0, 3.16) where κ > 0 an λ κ tan πα 2 ; furthermore, a strictly stable istribution exists for any such κ an λ. For α = 1, tan πα 2 =, so any real λ is possible. For α = 2, tan πα 2 = 0, so necessarily λ = 0.) The istribution S α γ, β, 0) α 1) or S 1 γ, 0, δ) α = 1) satisfies 3.16) with κ = γ α βκ tan πα an λ = 2, α 1, 3.17) δ, α = 1. Conversely, if 3.16) hols, then the istribution is S α γ, β, 0) with γ = κ 1/α, β = λ κ cot πα 2, α 1 S 1 κ, 0, λ), α = ) Remark 3.9. For a strictly stable ranom variable, another way to write the characteristic function 3.6) or 3.16) is ϕt) = exp ae i sgnt)π γ/2 t α), 3.19) with a > 0 an γ real. A comparison with 3.6) an 3.17) shows that a cos π γ 2 = κ = γα, 3.20) tan π γ 2 = λ κ = β tan πα 2, α 1, δ γ, α = 1. α 3.21) If 0 < α < 1, we have 0 < tan πα 2 < an γ α, while if 1 < α < 2, then tan πα 2 < 0 an tan π γ 2 = β tan π2 α) 2 with 0 < π2 α)/2 < π/2; hence γ 2 α. Finally, for α = 1, we have γ < 1. These ranges for γ are both necessary an sufficient. For α = 1, γ = ±1 is possible in 3.19), but yiels a egenerate istribution X = γa.)

9 STABLE DISTRIBUTIONS 9 For α 1, note the special cases β = 0 γ = 0 an α, 0 < α < 1, β = 1 γ = 2 α, 1 < α < ) Remark For α = 1, the general 1-stable characteristic function 3.6) may be written, similarly to 3.16), ϕt) = exp κ iλ)t ibt log t ), t > 0, 3.23) where κ = γ, λ = δ an b = 2 π βγ. Thus, b 2κ/π.) Definition A stable istribution is spectrally positive if its Lévy measure is concentrate on 0, ), i.e., Λx) = cx α 1 x, x > 0, 3.24) for some c > 0 an α 0, 2). By 3.3) an 3.12), this is equivalent to c = 0 an to β = 1, see also 3.22). A strictly stable istribution with characteristic function 3.16) is spectrally positive if an only if α 1 an λ = κ tan πα 2. Theorem Let 0 < α < 2. An α-stable ranom variable X S α γ, β, δ) has finite Laplace transform E e tx for t 0 if an only if it is spectrally positive, i.e., if β = 1, an then ) exp γα E e tx cos = πα t α δt, α 1, 2 ) exp 2 π γt log t δt 3.25), α = 1, Moreover, then 3.25) hols for every complex t with Re t 0. Proof. The conition for finiteness follows by Theorem 2.12 an 3.3), together with Definition When this hols, the right-han sie of 3.25) is a continuous function of t in the close right half-plane Re t 0, which is analytic in the open half-plane Re t > 0. The same is true for the left-han sie by Theorem 2.12, an the two functions are equal on the imaginary axis t = is, s R by 3.6) an a simple calculation. By uniqueness of analytic continuation, 3.25) hols for every complex t with Re t 0. Theorem An stable ranom variable X S α γ, β, δ) is positive, i.e. X > 0 a.s., if an only if 0 < α < 1, β = 1 an δ 0. Equivalently, X > 0 a.s. if an only if X = Y + δ where Y is spectrally positive strictly α-stable with 0 < α < 1 an δ 0. Proof. X > 0 a.s. if an only if the Laplace transform E e tx is finite for all t 0 an E e tx 0 as t. Suppose that this hols. We cannot have α = 2, since then X woul be normal an therefore not positive; thus Theorem 3.12 applies an shows that β = 1. Moreover, 3.25) hols. If 1 < α < 2 or α = 1, then the right-han sie of 3.25) tens to infinity as t, which is a contraiction; hence 0 < α < 1, an then 3.25) again shows that δ 0. The converse is immeiate from 3.25).

10 10 SVANTE JANSON Example If 0 < α < 1 an > 0, then X S α γ, 1, 0) with γ := ) cos πα 1/α is a positive strictly stable ranom variable with the Laplace 2 transform E e tx = exp t α). 3.26) Example If 1 < α < 2 an > 0, then X S α γ, 1, 0) with γ := cos πα 2 ) 1/α is a strictly stable ranom variable with the Laplace transform E e tx = exp t α). 3.27) Note that in this case cos πα 2 < 0. Note also that E e tx as t, which shows that PX < 0) > Other parametrisations. Our notation S α γ, β, δ) is in accorance with e.g. Samoronitsky an Taqqu [12, Definition an page 9]. Although they use the letters S α σ, β, µ).) Nolan [11] uses the notation Sα, β, γ, δ; 1); he also efines Sα, β, γ, δ 0 ; 0) := Sα, β, γ, δ 1 ; 1) where δ 0 βγ tan πα δ 1 := 2, α 1, δ 0 2 π βγ log γ, α = ) Note that our δ = δ 1.) This parametrisation has the avantage that the istribution Sα, β, γ, δ 0 ; 0) is a continuous function of all four parameters. Note also that Sα, 0, γ, δ; 0) = Sα, 0, γ, δ; 1), an that when γ = 0, 3.13) becomes γx + δ S α γ, β, δ; 0). Cf. the relate parametrisation in [12, Remark 1.1.4], which uses δ 1 + βγ α tan πα µ 1 = 2 = δ 0 + βγ α γ) tan πα 2, α 1, δ 1 = δ 0 2 π βγ log γ, α = 1; 3.29) again the istribution is a continuous function of α, β, γ, µ 1 ). Zolotarev [13] uses three ifferent parametrisations, with parameters enote α, β x, γ x, λ x ), where x A, B, M}; these are efine by writing the characteristic function 3.6) as where ϕt) = exp λ A itγa t α + itω A t, α, β A ) )) 3.30) = exp λ M itγm t α + itω M t, α, β M ) )) 3.31) = exp λ B itγb t α ω B t, α, β B ) )), 3.32) ω A t, α, β) := ω M t, α, β) := ω B t, α, β) := t α 1 β tan πα 2, α 1, β 2 π log t, α = 1; 3.33) t α 1 1)β tan πα 2, α 1, β 2 π log t, α = 1; 3.34) exp i π 2 βkα) sgn t), α 1, π 2 + iβ log t sgn t, α = 1, 3.35)

11 STABLE DISTRIBUTIONS 11 with Kα) := α 1 + sgn1 α), i.e., α, 0 < α < 1, Kα) := α 2, 1 < α < ) Here α is the same in all parametrisations an, with β, γ, δ is as in 3.6), β A = β M = β, 3.37) γ A = δ/γ α, 3.38) γ M = µ 1 /γ α = γ A + β tan πα 2, α 1, γ A, α = 1, 3.39) λ A = λ M = γ α, 3.40) an, for α 1, πkα) ) tan β B = β A tan πα 2 2 = β tan πα 2, 3.41) πkα) ) γ B = γ A cos β B, 3.42) 2 / πkα) ) λ B = λ A cos β B, 3.43) 2 while for α = 1, β B = β A = β, 3.44) γ B = π 2 γ A = πδ 2γ, 3.45) λ B = 2 π λ A = 2γ π. 3.46) Note that, for any α. β B = 0 β = 0 an β B = ±1 β = ±1, an that the mapping β = β A β B is an increasing homeomorphism of [ 1, 1] onto itself. In the strictly stable case, Zolotarev [13] also uses ϕt) = exp λ C e i sgnt)παθ/2 t α), 3.47) which is the same as 3.19) with λ C = a 3.48) θ = γ/α; 3.49) thus θ 1 for α 1 an θ 2/α 1 for α > 1. We also have Kα) β B θ = α, α 1, 2 π arctan2γ 3.50) B/π), α = 1. λb, α 1, λ C = λ B π 2 /4 + γb 2 ) 1/2, 3.51) α = 1.

12 12 SVANTE JANSON He also uses the parameters α, ρ, λ C where ρ := 1 + θ)/2. Zolotarev [13] uses Y α, β x, γ x, λ x ) as a notation for a ranom variable with the characteristic function 3.30) 3.32). The istribution is a continuous function of the parameters α, β B, γ B, λ B ). 4. Stable ensities A stable istribution has by 3.6) a characteristic function that ecreases rapily as t ±, an thus the istribution has a ensity that is infinitely ifferentiable. In the case α < 1 an β = 1, S α γ, β, δ) has support [δ, ) an in the case α < 1 an β = 1, S α γ, β, δ) has support, δ]; in all other cases the support is the entire real line. Moreover, the ensity function is strictly positive in the interior of the support, se Zolotarev [13, Remark 2.2.4]. Feller [3, Section XVII.6] lets, for α 1, px; α, γ) enote the ensity of the stable istribution with characteristic function 3.19) with a = 1. A stable ranom variable with the characteristic function 3.19) thus has the ensity function a 1/α pa 1/α x; α, γ). The ensity of a ranom variable X S α γ, β, δ) with α 1 is thus given by a 1/α p a 1/α x δ); α, γ ), 4.1) with a an γ given by 3.20) 3.21). Feller [3, Lemma XVII.6.1] gives the following series expansions. Theorem 4.1. i) If x > 0 an 0 < α < 1, then px; α, γ) = 1 πx k=1 ii) If x > 0 an 1 < α < 2, then px; α, γ) = 1 πx k=1 Γkα + 1) x α ) k sin kπ k! 2 Γ1 + k/α) k! For x < 0 we use p x; α, γ) = px; α, γ). In particular, if 1 < α < 2, then p0; α, γ) = 1 π γ α). 4.2) x) k sin kπ γ α). 4.3) 2α πα γ) Γ1 + 1/α) sin. 4.4) 2α In the special case β = 1 we have γ = 2 α by 3.22) an p0; α, 2 α) = 1 πα 1) Γ1 + 1/α) sin = 1 π α π Γ1 + 1/α) sin π α Γ1 + 1/α) = Γ1/α)Γ1 1/α) = 1 αγ1 1/α) = 1 Γ 1/α). 4.5)

13 STABLE DISTRIBUTIONS 13 For 1 < α < 2, the istribution S α γ, 1, 0) thus has, by 4.1) an 3.20), the ensity at x = 0 a 1/α a p0; α, 2 α) = Γ 1/α) = γ 1 cos πα 1/α Γ 1/α) ) 2 5. Domains of attraction Definition 5.1. A ranom variable X belongs to the omain of attraction of a stable istribution L if there exist constants a n > 0 an b n such that S n b n a n L 5.1) as n, where S n := n i=1 X i is a sum of n i.i.. copies of X. We will in the sequel always use the notation S n in the sense above as we alreay have one in Section 3). All unspecifie limits are as n. Theorem 5.2. Let 0 < α 2. A non-egenerate) ranom variable X belongs to the omain of attraction of an α-stable istribution if an only if the following two conitions hol: i) the truncate moment function µx) := E X 2 1 X x} ) 5.2) varies regularly with exponent 2 α as x, i.e., where L 1 x) varies slowly; ii) either α = 2, or the tails of X are balance: for some p + [0, 1]. Proof. Feller [3, Theorem XVII.5.2]. µx) x 2 α L 1 x), 5.3) PX > x) P X > x) p +, x, 5.4) For the case α < 2, the following version is often more convenient. Theorem 5.3. Let 0 < α < 2. A ranom variable X belongs to the omain of attraction of an α-stable istribution if an only if the following two conitions hol: i) the tail probability P X > x) varies regularly with exponent α as x, i.e., P X > x) x α L 2 x), 5.5) where L 2 x) varies slowly; ii) the tails of X are balance: for some p + [0, 1]. PX > x) P X > x) p +, x, 5.6)

14 14 SVANTE JANSON Proof. Feller [3, Corollary XVII.5.2]. We turn to ientifying the stable limit istributions in Theorems explicitly The case α < 2. If the conitions of Theorem 5.2 or 5.3 hol for some α < 2, then the conitions of the other hol too, an we have, by [3, 5.16)], L 2 x) 2 α α L 1x), x. 5.7) Furthermore, by [3, 5.6)], with a n, b n as in 5.1) an M an Λ the canonical measure an Lévy measure of the limit istribution L, an, by symmetry, n PX > a n x) Λx, ) = n PX < a n x) Λ, x) = In particular, x x y 2 My), x > 0, 5.8) y 2 My), x > ) n P X > a n ) Λy : y > 1} 0, ); 5.10) conversely, we may in 5.1) choose any sequence a n ) such that n P X > a n ) converges to a positive, finite limit. Any two such sequences a n ) an a n) must satisfy a n /a n c for some c 0, ), as a consequence of 5.5).) If n P X > a n ) C > ) an 5.5) 5.6) hol, then 5.8) 5.9) hol with Λx, ) = p + Cx α an Λ, x) = p Cx α, where p := 1 p +. Hence, 3.2) 3.3) hol with c + = p + Cα, c = p Cα. 5.12) Consequently, the limit istribution is given by 3.6) where, by 3.11) 3.12), γ = Cα Γ α) cos πα )) 1/α ) 2 = CΓ1 α) cos πα 1/α, ) β = p + p. 5.14) For α = 1 we interpret 5.13) by continuity as γ = C π ) Theorem 5.4. Let 0 < α < 2. Suppose that 5.5) 5.6) hol an that a n are chosen such that 5.11) hols, for some C. Let γ an β be efine by 5.13) 5.14). i) If 0 < α < 1, then S n a n S α γ, β, 0). 5.16)

15 ii) If 1 < α < 2, then iii) If α = 1, then STABLE DISTRIBUTIONS 15 S n n E X a n S n nb n a n where γ is given by 5.15) an S α γ, β, 0). 5.17) S 1 γ, β, 0), 5.18) b n := a n E sinx/a n ). 5.19) Proof. Feller [3, Theorem XVII.5.3] together with the calculations above. Example 5.5. Suppose that 0 < α < 2 an that X is a ranom variable such that, as x, PX > x) Cx α, 5.20) with C > 0, an PX < x) = ox α ). Then 5.5) 5.6) hol with L 2 x) := C an p + = 1, an thus p := 1 p + = 0. We take a n := n 1/α ; then 5.11) hols, an thus 3.2) 3.3) hol with hence, 5.13) 5.14) yiel c + = Cα, c = 0; 5.21) γ = CΓ1 α) cos πα 2 an β = 1. Consequently, Theorem 5.4 yiels the following. i) If 0 < α < 1, then S n n 1/α ) 1/α, 5.22) S α γ, 1, 0). 5.23) The limit variable Y is positive an has by Theorem 3.12 an 5.22) the Laplace transform ii) If 1 < α < 2, then E e ty = exp CΓ1 α)t α), Re t ) S n n E X n 1/α S α γ, 1, 0). 5.25) The limit variable Y has by Theorem 3.12 an 5.22) the finite Laplace transform E e ty = exp C Γ1 α) t α), Re t ) By 4.6) an 5.22), the ensity function f Y of the limit variable satisfies f0) = C 1/α Γ1 α) 1/α Γ 1/α) )

16 16 SVANTE JANSON iii) If α = 1, then S n nb n = S n n n b n S 1 γ, 1, 0), 5.28) where, by 5.15), γ = Cπ/2 an b n := n E sinx/n). 5.29) We return to the evaluation of b n in Section 5.2. Example 5.6. Suppose that 0 < α < 2 an that X 0 is an integer-value ranom variable such that, as n, PX = n) cn α ) Then 5.20) hols with C = c/α 5.31) an the results of Example 5.5 hol, with this C. In particular, 5.22) yiels an both 5.24) an 5.26) can be written γ α = c Γ α) cos πα 2, 5.32) E e ty = exp c Γ α)t α), Re t 0; 5.33) note that Γ α) < 0 for 0 < α < 1 but Γ α) > 0 for 1 < α < 2. Taking t imaginary in 5.33), we fin the characteristic function E e ity = exp c Γ α) it) α) = exp c Γ α)e i sgnt)πα/2 t α), t R. 5.34) 5.2. The special case α = 1. Suppose that, as x, PX > x) Cx ) an PX < x) = ox 1 ), with C > 0. Then Example 5.5 applies, an 5.28) 5.29) hol. We calculate the normalising quantity b n in 5.28) for some examples. Example 5.7. Let X := 1/U, where U U0, 1) has a uniform istribution. Then PX > x) = x 1 for x 1 so 5.35) hols with C = 1 an 5.15) yiels γ = π/2. Furthermore, X has a Pareto istribution with the ensity x 2, x > 1, fx) = 5.36) 0, x 1. Consequently, by 5.29), b n = n sinx/n) = n 1 = log n + = log n + 1/n 0 1 sin y y y 2 y + sinx/n)x 2 x = 1 sin y y1y < 1} y 2 sin y y 2 y 1/n siny)y 2 y y + o1) = log n + 1 γ + o1),

17 STABLE DISTRIBUTIONS 17 where γ is Euler s gamma. For the stanar evaluation of the last integral, see e.g. [7].) Hence, 5.28) yiels or S n n log n + 1 γ ) S 1 π/2, 1, 0). 5.37) S n n log n S 1 π/2, 1, 1 γ). 5.38) Example 5.8. Let X := 1/Y, where Y Exp1) has an exponential istribution. Then PX > x) = 1 exp 1/x) x 1 as x so C = 1 an 5.15) yiels γ = π/2. In this case we o not calculate b n irectly from 5.29). Instea we efine U := 1 e Y an X := 1/U an note that U has a uniform istribution on [0, 1] as in Example 5.7; furthermore X X = e Y 1 Y = e Y 1 + Y 1 e Y )Y. 5.39) This is a positive ranom variable with finite expectation EX e y 1 + y e y X) = 1 e y )y e y y = see e.g. [2, ] or [7]. 0 e y 1 e y y ) y = γ, 5.40) Taking i.i.. pairs X i, X i ) = X, X ) we thus have, with S n := n i=1 X i, by the law of large numbers, S n S n n Since Example 5.7 shows that S n/n log n that We thus have 5.28) with S n /n log n p EX X) = γ. 5.41) S 1 π/2, 1, 1 γ), it follows S 1 π/2, 1, 1 2 γ). 5.42) b n = log n γ + o1). 5.43) 5.3. The case α = 2. If α = 2, then a n in 5.1) have to be chosen such that nµa n ) a 2 C 5.44) n for some C > 0, see [3, 5.23)]; conversely any such sequence a n ) will o. Theorem 5.9. If µx) is slowly varying with µx) as x an 5.44) hols, then S n E S n a n Proof. Feller [3, Theorem XVII.5.3]. N0, C). 5.45)

18 18 SVANTE JANSON Example Suppose that α = 2 an that X is a ranom variable such that, as x, PX > x) Cx 2, 5.46) with C > 0, an PX < x) = ox 2 ). Then 5.4) hols with p + = 1, an thus p := 1 p + = 0. Furthermore, as x, X ) x µx) = E 2t t 1 X x} = E 1t X x}2t t = 0 x 0 2t Pt X x) t = x 0 0 2t P X > t) t x 2 P X > x) = 1 + o1) ) x 2tCt 2 t + O1) 2C log x. 5.47) 1 Thus 5.3) hols with L 1 x) = 2C log x. We take a n := n log n. Then µa n ) 2C 1 2 log n = C log n, so 5.44) hols an Theorem 5.9 yiels S n E S n n log n N0, C). 5.48) 6. Attraction an characteristic functions We stuy the relation between the attraction property 5.1) an the characteristic function ϕ X t) of X. For simplicity, we consier only the common case when a n = n 1/α. Moreover, for simplicity we state results for ϕ X t), t > 0 only, recalling 3.15) an ϕ X 0) = 1. Theorem 6.1. Let 0 < α 2. The following are equivalent. i) S n n 1/α Z for some non-egenerate ranom variable Z. ii) The characteristic function ϕ X of X satisfies ϕ X t) = 1 κ iλ)t α + ot α ) as t 0, 6.1) for some real κ > 0 an λ. In this case, Z is strictly α-stable an has the characteristic function 3.16). Hence, λ κ tan πα 2.) Proof. If i) hols, then for every integer m, S mn mn) 1/α = 1 m 1 n m 1/α n 1/α X k 1)n+j 1 m 1/α k=1 j=1 m Z k, as n, with Z k = Z i.i.. Since also mn) 1/α S mn Z, we have m 1/α m k=1 Z k = Z, an thus Z is strictly α-stable. We use Corollary 3.8 an suppose that Z has characteristic function 3.16). Then the continuity theorem yiels k=1 ϕ X t/n 1/α ) n ϕ Z t) = exp κ iλ)t α), t 0; 6.2) moreover, this hols uniformly for, e.g., 0 t 1.

19 STABLE DISTRIBUTIONS 19 In some neighbourhoo t 0, t 0 ) of 0, ϕ X 0 an thus ϕ X t) = e ψt) for some continuous function ψ : t 0, t 0 ) C with ψ0) = 0. Hence, 6.2) yiels for n > 1/t 0 ) t ) exp nψ n 1/α + κ iλ)t α) = 1 + o1), as n, uniformly for 0 t 1, which implies t ) nψ n 1/α + κ iλ)t α = o1), as n, since the left-han sie is continuous an 0 for t = 0, an thus ) t ψ n 1/α + κ iλ) tα = o1/n), as n, 6.3) n uniformly for 0 t 1. For s > 0, efine n := s α an t := sn 1/α 0, 1]. As s 0, we have n an 6.3) yiels ψs) = κ iλ)s α + o1/n) = κ iλ)s α + os α ). 6.4) Consequently, as s 0, ϕ X s) = e ψs) = 1 κ iλ)s α + os α ), 6.5) so 6.1) hols. Conversely, if 6.1) hols, then, for t > 0, E e itsn/n1/α = ϕ X t/n 1/α ) n = 1 κ iλ + o1)) tα n ) n exp κ iλ)t α ), as n, an thus by the continuity theorem S n /n 1/α Z, where Z has the characteristic function 3.16). For α = 1, it is not always possible to reuce to the case when b n = 0 in 5.1) an the limit is strictly stable. The most common case is covere by the following theorem. Theorem 6.2. The following are equivalent, for any real b. i) S n n b log n Z for some non-egenerate ranom variable Z. ii) The characteristic function ϕ X of X satisfies ϕ X t) = 1 κ iλ)t ibt log t + ot) as t 0, 6.6) for some real κ > 0 an λ. In this case, Z is 1-stable an has the characteristic function 3.23). Hence, b 2κ/π.) Proof. ii) = i). If 6.6) hols, for any κ R, then, as t 0, log ϕ X t) = κ iλ + o1))t ibt log t 6.7)

20 20 SVANTE JANSON an thus, as n, for every fixe t > 0, E e itsn/n b log n) ) ne = ϕ X t/n ibt log n = exp n κ iλ + o1)) t n ib t n log t ) ) ibt log n n exp κ iλ)t ibt log t ) which shows i), where Z has the characteristic function 3.23). Furthermore, for use below, note that 3.23) implies ϕ Z t) = e κt for t > 0. Since ϕ Z t) 1, this shows that κ 0. Moreover, if κ = 0, then ϕ Z t) = 1 for t > 0, an thus for all t, which implies that Z = c a.s. for some c R, so Z is egenerate an b = 0. Hence, 6.6) implies κ 0, an κ = 0 is possible only when b = 0 an S n /n p λ. i) = ii). Let γ 1 := b π/2 an β 1 := sgn b. Let Y an Y i be i.i.., an inepenent of X j ) 1 an Z, with istribution S 1 γ 1, β 1, 0). If b = 0 we simply take Y i := 0.) Then Y i has, by 3.6), the characteristic function ϕ Y t) = exp γ 1 t + ibt log t ), t > ) By Theorem 3.3ii), n Y i = ny bn log n. 6.9) i=1 Define X i := X i + Y i. Then, 1 n n i=1 X i = 1 n n X i + 1 n i=1 n i=1 Thus, by Theorem 6.1, for some κ 2 > 0 an λ 2, S n Y i = n + Y b log n Z + Y. 6.10) ϕ X t)ϕ Y t) = E e it X i = 1 κ 2 iλ 2 )t + ot) as t 0, 6.11) an hence, using 6.8), ϕ X t) = E e it X i /ϕ Y t) = 1 κ 2 iλ 2 γ 1 )t ibt log t + ot), 6.12) which shows 6.6), with κ = κ 2 γ 1 R. Finally, we have shown in the first part of the proof that 6.6) implies κ > 0, because Z is non-egenerate. We can use these theorems to show the following. Theorem 6.3. Let 0 < α 2. Suppose that X is such that n 1/α n i=1 X i Z, 6.13)

21 STABLE DISTRIBUTIONS 21 where Z is an α-stable ranom variable with characteristic function 3.16) an that Y 0 is a ranom variable with E Y α <. Let Y i ) 1 be inepenent copies of Y that are inepenent of X i ) 1. Then n n 1/α X i Y i Z := E Y α) 1/α Z, 6.14) i=1 where the limit Z has the characteristic function ϕ Z t) = exp E Y α κ i E Y α λ)t α), t ) If Z S α γ, β, 0) where β = 0 if α = 1), then Z S α E Y α ) 1/α γ, β, 0). Proof. By Theorem 6.1, for t 0, ϕ X t) = 1 κ iλ)t α + t α rt), 6.16) where rt) 0 as t 0. Furthermore, 6.16) implies that rt) = O1) as t, an thus rt) = O1) for t 0. Consequently, for t > 0, assuming as we may that Y is inepenent of X, ϕ XY t) = E e itxy = E ϕ X ty ) = E 1 κ iλ)t α Y α + t α Y α rty ) ) = 1 κ iλ)t α E Y α + t α E Y α rty ) ), 6.17) where E Y α rty ) ) 0 as t 0 by ominate convergence; hence ϕ XY t) = 1 κ iλ)t α E Y α + ot α ) as t ) Theorem 6.1 applies an shows that n 1/α n i=1 X iy i Z, where Z has the characteristic function 6.15). Moreover, by 3.16), E Y α ) 1/α has this characteristic function, so we may take Z := E Y α ) 1/α. The final claim follows by Remark 3.6. Theorem 6.4. Suppose that X is such that, for some real b, n n 1 X i b log n Z, 6.19) i=1 where Z is a 1-stable ranom variable, an that Y 0 is a ranom variable with E Y log Y <. Let Y i ) 1 be inepenent copies of Y that are inepenent of X i ) 1. Then, with µ := E Y, n n 1 X i Y i bµ log n Z := µz b EY log Y ) µ log µ ). 6.20) i=1 Z has the characteristic function 3.23) for some κ an λ, an then the limit Z has the characteristic function, with ν := EY log Y ), ϕ Z t) = exp µκ + ibν µλ)t ) ibµtlog t ), t > ) If Z S 1 γ, β, δ), then Z S 1 µγ, β, µδ bν).

22 22 SVANTE JANSON Proof. By Theorem 6.2, for t 0, ϕ X t) = 1 κ iλ)t ibt log t + trt), 6.22) where rt) 0 as t 0; moreover Z has the characteristic function 3.23). Furthermore, 6.22) implies that rt) = Olog t) as t, an thus rt) = O1 + log + t) for t 0. Consequently, for t > 0, assuming as we may that Y is inepenent of X, ϕ XY t) = E ϕ X ty ) = 1 κ iλ)t E Y ibt E Y logty ) ) + t E Y rty ) ), = 1 µκ iµλ + ib EY log Y ) ) t ibµtlog t + t E Y rty ) ), where E Y rty ) ) 0 as t 0 by ominate convergence; hence ϕ XY t) = 1 µκ iµλ + ibν ) t ibµtlog t + ot) as t ) Theorem 6.2 applies an shows that n 1 n i=1 X iy i bµ log n Z, where Z has the characteristic function 6.21). Moreover, it follows easily from 3.23) that µz b EY log Y ) µ log µ ) has this characteristic function, an thus 6.20) follows. Finally, if Z S 1 γ, β, δ), then b = 2 π βγ by Remark 3.10 an it follows easily from Remark 3.6 that Z S 1 µγ, β, µδ bν); alternatively, it follows irectly from 6.20) an 3.6) that Z has the characteristic function ϕ Z t) = ϕ Z µt) exp ibtν µ log µ) ) = exp γµ t 1 + iβ 2 ) ) π sgnt) log t + iδµt ibtν. 6.24) Example 6.5. Let X := U/U, where U, U U0, 1) are inepenent. By Example 5.7 an Theorem 6.4, with Z S 1 π/2, 1, 1 γ), b = 1, µ := E U = 1/2 an ν := E U log U = 1 0 x log x x = 1 4, 6.25) we obtain S n n 1 2 log n 1 2 Z ν log 1 2 = 1 2 Z π 2 log 2 S 1 4, 1, 3 4 γ ) ) Example 6.6. Let X := Y/Y where Y, Y Exp1) are inepenent. Thus X has the F -istribution F 2,2.) By Example 5.8 an Theorem 6.4, with Z S 1 π/2, 1, 1 2 γ), b = 1, µ := E Y = 1 an ν := E Y log Y = 0 x log x e x x = Γ 2) = 1 γ, 6.27) we obtain S n n log n Z ν = Z 1 + γ S 1 π/2, 1, γ). 6.28)

23 STABLE DISTRIBUTIONS 23 This is in accorance with Example 5.7, since, as is well-known, U := Y /Y +Y ) U0, 1), an thus we can write X = Y +Y )/Y 1 = 1/U 1. Example 6.7. Let X := V 2 /W where V U 1 2, 1 2 ) an W Exp1) are inepenent. By Example 5.8 an Theorem 6.4, with Z S 1 π/2, 1, 1 2 γ), b = 1, µ := E V 2 = 1/12 an 1/2 [ ] x ν := 2 E V 2 log V = 4 x 2 3 1/2 x3 log x x = 4 log x = 3 log 2 + 1, 6.29) we obtain S n n 1 12 log n Z ν log 1 π 12 S 1 24 Equivalently, using Remark 3.6, 5 6 γ + 6 log 2 ), 1,. 6.30) 36 24S n πn 2 π log n 2 π Z 24ν π 2 π log 12 S 1 1, 1, 2 5 π 3 2 γ + log π )) ) This is shown irectly in Heinrich, Pukelsheim an Schwingenschlögl [5, Theorem 5.2 an its proof]. Example 6.8. More generally, let X := V 2 /W where V Uq 1, 1) an W Exp1) are inepenent, for some fixe real q. By Example 5.8 an Theorem 6.4, with Z S 1 π/2, 1, 1 2 γ), b = 1, an µ = E V 2 = E V ν := 2 E V 2 log V = 2 ) 2 + Var V = q 1 2 q q 1 = 2 q3 log q + 1 q) 3 log 1 q 3 we obtain S n n µ log n Equivalently, using Remark 3.6, ) = 3q2 3q [ ] x x 2 3 q x3 log x x = 2 log x 3 9 q ) 2 3q2 3q + 1, 6.33) 9 µz ν + µ log µ S 1 µ π ). 2, 1, 1 2 γ)µ ν 6.34) S n ne V ) 2 log n Z ν E V )2 + log µ µn µ µ with b q := γ 2q3 log q + 1 q) 3 log 1 q 3q 2 3q + 1 S 1 π 2, 1, b q + log 3q2 3q ), 6.35) µ. 6.36) This is shown in the case 0 q 1) irectly in Heinrich, Pukelsheim an Schwingenschlögl [6, Theorem 4.2 an its proof].

24 24 SVANTE JANSON References [1] J. Bertoin, Lévy Processes. Cambrige University Press, Cambrige, [2] Digital Library of Mathematical Functions. Version 1.0.3; National Institute of Stanars an Technology. gov/ [3] W. Feller, An Introuction to Probability Theory an its Applications, Volume II, 2n e., Wiley, New York, [4] A. Gut, Probability: A Grauate Course. Springer, New York, [5] L. Heinrich, F. Pukelsheim & U. Schwingenschlögl, Sainte-Laguë s chisquare ivergence for the rouning of probabilities an its convergence to a stable law. Statistics & Decisions ), [6] L. Heinrich, F. Pukelsheim & U. Schwingenschlögl, On stationary multiplier methos for the rouning of probabilities an the limiting law of the Sainte-Laguë ivergence. Statistics & Decisions ), [7] S. Janson, Some integrals relate to the Gamma integral. Notes, [8] S. Janson, Simply generate trees, conitione Galton Watson trees, ranom allocations an conensation. Preprint, math.uu.se/~svante/papers/#264 [9] O. Kallenberg, Founations of Moern Probability. 2n e., Springer, New York, [10] J. Marcinkiewicz, Sur les fonctions inépenants III. Fun. Math ), [11] J. P. Nolan, Stable Distributions - Moels for Heavy Taile Data. Book manuscript in progress); Chapter 1 online at acaemic2.american.eu/ jpnolan. [12] G. Samoronitsky & M. S. Taqqu, Stable Non-Gaussian Ranom Processes. Chapman & Hall, New York, [13] V. M. Zolotarev, One-imensional Stable Distributions. Nauka, Moscow, Russian.) English transl.: American Mathematical Society, Provience, RI, Department of Mathematics, Uppsala University, PO Box 480, SE Uppsala, Sween aress: svante.janson@math.uu.se URL: svante/