Independent And Stationary Sequences Of Random Variables - Chapter 9 ppsx

Zones of normal attraction In this chapter it will be assumed that the independent random variables X; satisfy EX;=0, VXX=62>0, and that Sn = X1 +X2+.. The method discussed in this chapt

Chapter 9

MONOMIAL ZONES OF LOCAL NORMAL A 11 RACTION

„ 1 Zones of normal attraction

In this chapter it will be assumed that the independent random variables X; satisfy

E(X;)=0, V(XX)=62>0, and that

Sn = X1 +X2+ +X,, Zn = Sn / an-1

We shall also suppose that the XXbelong to the class(d) ofvariables having

a bounded continuous probability density g (x) The method discussed

in this chapter may be used under less stringent conditions on g (x), and also for lattice variables, but we restrict attention to (d)for simplicity

ofpresentation Let 0(n) be any function increasing to infinity The seg-ments [0, 0(n)] will be called a zone of (integral) normal attraction if, uniformly in xc- [0, 0 (n)] as n-* oo,

P (Zn >x )/( 27r )-4 J§§ e j"2 du +I

x

(9 1 1)

If it is desired to emphasise the uniformity, the phrase "zone ofuniform normal attraction" may be used A similar definition holds for zones of normal attraction of the form [ - 0 (n), 0]

When Z© has a probability density pn (x), we can similarly define a zone

of local normal attraction as a sequence of segments [0, c (n)], in which P1 (x)/(27r) -1e 2x2 -' 1

(9 1 2) uniformly in x

It will be seen later that a special role is played by the zones delimited by

MONOMIAL ZONES OF LOCAL NORMAL ATTRACTION

Chap 9

J,(n) = o (n ) ;

(9 1 3) such zones are said to be narrow. Zones of the form [0, n"] (or [ -n", 0] ) are called monomial.

In what follows, 6 1, 62 , ; E1, E2, ; y11, r12,

CO, ~1, C2, are small and positive, each one depending on its predecessors, c o , c1 , ;

C0, C1 , ; K o , K 1 , are positive constants similarly chosen, B is bounded and varies from one expression to the next, and p( n), p 1 (n), p 2 (n) are positive functions converging to 0o as n-* oo

In this chapter we study monomial zones of local normal attraction, both narrow and wide

„ 2 The fundamental conditions Theorem 9 2 1 Let 0 < a < 2 Then the condition

E exp(IX;14a/(2 a+1) ) < 00

(9 2 1)

is necessary for [0, no 'p (n)], [ - no 'p (n), 0] to be zones of local normal attraction

Proof. Write i3=4a/(2a+ 1) Suppose that (9 2.1) does not hold Then there exists a sequence x,©-+ oo such that

P (X1 > x,©) > exp (- 2x©,)

(9 2 2) for all m, or

P (X1 < - x©,) > exp (-2xm)

(9 2 3) for all m Suppose that (9.2 2) holds For sufficiently large m, choose n so that

x,© = an-j + "p(n)+0 ,

101 < 1 Since [0, no'p (n)] is a zone of normal attraction,

P (Z© > 2 n" p (n)) < exp(-116 n 2a p (n)2)

.

(9 2 4) But the event {Z© > -in" P (n) } will certainly occur if the independent events {X1 > 6n " p(n)+01 and {1(X2+ X3 + + X©)/ oni- 1 < 1 } both occur Hence, by the central limit theorem and (9 2 2),

9 2

THE FUNDAMENTAL CONDITIONS

1 79

P(Z,, > in"p(n)) > c o P(X I > xm ) > c o exp(-c1 n 2ap(n) " )

(9 2 5) Since a< Z,/3 < 1 and (9 2 5) contradicts (9 2 4) The case of (9 2 3)istreated similarly

Theorem 9 2 2 For random variables of class (d) the condition (9 2 1) is necessary in order that [0, n" p (n)] and [- n" p (n), 0] should be zones of local normal attraction

We remark that this result is not an immediate consequence of the last theorem since uniform convergence of densities does not at once imply anything about P (Z© > x)

Proof. Suppose that (9 2 1) is not fulfilled We show that there is either a sequence x •+oo such that

f

2x.

g (x) dx > exp (- 4xm) ,

(9 2 6)

xm

or one such that

x m

g (x) dx > exp (- 4xm)

(9 2 7)

- 2x.

Indeed, if there is no such sequence, then

f

2x

g (x) dx = B exp (- 4xQ)

(9 2 8)

x

for x > 0, and a similar condition for x < 0 Hence

2x

exp (xfl) g (x) dx = B exp (- xQ)

x

in x > 0 Taking x =1, 2, 4, and adding, we get

f

00 exp(xQ) g (x)dx < oo , i

and combining this with the corresponding argument we get (9 2 1) Thus if (9 2 1) does not hold, either (9 2 6) or (9 2 7) does ; suppose the former, and write

xm = an-j + "p (n) + 9 ,

1

MONOMIAL ZONES OF LOCAL NORMAL ATTRACTION

Chap 9

Since [0, n' p (n)] is a zone of local normal attraction, we have

P {in" p (n) < Z, < 2n" p (n) } < exp { - n" p (n) 2c3} The event on the left-hand side certainly occurs if x,,, < X1 < 2x m and 1(X2+ +X©)/cn' I < 1 The argument then proceeds as for the last theorem

„ 3 Fundamental theorems

Theorem 9 3 1 If [0, n"] and [ - na, 0] are zones of local normal attraction for all a<-L, then the X, have a normal distribution

In Chapters 10 and 12 analogous theorems will be proved for integral normal attraction Thus only bands with fixed a < iare interesting The theorem is a corollary of the following more complex result

Theorem 9 3 2 Let0 < a<1, and let p (n) be any increasing function tending

to infinity and slowly varying as n- ; oo If a < 6,then the condition (9 2 1)

E {exp 1X;I4a/(2a+ 1) } < 00 ,

which is necessary for [0, no 'p (n) ] and [ - n" p (n), 0] to be zones of local normal attraction, is also sufficient for [0, n"/p (n)] and [ - n"/p (n), 0] to

be zones of local normal attraction

If on the other hand 6< a < 2, we considera relative to the series of "critical numbers"

1 1 3

1 s+1 1

6' 45 105 • • •~ 2

s+3 I Let s be the unique integer with

1 s+1

s+2

2 < a<2 s+3

s+4 Then for [ - n" p (n), 0] and [0, n" p (n)] to be zones of local normal attrac-tion it is necessary that (9 2 1) holds and that the moments of X X , up to order (s+3), should coincide with those of a normal distribution Conversely, these two conditions suffice for [ - n"/p (n), 0] and [0, n"/p (n)] to be zones

of local normal attraction

(9 2 9)

(9 3 1)

9 3

THE FUNDAMENTAL THEOREMS

1 8 1

Proof. In view of Theorem 9 2 2, it is sufficient to consider variables satisfying (9 2 1), but we shall use only the weaker assumption that

E{exp (AIX;I '3)} < co ,

(9 3 2) whereA < 1 is a constant, and / = 4a/(2a +1) From (9 3 2) all the moments

°k = E X;exist, and there is no loss of generality in taking c2=1 Suppose that a < i is fixed and, if a >,6, take the integer s to satisfy

1 s+1

1 s+2

2 < a<2 s+3

s+4

If a > i(s +1)/(s +3), we consider the moments °3, °4,

°s+ 3 and the 'cumulants K3 = 11 3, K4=°4-3, Ks =°s - lO°3, .•, KS+3• If on the other hand a=Z(s+1 )/(s+3 ) we consider only °3, , °s+2, K3, , KS+2 For the moment, however, we remain with the former case of strict inequality Assume that the first non-zero cumulant is x4 , so that

K r =O (r<a=so+3), K Q zAO Suppose that s o < s (We must return later to the case in which a< 6,

or in which there is equality in (9 3 3) ) Since the X; belong to class (d), they have a bounded continuous prob-ability density g (x), and their characteristic function is

00 =

f

00

eitxg (x) dx

Then 1 0(t)1 2 is the characteristic function of (X1 -X2 ), which has a pro-bability density, whence 10 (t)1 2 has a non-negative Fourier transform From the lemma quoted in „ 7 2, 10(t)2EL1(- oo, cc), so that

f

00

10(t)12 dt <

oo

(9 3 3)

(9 3 4)

(9 3 5) The normalised sum Z n= n- fSn (a= 1) has a probability density

pn (x) =

2rc f~~

(t)" exp(- in' tx) dt

(9 3 6)

Moreover, 10(t)I< Ifort:AOand4(t)-+O as ltl-+oo Thus,forany0« o <1, (9 3 5) implies that

Ev

Pn(x) = n J-_ / ( t)" exp(-in 2 tx)dt+R 1 ,

(9.3.7) 2m

MONOMIAL ZONES OF LOCAL NORMAL ATTRACTION

Chap 9

Because of (9 3 2) 0(t) is infinitely differentiable for all t. For positive integers T, p and I tl < T,

tP - 1 O(P-1)(0)

tP 0(t) = 0(0)+ to'(0) + +

(pal- i) i

+ -1 RP (t) ,

(9.3.9) where

1 RP (t)1 < 2 sup 14(P)( t )I

(9.3 10) Itk< T

Further, 0'(0) =0, 0"(0) _ -1, so that, for suitable so and ltl < s o , (9 3 9) implies that

10(t)1 < 1-4t2

(9 3 11)

If we write

1

so + 3 < a

(9 3 12) and

°1 - 2-a1

- s0+3'

(9 3 13) then (9 3 11) shows that, for n -"`' < Itl <so,

1 ~(t)1" < (1-n-2"`')"= B exp(-c 4 n2 a')

(9 3.14)

4

„ 4 Approximation of the characteristic function by a finite Taylor series

The function 0 (t) is infinitely differentiable, but not in general analytic, and to estimate the remainder in (9 3 9) we need bounds for 0 (q) for large

q Now

00 I~,c4)(t)) < J

If

then (9 3 2) implies that

00

J -exp (A1x1 11k)g(x)dx < oo (9 4 3)

9 4

APPROXIMATING THE CHARACTERISTIC FUNCTION

1 8 3

Thus, for x >, 1, exp (Ax'Ik )

g(u)du = B,

fx

and a similar condition holds for x < 1 It follows easily that

10(q) (t) I = Bq F (kq)

(9 4 4)

In Itl <n - "`', write K(t) = log 0(t) ,

K(0) =0 Then, from (9 3.7) and (9.3 14),

n

n-°

p n (x) _

-

exp(nK (t) - in -1 tx) dt + B exp(-c 4 n 2 a1) ,

( 9 4 5)

2n -n-°

and from (9 3.11),

K (t) = B,

(ItI < g o )

( 9 4.6)

Write

,/,t4)

q (t + to)_ 4(to) +

to,

(to)+ + t

q

(9 4.7)

q !

Then, sinceK ( q ) (t o ) depends only on O ( P) (t o )forp < q,and since c(P)(to)_

~1P )(0), we have

K(q)(to) = (log '(t+to))(q)It=0 ( 9 4.8)

For sufficiently small p, log ~(t + to) is an analytic function of the complex variable t in the disc bounded by C P ={z ; IzI=p}, so that

K( q)

q!

f

log c¢(t+t)

2ni cp

t q

From (9 4 4), 0(P)

P !

= B exp(Bp+ (k-1) p log p)

(9 4 9)

(9 4 10)

Choose

p = exp (-K o - (k-1) log q),

(9 4.11)

K o being sufficiently large Then

E

0(p) (to) tP

= B E exp(Bp-K o p+(k-1)p(log p-log q))

P =i

P!

P =1

(9 4 12)

184

MONOMIAL ZONES OF LOCAL NORMAL ATTRACTION

Chap 9

For sufficiently large Ko, the absolute value of this expression is less than 4, and (9.3 9) gives, for Itj <2EOi

where K1 is a large positive constant to be chosen later Using the values

of k and °1, we have

B exp (Bm +( k- 1) m log m - ° 1 m log n) _

= B exp (Bm+(k-1) m log m-(2-a 1 ) m log n) _

=B exp Bm+m 1 -2a

4a

= B exp m B-(a-a1) log n

-1-2a 4a

(9 4 17)

But a >, a, 1- 2a > 0, so that (9 4 17), and thus (9 4 15), are bounded by

B exp.(-E 1 n 2a '),

(9 4 18)

if K 1 is chosen sufficiently large (E 1 =E1 (K1 ))

„ 5 Derivation of the basic integral Write clr= K( r )( 0) = irx r , so that I r =0 for 3 < r < sO +3 Then

m

tr

nK(t)=-2nt 2 +n E Y,.

+Bexp(-E 1 n 2 a')

(9 5 1)

r=so+3

r

K (" ) (to) = B exp(Bq +kq log q) (9 4 13) Moreover, for Itj <n - °',

K (t) = K (0) + tK' (0) + tmK(m)(0)

tm+l

(9 4 14)

Rm(t) , +

m!

+ (m+ 1)!

where

tm+ 1

Rm (t)

= B exp m (B+ (k - 1) log m -,u1 log n) (9 4 15) (m + 1) 1

We now take

9 5

DERIVATION OF THE BASIC INTEGRAL

Now Re nK (t) < 0 for I tl < n - 41, and writing

m

Kso+3 = E

r=so+3

t r

,/,

Yr r -, ,

r

we have, from (9 4 5), (9 4 14) and (9.4.18),

-27r -n-14

p©(x) = n J

exp(-Znt 2 +nK s0+3 (t)-in+tx)dt+

Now consider the entire function

r

exp{nKso+ 3 (t) } = 1 + E

Xr 1 ,

( 9 5 4)

r=so+3 r

where, from (9.5.2), Xso+3 = mr-so+3

( 9 5 5)

From (9 4 13) with t o = 0, we have for ItI< nr < m,

r

r r ~ = B exp r (B +

1-2a 4

log r-p1 log n

(9 5 6)

For r < C 1 , ( 9 5 6) is, for I tl < n equal to

Bn-r,u'

,

-

(9 5 7)

and for r > C1, if log r <E 1 log n, to

B n - r a t

( 9 5 8)

If log r>6 1 log n, then (9 4 17) and (9 4 18) show that (9 5 6) is equal to Be-6'r

(nal < r < m)

(9 5 9)

Thus, in ftl < n-°", we have

c l

nKso+ 3(t) = Bn I n - r"`' + Bn -1 =

r=so+3

(9 5 10)

r=so+3

using (9 3 13) and (9 3 12)

We may express Zr as a Cauchy integral around jtj =n - "` 1, and then

(9 5 10) shows that

1 8 5

(9 5 2)

+B exp(-E 2n 2 a1) (9 5 3)

MONOMIAL ZONES OF LOCAL NORMAL ATTRACTION

Chap 9

yr = Br! n''`'

(9 5 11)

We also remark that, because of (9.5.5),X,,,+3 is of exact order n Now let

Itl < 11 1 11

' ,

( 9 5 12)

so that

xr t' Y

!

= Bl71

(r>,so+3)

If m is chosen in accordance with (9.4 16), then

Bil = B exp' n2a' logq1 = B exp (-E2n2al) ,

(9 5 13)

\

, where

Now turn to (9 3 14) and to (9 4 5) with n - "`' replaced by >y l n - u ' ; ( 9 5.3) gives

n

n nin -°

p n (x) =

exp (-2nt 2 )exp(nK so+3(t)) exp (-in -1 tx) dt +

+B exp(-c 4qin 2 a') , (9.5 15)

or, taking into account (9 5 13) and (9 5 14),

pn (x)

= 2n

0 f nl 7

n_°

exp(-2nt2) 1 Z X'

t' exp(-i0In-°

r=,,,+3

(9.5 16) where

R 1 = B exp (-c4 711n2 "')+B exp(-E3 n 2a'),

(9 5 17) and

E3 = Y = 2 log( 1 /q 1)/K1

(9 5 18) Making the substitution ~ = tot, we find that

1

~/ In'/z - ° i -~_

e_2X2

1 + E X'

r

2r e-`~- d~+R1 2n

n,nl/z-°~

S

r-s

m

o+3 r!n Pn(x) =

For r < m,

(9 5 19)

9 6

and

-1

00

nin'/z - °,

B'F(2r)n-~i-°1)r

= B exp (Br+2r log r-r(z -,u 1 ) log n) _

= B exp (Br +r(a l log n-2 log K 1 -cc, log n)) _

= B exp (Br- 2r log K1 )=

=Be-c5r

(9.5 21) Therefore, summing (9 5.20) over r < m, we get an expression equal to

B exp (-11712n2«1) ;

( 9.5 22)

a similar argument obtains for the integral over (-co, -r~ln=-°1) Thus (9 5 16) may be written

Pn(x) =

J ee - z ~2 1 +

'n

Zr ~ r

27c _"

r-I

3

+R2 ,

(9 5 23) where R2 satisfies the same equation (9 5.17) as R1

„ 6 Completion of the proof

In view of (9.5.23), we need to study the integrals

00 n-Zr

a-z ee -it rd~

= n +rHr§)(x)e 2x2 ,

-00

COMPLETION OF THE PROOF

_

r

r .00 e

2

e _1~2

r d r!n J

nine/Z-°

= Br exp (-+ i n1-Z°1) j, (i r)

n-cZ-°1)r

9 5 20 (

)

(9 6 1)

where H;§)(x)=bHr(ax) for suitable constants a, b, and the Hr (r< m) are the Hermite polynomials [164]

C -1r]

(2x)q -2s

Hr (x) = r ! Z ( -1)s

S=0

s ! (q - 2s) ! '

(9 6 2)

We suppose that 0<x<~n"' _ ~n2-°1 ,

(9 6 3) and estimate (9.6.1) when C2 < r < m From (9.6.2), for x >- 0,

1 8 7

188

MONOMIAL ZONES OF LOCAL NORMAL ATTRACTION

Chap 9

x q-2s

H (o) (x) = Bq q ! max ,

(9 6 4) s<Zq 6!(q-2s)!

so that, writing s = pq (0 <'P<12 ),

H9 §) (x)= B exp [q(B+1-2p){2-p l log n+log c} +

-p log q-(1-2p) log q+log q-p log p+

-(1-2p) log (1-2p)] =

=B exp [q(B+1-2p){2-pl) log n+log C1 +p log q] Multiplying this by

If p < 4, Ilog C (1- 2p) I may be made arbitrarily large by taking small ;

if P> 1 and K 1 sufficiently large, p log K 1 may be made arbitrarily large Thus, taking q = r, the sum of the terms in (9 5 23) for C 2 < r < m is of order

B e-2x2e-C,

(9 6 8)

We now turn to the terms in (9 5 23) with

so +4 < r < C 2 ,

(9 6 9) whose sum is of order

Cz

r

B e-2x2

XrX

= Be- 2x2(xn-2+°,)so+4 .

( 9 6 10)

r=so+4 r1 n zr Moreover, the term with r = s o + 3 is Xso+3

(o)

-1x2 -2(so+3) _ (so+3)! Hso+3(x)e

n

= ao~so+3

e - 'X 2n(x/n 2)so+3(1+0(1))

(9 6 11)

(S O + 3) !

n- -!q

and by

q!

we obtain the expression

B exp[q(B+(1-2p)(2-p 1 ) log n-(2-p 1 ) log n+

+p log q+log C(1-2p))] =

= B exp [q(B+log c(1-2p)-p log K 1 )] (9 6 7)