Chapter STABLE DISTRIBUTIONS ; ANALYTICAL PROPERTIES AND DOMAINS OF ATTRACTION § Stable distributions Definition A distribution function F is called stable if, for any a , a2 >0 and any b1 , b2 , there exist constants a > and b such that F(a l x+b l ) * F(a x+b2) = F(ax+b) (2.1 1) It clearly suffices to take b =b =0 Then in terms of the characteristic function f of F, (2.1 1) becomes f(t/al)f(t/a2) = f(t/a)e-`6` (2.1 2) Interest in the stable distributions is motivated by the fact that, under weak assumptions, they are the only possible limiting distributions of normed sums Zn= Xl+X2+ + Xn _ An Bn (2.1 3) of stationarily dependent random variables In this section we establish this result for independent random variables ; the general case is dealt with in Theorem 18 1 Theorem 1 In order that a distribution function F be the weak limit of the distribution of Z n for some sequence (Xi) of independent identically distributed random variables, it is necessary and sufficient that F be stable If this is so, then unless F is degenerate, the constants B n in (2.1 3) must take the form B n = n' lx h (n), where 0) (2.2.7) * Equation (2 3) in the original is identical to (1 1) 2.2 CANONICAL REPRESENTATION OF STABLE LAWS 41 Suppose that M is not identically zero, and write m(x) =M(e - "), (-co From (2.2 10) these are equal, and A (s) is defined as a non-increasing continuous function on s>0, satisfying m {x + (s) } = sm (x) (2.2.11) Moreover, it follows from this equation that lim (s) = oo , lim A (S) S-0 S-00 _ - co Since m is not identically zero, we may assume that m (0) =A (otherwise shift the origin), and write m1(x)=m(x)/m(0) Let X1, x2 be arbitrary, and choose s , s2 so that 42 STABLE DISTRIBUTIONS )'(s1) Then = x1 , ~,( s ) Chap = x2 s1 m(0) = m(x 1), s2 m(0) = m(x 2), s2 m(x ) = m(x +x2 ) , so that (2.2.12) m1(x1+x2) = m1(x1)m1(x2) Since m is non-negative, non-increasing and not identically zero, (2.2.12) shows that ml > 0, and then m = 109m, is monotonic and satisfies M2 (X1 +x2) = m2(x1)+m2(x2) (2 2.13) It is known (see for example [50], page 106) that the only monotonic functions satisfying this equation are of the form m (x) = ax Since M (- oo) = 0, this implies that m1(x) = e -"x a>0, c1 >0 M(u)=c1(-u)-", As the integral ~- u2dM(u)=cla j0 u 1- "du must converge, we have a < Thus finally M(u)=c1(-u)-", 0 and, for large y > 0, take n so that Bnx, x z h (x) dx > i h (z) log (z/z 1) , ~zj x and so Z H (z) = h(x) dx(1+o(1)) Jo For any k > 0, as z-± oc, kz z h (x) x dx < log kh (z) _ = o0z h (z) dx o x so that lim Z- 00 H( zz) = () Collecting these results together, the theorem is proved It is clear that, when the variance is finite, H (z) will be slowly varying, and thus the theorem may be expressed in the following way The distribution function F (x) belongs to the domain of attraction of a normal law if and only if z H (z) = c2 dF (x) (2.6.21) -Z is a slowly varying function STABLE DISTRIBUTIONS 84 Chap I t may be shown in a similar way that the conditions of Theorem 6.1 are satisfied if and only if the function H(z) IxladF(x) (2.6.22) = Sy -Z is slowly varying, and lim z xa dF (x) Jo o (-x)' dF (x) cl (2.6 23) C2 -z These conditions imply that a h(x) ti x cl +C2 x(x) = o {H(x)} (2.6 24) Conversely, the methods used in the proof of Theorem can be used to show that (2 6.24) implies (2.6.2), (2.6.22) and (2.6.23) This permits a unification of Theorems 6.1 and 2.6.2 Theorem 2.6.3 In order that a distribution function F (x) belong to the domain of attraction of a stable law with index a, it is necessary and sufficient that Z Jim Z" GO z2 x (z) o x dx (z) = 2-a a Theorem 6.4 If F(x) belongs to the domain of attraction of a stable law with index a, then for any b (0 < < a), ~00 Ixj a dF(x) < oo - 00 Proof The result is obvious if the variance is finite If it is infinite and a < 2, then Theorem 6.1, together with the results of Appendix 1, shows that x(x) = 1-F(x) +F(-x) = o0xI - a + E} for any E>0 Taking is sufficiently small, we have DOMAINS OF ATTRACTION c Go o Ixla dF (x) = 2b xb -1 x (x) dx Go o (x < o 85 E) dx + < oo x fo If d = 2, use the formula (2 21) Theorem 2.6.5 In order that the distribution with characteristic function f (t) belong to the domain of attraction of the stable law whose characteristic function has logarithm -cItl" w(t, a) , ~tI where a, j3, c, w (t, a) are as in Theorem 2.2.1, it is necessary and sufficient that, in the neighbourhood of the origin, log f (t) = iyt - c I tI" h (t) 1- i(3 C tI (wt, a) , where y is a constant, and h(t) is slowly varying as t-0 Proof To prove necessity, first note that, in the neighbourhood of the origin, logf(t)=log{1+(f(t)-1)}= = If (t)_11 +0(If(t)-ll2), where that branch of the function log is taken with log = If, for x >, 0, G1 (x) = 1-F(x) , G2 (x) = F(-x), then (1-e."x)dF(x)= 1-f(t)= 00 = So 00 (e"x -1)dG (x) + J o (e-"x-1)dG2(x) (2 25) The asymptotic behaviour of G and G2 for large x is given by (2 6.2) ; from this we deduce the behaviour of their Fourier transforms, and thus that of 1- f(t), as t +O Further calculations depend on the value of a ; we distinguish four cases (1) < a < If suffices to examine the first integral on the right-hand side of (2.6.25) 86 STABLE DISTRIBUTIONS Chap Integrating by parts, we have 00 e"xG (x)dx= it ` ~0 (eitx_l)dGi(x)= 00 _ - Iti" h, sin x x /ItI 00 sgn t I ti- + i cos x o dx + hl x/ I tl) dx x where, by Theorem 1, h (x) is slowly varying as x ; oo (We are assuming, without loss of generality, that c =A 0.) The analysis of these integrals requires the following lemma Lemma 6.1 If h(x) is a positive slowly varying function (as x-+oo), and x - "h(x) is monotone decreasing, then as t-+0, °° o °° I sin x dx sin cos x x sin x h (x/t) " dx - h(t-1) x t cos x (2.6.26) Proof of the lemma Consider for example the integral involving sin x, and split it into four parts a dl f JAl +S + a dzr + dzt sin x h (x/t) dx x By the second mean value theorem, OC) Jd h (x/t) sin x dx x A lim A-• o0 sin x d h (x/t) dx x h(1/t) h(dlt) d -" sup ( /) A' 4h(1/t)A - " , since as t >O for fixed d, ~(dl~) (/) ~ A' sin x dx d (2 6.27) DOMAINS OF ATTRACTION 87 From (2.6.19), for all x E (b, d20, h(x/t) E 0 ; 2(t) = c/n} , (this definition being meaningful for large neighbourhood of t=0) Then n, since (t) is continuous in a lim f (t/B ,)n = n- o0 = lim exp n -~ 00 - n2 ~ ( )A ( (1/)) /B n Bn t + i/3 ( ) w (t, a) _ II =exp - c I t I" (1+iflw(t)) I ,a I t (2 6.40) t, and the theorem is proved It was shown in § 2 that the normalising constants B n determining attraction to a stable law of index a were necessarily of the form Bn = n is h (n) , where h (n) is slowly varying The classical theorems of probability (de Moivre-Laplace-Levy) show that, for convergence to the normal law, the most interesting case is that in which Bn =an= for a constant 92 STABLE DISTRIBUTIONS Chap On the other hand, any stable law G of exponent a belongs to its own domain of attraction, with Bn =an l« This suggests the following definition A distribution belongs to the normal domain of attraction of a stable law G with exponent a if it is in the domain ofattraction ofG and if the normalising constants are given by Bn = an l/ « ~ where a is a constant Normal domains of attraction are characterised by the following theorems Theorem 2.6 In order that the distribution F(x) belong to the normal domain of attraction of the normal distribution x O(x) _ (2~)- ~ - e - - "2 du , 00 it is necessary and sufficient that it have finite variance a , and then B n = a wi- Proof The sufficiency follows from Levy's theorem To prove the necessity take Bn = ani- and assume without loss of generality that cc ~- x dF (x) = It then follows from Theorems 2.6.2, 6.5 and equation (2.6.39) that Jim n oo t2 an1 1t2 H Il ) - t 2a2 ( This is only possible if 00 00 x dx (x) = H (oo) _ o x2 dF (x) = a2 < oo , -00 and a=u Theorem 2.6.7 I n order that the distribution F (x) belong to the normal domain of attraction of the stable law G (x) with exponent a (0 < a < 2) and given constants c , c2 , with B n =and, it is necessary and sufficient that DOMAINS OF ATTRACTION F(x) =(c1a"+a1(x))Ixl-", F (x) =1 - (c2 a" + a2 (x)) x -" , (x ) , 93 (2 41) where (x) >0 as jxI-* oo Proof The sufficiency is immediate To prove the necessity, note that from (2.6.35), for small t, limx(an l1"Itt -1 )a - "jtj"n=It{", t-0 which is only possible if (2 6.41) holds ... a(0)=(a- 1-1 ) +2( 1-a) 02+ b(0), where 00 ( b(~) _ - n )n n-3 {r(1-a)} In (2 35) substitute -z for to give -1 -1 )} Re )_(1-a) -2 r exp{-r(a Y''0 i rz{r(1 - a)} x xexp { -2 4 '' 2- rb[{r(1-a)}-Z~]} e -ir - (2 4.36)... (2. 4 .29 ) p(x ;a, -1 )=0, and p(x ; a, 1)'' {2n (1 -a)}-gal /2( 1-a)x- 1-1 /2( 1 -a) exp {-( 1-a)(a/x)"/( 1-? ?)} x z (1+ a -n /2( 1-a) anx an /2 (1-a) '' X C2 - (2. 4.30) n=1 where 00 an = Re f0 n(o) e 10zdo , and. .. - 20 on do+0 (e -grz2/s) so that N 9M = -2 0e -q ~ Z n=0 q- -2 n L Cn J0 00 n=1 (0) e -2 , ,2 + = (Yj -2 ( N+ 1)) anq -2 n +0( ~ -2 (N+1)) , 7~) (2 I (2. 4 .23 ) Collecting together (2 13), (2 4.14) and