Independent And Stationary Sequences Of Random Variables - Chapter 12 ppsx

Chapter 12WIDE MONOMIAL ZONES OF INTEGRAL NORMAL ATTRACTION „ 1.. Formulation In this chapter, as before, we study the independent, identixally distributed random variables X1, X2,.. In

Chapter 12

WIDE MONOMIAL ZONES OF INTEGRAL NORMAL ATTRACTION

„ 1 Formulation

In this chapter, as before, we study the independent, identixally distributed random variables X1, X2, with

E (Xi) = 0,

V (Xl) = 1

We shall study the zone [0, n"] where a > 6 ; we recall that this is said to

be a zone of normal attraction if, uniformly in 0 < x < n" as n + oo,

(2n)- J e-+"Zdu-* 1

(12 1 1) x

P (Zp >

An analogous definition holds for [ - n", 0]

As before, the symbols p(n), p1(n), , pk(n) will denote functions mono-tonically increasing to infinity, each one usually defined in terms of its predecessors In this chapter we prove the following theorems

Theorem 12 1 1 If [ - n", 0] and [0, n"] are zones of normal attraction for all a < 2, then the variables X, are normally distributed

It follows that we need only consider values of a < 2 This theorem is a corollary of the following more precise result

Theorem 12 1 2 If 6 < a < 2, consider the series of critical numbers

1 1 3

1 s+1

1

654 105 52 s+3, 2 Let s be the unique integer with

1 s+1

1 s+2 2

<<' a < 2 s+3

s+4

(12 1 2)

12 2

THE PROBABILITY OF A LARGE DEVIATION

I n order that [0, no'p (n)] and [ - no 'p (n), 0] be zones of normal attraction,

it is necessary that E(exp A IXil4a/(2a+ 1)) < oc

(0 < A < 1) ,

(12 1 3) and that the moments of X,, up to order (s+3), should coincide with those

of a normal distribution These conditions are moreover sufficient for [0, n"/ p (n)] and [ - n"/ p (n), 0] to be zones of normal attraction

The reason why it is necessary to include A in (12 1 3) is that we have used

a change of scale already to set a= 1 We remark that the necessity of (12 1 3) has already been proved in „ 9 2

Theorem 12 1 2 is completely analogous to the corresponding local Theorem 9 3 2, but the method used in Chapter 9 is not sufficiently power-ful to prove the present theorem except under more restrictive conditions ore precisely, it requires that ~~ (t) I :1 for t 00 and that ~~(t) I < c< 1 for I t I > C This will be true, for example, if F (x) = P (XX< x) contains an absolutely continuous-component, in which case Theorem 12 1 2 can be proved by the methods of Chapter 9

„ 2 An upper bound for the probability of a large deviation

We now proceed to the proof of the sufficiency part of Theorem 12 1 2, assuming (12 1 3) and

227

and an analogous inequality for x < - n"/ p 1 (n)

This is a weaker inequality than would be implied by the integral limit theorem that we are trying to prove, and the methods of Chapter 11 are

~f 3=`r4= = O s+3=0, where or is the r th cumulant of X; Lemma 12 2 1 For

(12 2 1)

x >, no'/ p, (n) ,

we have

(12 2 2)

P(S©>xn -1 ) < C 1 exp [-c 1 n 2 "/p 1 (n) 2 ] , (12 2 3)

22 8

INTEGRAL NORMAL ATTRACTION : WIDE MONOMIAL ZONES Chap 12

appropriate We shall indicate the necessary changes in „„ 11 11-13 which are necessary to arrive at (12 2.3)

It is clearly sufficient to take

x=x1 =n"/2P1(n)

as in „ 11 11 we introduce an auxiliary normal variable Y© e N (0, n"),

with characteristic function

~© (t) = exp (_-1112"t2)zn2" t 2)

and write

Z ;,=n-2(Sn+Yn)

Then V(Z ;,)= 1+n2«-1 , and the other cumulants of Z;, coincide with those of Zn = n-2 S,

Suppose that we can prove that, for

(12 2 4)

na

na

4 P1(n) < x < 2 p1(n) P(Z ;,>x) < C 2 exp[-c2 n 2 a/p 1 (n) 2 ] ;

(12 2 5)

we show that this implies (12 2 3) From (11 11 4)-(11 11 6) we conclude that

P(Z©>xln_ZIYn ' < n 2 a -2)= (1+o(1))P(Z;,>x)+B exp(-2n2«)

(12 2 6)

From (12 2 5),

P(Z;,>xln-1 -j1' l n2a-j-)=Bexp[-c2n2"7P1(n)2]

( 12 2.7)

Taking x=3na/8p 1 (n) and following (11 11 7), we find that

P(Z n >x) >,P(Z ;,>x+n 2a - Zln- ZIYnl < n 2 a - )

(12 2 8)

Since a<2, 2a - i < a, so that for sufficiently large n, x+n 2 a -2 < 1 01x ,

and (12 2 50 2 2 7) combine to prove (12 2 3)

It therefore suffices to establish (12 2 5), and we do this by following closely the argument of „„ 11 12, 13 The only difference is that now a> 6, so

12 3

INTRODUCTION OF AUXILIARY VARIABLES

229

that in the derivation of (11 13 2) we cannot use the inequality 3a-1 < 0 However, since a<121 3a -1 < a, and we can replace (11 13 2) by the esti-mate

00

~

B

e-2 " Zdu = B exP[-czn2a/P1 (n) 2 ]

(12 2 9)

Zz1

The later arguments of „„ 11 13, 14 go through unchanged, and we arrive

at (12 2 5), which proves (12.2 3) Replacing X; by -X; we get the cor-responding inequality for negative x

We need a slight strengthening of Lemma 12 2 1

Lemma 12.2 2 Let n 1 be an integer in 1 <n, <n, and let x satisfy (12 2 2) Then

P(S 111 > xn2) < C3 exp [-c 3 n 2a /P 1 (x) 2 ] ,

(12 2 10)

with a similar result for negative x

Proof Write S© = S© 1 + T©

1 , where

T1 , = X©1+1+ +Xn

Then

P(S©>Zxn2) % P(Sn1> xn 2) P(IT©1I <3xn2)> c 4P(Sn1 > xn 2) ,

whence (12 2 10) follows from (12 2 3)

„ 3 Introduction of auxiliary variables

We now write XI = Xi +,J i , (i < n), where the d ; are independent, small normal variables,

z1 EN(0, n -10 ),

(12 3 1) and

S©=Xi+Xz+ +Xn

Then

V (S©) = V (S©) + n-19

and the other cumulants of S ;1 coincide with those of S, It is easy to see

2 3 0

INTEGRAL NORMAL ATTRACTION : WIDE MONOMIAL ZONES Chap 12

that if (12 1 1) is true with S© replaced by Sn then it is true without this substitution, so that we can work with the sequence S;, This is conve-nient, since X ; has a continuous probability density p(x) with

0<p(x),<n10

(i,<n)

( 12 3.2) (It is to be noted that p (x) depends on n,but not on i <n

We shall use a modification of Cramer's method ([156] and Chapter 8) For functions p2 (n), p 3 (n), to be specified later, take h in

n 2 h n" /p2(n) ,

( 12 3 3) and define

el (~) = el , (ICI,< n 2a/P3 (n))

= 0 , ( kkI > n 2,/P3 (n))

(12.3 4) Let X1 be independent random variables with probability density

d17(x)

= R -l e1 (hx)p(x) ,

(12.3 5) dx

where

R = _ e 1(hy)p(dy)

(12 3 6) For nl n, write f©1(x) for the probability density of

S;,1 = X1+ -+X;,1 , andf,1(x) for the probability density of Sn1 = X1+ +Xn l

For all we then have f1() = R -1 el (h~)f1()

(12 3 7)

We shall seek estimates of the form

.fin, (~) = R -nel (h~)fn, (~) + Op n , ,

( 12 3 8) where 101 < 1 (and 0 ; like B, may vary from place to place), and pn1 is

some error bound, attempting to establish these by induction on n 1

12 4 STUDY OF THE BASIC RELATION

„ 4 Study of the basic relation

We have

fn1+1(~) = R -1 f"O

- fn,(~-z)ej(hz)P(z)dz ,

(12 4 1) and supposing that (12 3 8) holds, this gives

fn1+1(~) = R-n1-1

fn1+~()=R-n1-1 J

R-n1-1

f

00

-00

e, (h (~ - z)) e l (hz) fn , (~ - z) p (z) dz +

231

00

+R-1

J

8Pn1 e1(hz)p(z)dz

(12 4 2)

In view of (12 3 6), this may be rewritten as

e l (h(c -z)) el (hz)fn 1(~ -z) P (z)dz+Op n1

(12 4 3)

We now compare the first term

00

R-n1-1

f- el(h( _ z))el(hz)fn,(~-z)p(z)dz

(12 4 4) with the integral

-00

J

00

eI(h~)fn1(-z)P(z)dz=R-n1-Ie1(h~)fn1+I( ) ( 12 4 5)

-00

We separate the domain of integration into two sets 2C={z ; Iz1 <n"+?/P4(n),

I~-zj<n"+11P4(n)1

(12 4 6) and its complement 2C We first estimate

R-n1-1

f Ie1(h(~ - z))eI(hz) -e 1( h ~)I fn1( -z)p(z)dz,

(12 4 7)

assuming that

P4 (n) < {P3 (n)}

From (12 3 4),

e 1 (hu) < exp(n 2a /P3(n))

(12 4 8)

2 3 2

INTEGRAL NORMAL ATTRACTION : WIDE MONOMIAL ZONES Chap 12

for all u, so that

(e, (h(~-z))e, (hz)-e, (h~) I < exp(2n2a/P3(n))

(12 4 9) Moreover, as in (12 3 2),

0 <f",(~ - z) < n10 ,

(12 4 10)

so that (12 4 7) is bounded by

R -n1-1

exp (2n2a/P3 (11))

n'01

f",(~ -z) dz +

14 - z 1 > "-+ 1j2/Pd (")

+

P(z)dz

(12 4 11)

J IjzI > "§ + '/2/P4(")

It is easy to see that Lemma 12 2.2 (with c 3 replaced by 2c 3 ) applies to the sums S;, 1 , so that the sum of the integrals in brackets does not exceed 2C3 exp(-C4n2a/p 4 (n)2) < 2C 3 exp(-c4n2a/P3(n)3) ,

(12 4 12) using (12 4 8) Thus (12 4 11) does not exceed

2C 3 R-"'-1 exp [ -c5n2a /P3(n)']

(12 4.13) for sufficiently large n, with c 5= 2c4

We now assume that in (12 3 3) P2 (n) has been defined by P2(n)= {P3(n)}1§ .

(12 4 14) Then, for z c 21,

JhzI < n2a/P3(n)'§, lh(~-z)j < n2"/P3(n) 10 ,

(12 4 15)

so that (12 3 4) shows that

el (h (~ - z)) e, (hz) - e, (h~) = 0

(12 4 16) Thus, (12 4 4) is

R- "' - l

e,(h5)f,,+1(~)+20C3R-"'-lexp[-c5n2a/ P3(n)4] (12 4.17)

„ 5 Derivation of the fundamental formula

We have shown in „ 4 that (12 3 8) implies that

f",+1()=R-",-1el(h~)f",+1()+

+2OC 3 R -n,-1 exp[-c 5 n2a/P 3

(n)*]+OP",

(12.5 1)

12 5

DERIVATION OF THE FUNDAMENTAL FORMULA

233

We have therefore proved by induction on n the formula f,

= R - "el (h')f"( )+8p" ,

(12 5 2)

where

p" = 2(R -1 +R-2 + .+R - ") exp[c5n`"/ P3(n)']

( 12 5 3)

Thus, for values of c such that e 1 (h~) ; 0, f"(~) = R"el ( - h~) ~"( )+OR"el (-hS)P"

(12 5 4)

Since for all u,

e l (hu) < exp(n 2a /P3(n))

(12 5 4) and 12 5 3) yield, when e 1 (hc) 5 0,

f"(~) = R"e1(-h~)f"(c)+8p ;, ,

(12 5 5)

where

p ;,= 2(1+R+ +R") exp[c 6 n 2 a/P3(n)f]

( 12 5 6)

If e l (h~) 0 0, then l h~l n 2 a/ p3 (n), so that by (12 3 3) and (12 4 14),

P2(n)n2a+?

ICI

3(n)

n 2a+-t -P3(n) 9 < n2

( 12 5 7)

P3

Now consider the function

W" (u) _ f u f" (~) d~ ,

and write

2T={~<u ; e 1 (h~)~0}, 2T={c<u ; e 1 (h~)=0}

From (12 5 5) and (12.5 7),

J,i~f"( )d = R"

e 1 (-h )f"( )d +9n2pn _

( 12 5 9)

u

=R " l

e 1 (-h') f"(~)d'+9n2 p ,, ,

( 12 5 10)

J- x since the integrand vanishes on 2t We therefore estimate

(12 5 8)

2 34

INTEGRAL NORMAL ATTRACTION : WIDE MONOMIAL ZONES Chap 12

.ff()d"' .

If e1 (h~)=0, then by (12 3 3) and (12 4.14),

n§`+2

P3 (11)

P2(n) =112+ 2 P 3 (n) 9 ,

( 12 5 12) and thus by Lemma 12 2 2,

Lfn(~)d~ <C3 exp [-c 3 n 2a/p 1(n) 2 ] ,

(12 5 13)

for arbitrary PI We takep 1(n) =P3 (n), and then (12 5 13) can be seen, for sufficiently large n, to be smaller than p ;, Thus we arrive at the equation

Wn(u) =

J u

fn(~)d~= Rn

fU

el (-h~)fn(~)d~+20n2 p;, (12 5 14)

-00

- 00

„ 6 The fundamental integral formula

We consider the distribution function Wn(~)=P(X1+ +XX<~) =

fn(v)dv

-00

The random variables Xi (i < n) have distribution function

x

V(x)=P(Xl<x)=R-1 f

el(hy)P(y)dy

( 12 6 1)

It is clear that for sufficiently large n, and i < n,

U 2 = V (X) > 0 ,

m = E (X~) o0

Writing

F,.(u) = Wn(un 2) ,

Fn(u) = Wn(mn + bun')

(12 5 14) gives

un'/2

Fn(u) = Rn ~_ el (-h~)dWn(~)+OPn2 ,

(12 6 3) 00

where

(12.6 2)

12 7

STUDY OF THE AUXILIARY INTEGRAL

235

P,,2 = 2n 2 Pn

(12 6 4)

In (12 6 3) we set u = co and u = x > 1 and subtract one expression from the other, to get

00

1-Fn (x) = Rn

J

eI (-h~)dW,,(~) +20P n2

(12 6 5)

xn 1/z

If x >, 1, h~ >,0, then el (-h~)=e-h4 for h~< n2a/p 3 (n), and e l(- hc)=0

for larger values Thus

1-F,,(x) = Rn

j

e - " 4 dWn (~)+9Pn 3 ,

(12 6 6) xn'/z

where

Pn3 = 2Pn2+ Rn

xo

and

n 2a xo

nP3 (n)

From this

J ,0 xo e-h4dWn(~) < eXp( - n 2a /P3(n))

so that we can take Pn3 = 2Pn2 + Rn exp [-n 2 a/P 3 (n)]

(12 6 9)

„ 7 Study of the auxiliary integral

In (12 6 6) set = mn 2+ -vnZ, to obtain

f

00

1 _F,,(x) = Re-hmn'/z

e -hon'/zv dF,,(v)+9pn3

(12 7 1) (x-mn'/z)/an'/z

We now turn to the quantity

R = j ei ( hY)P(Y)dY ,

(12 7 2

x

and approximate it by means of a truncated power series Set p5 (n) =p3 (n) y© = na + '/ p 5 (n) , so that Lemma 12 2 2 and (12 3 4) give

e -h 'dW© W ,

( 12 6 7)

(12 6 8)

2 3 6

INTEGRAL NORMAL ATTRACTION : WIDE MONOMIAL ZONES Chap 12

f

R =

e1(hy)P(y)dy+ePn4,

(12 7 3)

~where

Pn4 = C4 exp [- c7 n2a/P3 (n)']

(12 7 4) Because of (12 3 4) and (12 3 3), e1(hy)=e'')' for Iyj <yn, so that

R

= f

eh, P (y) dy +ePn4

(12 7 5)

~I

Yn

We need an upper bound for R ;

R < c5

(12 7 6)

To prove this, set

P (y) = JP(z)dz

(y % 0) , Y

and use (12 1 3), (12 3 1 and the arguments of „ 9 4 (following (9 4 3)) to prove that

P(y) < C6 exp(-c8y4a/(2 a+1)) Use this and integrate (12 7 5) by parts, using the fact that, for 1 < y < yn, hy- c8 y4,/(2a + 1) < _ 2c8 y4a/(2a+ 1)

to obtain (12 7 6) More generally, if 0(y) is continuous on [- yn, yn], and I8 (y) I < 1, then

exp (hy 0 (y)) p (y)dy < C5

„ 8 Expansion of R as a Taylor series

For a fixed positive integer K, expand (12 7 5) as a Taylor series of (K + 2) terms

R =

hp

yp P (Y) d y + KK + 1 I

exp (hye (y)) P (y) dy + k=0 P lIY-<Yn

( + )

Yn

(12 7 7)

(12 7 8)

(12 7 9)

+OPn4

(12 8 1)

12 8

EXPANSION OF R AS A TAYLOR SERIES

237

We take

K = [10/(2 -a)] + 1 and note that, for p < K,

JIyI>yn for sufficiently large n Thus (12.8 1) becomes, using (12 8 2), (12 8 3) and (12 7 5),

K

hp R=1+ E a p -+ 6C 7 n10 ,

( 12 8 4) p=2

P!

where the ap are the moments of the Xj (i<n) Moreover,

00

m=E(Xj)=R-1

f - 00

yehyp(y)dy,

(12 8 5)

y"p(y)dy < C6 exp [-c9n27p5(n)2] < C6n-10

(12 8 3)

and the argument used to derive (12 7 5) and (12 7 6) gives

m= R-1

J

y e"yP (y) dy + 6Pn 5,

(12 8 6)

Ivi ~yn where

p, 5 = C4 exp[-c 9n2

§`/P 3 (n)']

( 12 8 7) Hence, arguing as for (12 8 4),

K

hp-1 m= R-1 E a + OC7n-9

.

( 12 8 8)

P (p-i) ~ p=2

We also need the variance

62 = V (Xi) = E (Xi2) - m 2

As in (12 8.8) we have

K

hp-2

E(XX2) = R-1 1 ap + 8C 7 n-8 ,

( 12 8 9) p=2

(P -2 )

so that a2 may be obtained from (12 8 8) and (12 8 9) This is most easily done by remarking that, as far as their principal terms are concerned,

m and a2 may be obtained from Ri/R 1 and (R" R 1 - R 12)/R i (cf Chapter 8), where

(12.8 2)

where the yp are the cumulants of Xi Under the conditions (12 8 1) and (12 3 1) we have

238 INTEGRAL NORMAL ATTRACTION : WIDE MONOMIAL ZONES Chap 12

R 1

= 1. Ivl,y© eh''P(Y)dy Thus

K hp

log R = 1

p=2 P •

m=

K

I

hn

y p

p=2 P •

K hp-2

(12 8.12)

Q Z - E

OC7n - '

y p

+

, p=2

(P-2) •

72 = 1+n -20 , y 3=O

(j=3, 4, , s+3), (12 8 13) where s is the greatest integer with

1 S+ l z

< a s+3

Thus

and from (12 8 10), (12 8 11) and (12 3 3) we have

(12 8 15) log R = 2Y2h 2 + OC8n-(s+ 4)/(s+3) ,

m = y 2 h+0C 8 n -1/p2(n)s+3 (12 8 16) Thus (12 8 13) implies that

(12 8 17)

h = m+OC8n-1P2(n)-s-3 .

„ 9 Further transformations

(12 9 1)

Turning to (12 7 1) we now choose h so that x=mn2 ,

where

(12 9 2)

1 < x< n"/P7(n) ,

P7 (n) % P2 (n) 20 •