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Chapter 14INTEGRAL THEOREMS HOLDING ON THE WHOLE LINE „ 1.. The role of the linear functionals aj, b j was played by moments of the random variables XX.. In the case 0 n = n" p n, the co

Trang 1

Chapter 14

INTEGRAL THEOREMS HOLDING ON THE WHOLE LINE

„ 1 Formulation

In the preceding chapters we have studied theorems of a collective type concerning large deviations in zones of the form [0, 0 (n)] and [ - (n), 0], where 0(n) = o (n 2) The role of the linear functionals aj, b j was played by moments of the random variables XX In the case 0 (n) = n" p (n), the con-dition

E{exp (A IXXI4a/(2a+ 1)) ) < 00 (14 1 1) appears as a condition for normal attraction ; this implies that all the moments ofXi exist and the probability of a large deviation in XX itself falls off very sharply

In this chapter we study theorems in which x is not restricted to any zone, but allowed to range over the whole real line Thus let X1 , X2 , be independent and identically distributed with

E(Xj) = 0,

V (Xj) = U2>0

(14 1 2)

We shall seek classes of such variables for which collective limit theorems hold which assert that, uniformly in x > 1 as n->00,

P(Z§>x)/0(x, a 1 , a,, n)-* 1

(14 1 3) and

P(Z§<x)/i(-x, b 1 , , b 1 , n)-*1

(14 1 4) Here the limiting tails c depend on linear functionals aj , b ; of F (x) =

P (Xl < x) We remark that the restriction x > 1 is harmless, since in Ix I < 1 the classical theorems hold

For simplicity we shall restrict attention to the case in which F is symme-tric, having a bounded continuous density g (x) such that, for x >, 1,

Trang 2

14 2 PROBABILITY OF VERY LARGE DEVIATIONS : ELEMENTARY RESULT 2 5 5

P(X1 >x) _

{ 00

g(u)du = r Ar + O( -6 a - E) ,

(14.1 5)

Jx

r-a

and thus

P(X 1< - x) =

-x

g(u)du = Z xr + O(x-6a-E)

(14.1 6)

-Co

r-a

Herea >,3 (since the variance exists), the A,are constants, with Aa> 0,and

s>0 The class of such probability densities we call (A) Such variables have only a finite number of moments, and the role of the linear functionals

aj , biis layed by pseudomoments defined in „ 5 below Theorem 14 1 1 For x >, 1 we have, uniformly in x as n -* co,

P(Z§>x)I (27c)-'

J e

- iu2du+r(x, n+) -+1

(14 1 7)

x

where r(x, n+) is a rational function in both arguments For x>,nZ+a-1+E, n>n o (8),

(2rc)-Z

e-Za2 du+r(x,ni)-nP(X j >axn+) ; x

r(x, n+) is determined by a finite number of linear functionals of the distri-bution of X1 , called pseudomoments

f

This theorem has a collective character since the asymptotic form is determined by a finite number of pseudomoments For x < -1, of course, another analogous relation holds, and for IxI< 1 the classical theorems hold

„ 2 An elementary result on the probability of very large deviations

We shall be concerned with the deduction of the asymptotic forms of P(X1+ +Xn > unZx)

for very large x We begin with the first of these expressions, setting

(14 1 8)

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256

INTEGRAL THEOREMS HOLDING ON THE WHOLE LINE

Chap 14

y = axn 2 and supposing that y > n Thus we consider the probability of the event

X, +X2+ +Xn>y

( 14 2 1)

This can only occur if at least one of the events

X, > y/n

(i 1, 2, , n)

(14 2.2)

occurs These events overlap, but their intersections have small probability

if Y's large More precisely, we shall find values of y for which the probab-ility that two or more of the events (14 2 2) occur is of order

Bi7nny -a ,

(14 2 3)

where i , -*0 as n-> oo For each k >, 2 the probability that exactly k of the events (14 2 2) occur is

P(Sn>YIH1),

(14 2 10)

n ) {

P (X >

} k = B n

k (2Aa) k n ka y/n)

-k

1

My ka

(2A )kn ka+k

The sum over k>,2 is bounded by (14 2 3) if

n ka+k

n

(14 2 5) yka 1< 'In Y

a

since (2A a ) k/k ! = Be -k.

This is equivalent to

y

qn1/(n-1 )ank/(k-1)+1/a

and is certainly satisfied if

y > f1n 1 n 2+a - 1

(14 2 7)

In particular, we may take

y

> Yn = n 2 + a - ' log n ,

' I n = (log n)-1 (14 2 8)

Let H1 be the event {X 1 > y/n} Then in view ofthe discussion,

P(S n > y) = nP(H1 ) P(Sn >yIH I ) +Bq n ny - a (14 2 9)

We now investigate the expression

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14 2.

PROBABILITY OF VERY LARGE DEVIATIONS : ELEMENTARY RESULT

257

which if

L = y,,/n-' log n

(14 2 11) may be written

P(Sn>ylH1)=P(X1+ +Xn>ylH1) =

For the first of these two expressions we have the inequalities

P(

P(

P(

X2 + 1 + Xn

< L P ( s>n Y

an ©

X2+ +Xnl ant

X2+ - + X n

an`

= P (Sn

>y,

+P(sn >y,

X2 + +Xn an2

ant

< L PCS n >y

X2+ -+Xn

H1,

L H1 +

> LI H i

( 14.2 12)

X2+ -+Xn an2 (1 +o(1)) P(X1 >y+Lan 4 l H i ) , (14 2 13)

H1, X, + +-xn

ant

is independent of H 1 , and implies that for some i,

Xl > yna/log n Arguing as before, and using (14 2 8) and (14 2 7), we have

<L ) <1

< (1 +o(1)) P(X1 <y-Lan 2 j Hi ), (14 2 14)

by virtue of the central limit theorem Further, under (14 2.7), P(X1 >y„Lan 2 I H 1 ) = P(X 1>y„Lan2)/P(X1 >y/n)

=n - all +0(1)) ,

( 14 2 15) because of (14 1 5), (14.2 8) and (14 2 11)

We now examine the second term in (14 2 12) The event X2+ +Xn > L

(14 2 16) ant

> L) < nP Xi > log

)

+ Bq n n y =

g n

= Bn 1-2 alog n=Bn - © - 1 , as a<3

(14.2 18)

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2 5 8

INTEGRAL THEOREMS HOLDING ON THE WHOLE LINE

Chap 14

We* now use (14 2 14), (14 2 15) and (14 2 18) to rewrite (14 2 9) in the form

P(Sn>y) = nP(HI) P(X1 Y) (1 +o(1)) _

(HI)

= nP(X1 >y)(1 +0(1))

(14.2 19) Thus, for y >, y,,,

P(S,,> y) = A,,ny - ©(1 +o(1)) ,

(14.2 10) where o (1) is uniform in y as n -+ cc

This simple result has an immediate probabilistic significance ; it asserts that if S§ takes a very large value this is most likely to be because exactly one of the summands is very large ; the probability of S§ being large as a result of an accumulation of moderately large summands is comparatively small

Since the underlying distribution is symmetric, we also have P(S§<-y)=nP(Xl<-y)(1+o(1)) =Aany-a(l+0(1)), (14 2 22) where o (1) is uniform in y >, y,,

± 3 Radial extensions

We now set

whereE< 10 -4 is a small positive constant Because of the previous results

we have, for x >, x,,,

P (Z§ > x) - nP (Xl > xan 1 )

(14.3 2) Since the range x < 1 is dealt with by the central limit theorem, it is suffi-cient now to examine the range

1<x<x,,

(14 3 3)

We shall do this with the help of the analytic method of Chapter 9 Con-sider the characteristic function

00

0(t) _

f- e`txg(x)dx ,

(14 3 4)

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14 3

RADIAL EXTENSIONS

2 5 9

which by (14.1 5) and (14 1 6) is differentiable for all t at least (a-1) times

We now introduce the concept of a radial extension of 0 (t) A function ,,/ (t) will be called a radial extension in t >, 0 if is it defined in some neigh-bourhood [ -to , to] of t =0 and coincides with 0(t) on [0, to] A radial extension in t < 0 is similarly defined For example, the characteristic function 0 (t) = e- Itt (ItI + 1) corresponding to the probability density

g (x) = 2/n (1 +x2)2 has radial extensions y(t)=e-'(t+1) in t>,0 and

y (t) =e` ( -t + 1) in t < 0 Both are entire, neither is even

We now prove that, under the conditions here assumed, 0 (t) has a radial extension y + (t) in t > 0 which is everywhere differentiable at least (4a + 2) times, and a similar radial extension y - (t) in t < 0 From (14 1 5) and (14 1 6)

it is immediately clear that it is sufficient to prove that, for any r >, 3, the expression

eit

-1

eir

J1

r

d

+~-

r d

(14 3 5) ma

has radial extensions which are differentiable any number of times It is clearly sufficient instead to consider

0o eit

- 3

o

eit

J br d + ~- br

d

3

a

since J

3 eit

0 r

d

is an entire function For j 3 we can expand -r as a power series in

W = 1+~2 = 1+

~-2 Thus

-r_

Z

00

K

ekWk-I' ek l+ 2

k

+0(~-K-1 )

(14.3 7)

k=r

k=r

The question of the differentiable of the radial extensions of (14 3 5) there-fore reduces to that of the radial extensions of

00

eit4 ~k -00 (1 +Ok

(14 3 8)

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2 60

INTEGRAL THEOREMS HOLDING ON THE WHOLE LINE

Chap 14

for r <k < K, for if the continuous function p () = 0( -s-1 ), its Fourier transform is differentiable at least (K - 1) times

In the integral (14 3 8) the integrand is rational, non-zero on the real axis, has poles „ i and is of order O (~-"`) at infinity Moving the contour of integration upwards for t>,0 and downwards for t,<0 we obtain radial extensions in the form of entire functions (as in the example) Hence (14 3 5) has radial extensions which are infinitely differentiable, and

° (t) has radial extensions which are differentiable at least (4a + 2) times

± 4 Investigation of the fundamental integral Because 0 (t) is real, the probability density p, (x) of Zn is given by

n

00

Pn(x) = - ~

o(t)ne-an'/zitxdt =

-00

n2

f

00

_

Re

0( t)n e-an'/zitxdt 7r

o Moreover, since there is a bounded continuous probability density g (x),

we have (cf (9 3 7))

2

E0

n

n -an'zitx

E n

pn (x) = -ReRe

(t)n

dt+Be- '

(14 4.2)

f0

We note that fort >, 0, 0 (t) = y (t),and that y (t) is differentiable in the neigh-bourhood at least b >, 6a- 3 times From (14 4.2) we have

Pn(x) =

n2

Re J

Eo

y(t)ne - an'/2i txdt+Be -E'" ,

( 14 4 3) 71

o where y (t) is differentiable b times in [0, g o] In t > 0,

Y (t) _ (t) = 1- it2 +o (t3) ,

( 0 < t <go)

In view of this we find that, for n- 2 log n< t< g o ,

y (t) < 1- 2n- 1 (log n) 2

Y (t)" = B exp (- E2 (log n)2) , and from (14 4 3) that

(14 4 1)

Trang 8

14 4

INVESTIGATION OF THE FUNDAMENTAL INTEGRAL

n -2

-n - 11ogn

p n (x) =

n Re

y(t)ne-and/2itxdt+B exp [-•2(log n) 2 ]

0

If for t <n- z log n we write

K (t) = log y (t) , then

nz

-n- flogn _ -Re

exp [nK (t) - an' itx] dt +

n

o

+ B exp [ -E2(log n) 2 ] (14 4.5) Note that y (t) is not necessarily even Since it is b times differentiable,

b - 1 (q)

y(t)= 1-2t2 + Y Y 0) tq+Bt b

(14 4 6)

q =3 q

in JtI 5 e o If ltl ,<n - 2log n, nBt b = nB(n-ib log n)b = BEn- 26+ i +E ,

( 14.4.7) and since

exp(B n

-,b+

1+B n -26+1+E E

E

~

we have, writing

Yo (t)

2 + b- 1 y(q)(0)

tq

= 1-2t1 ,

q=3 q

that

K(t) =log yo(t)+BEn-26+E ,

( 14.4.9) since in our interval 2< y o (t) <2

For !tl <n 2 log n,

b - 1

tq

log y o(t) _ -2t2

g q - + Btb ,

( 14 4 10)

q=3

q~

where

gq = [log Yo(t)](q)It-o Moreover,

Bntb = B n-2b+E +E

7

and substituting into (14 4 5) we find that

2 6 1

(14.4 4)

(14 4 11)

(14 4 12)

Trang 9

26 2

INTEGRAL THEOREMS HOLDING ON THE WHOLE LINE

Chap 14

We write

b- i

tq K3 (t) _

gq i , q=3

and examine the entire function exp [nK3 (t)]

(14 4.15) For Itl < n-2 log n,

InK3 (01 = BEn 2+` ,

(14 4 16) and if we work to accuracy

BEn-2b+1+E

(14 4.17)

we can ignore [nK3 (t)]b, and write exp [nK3 (t)] = 1 + K4 (t, n) + Bn-Zb+ 1 +E ,

(14 4.18) where.

K4(t, n) = b [nK3 (t) ]q

(14 4 19) Y

q=1

q Substituting in (14 4.5) we obtain

0

n-1/zlogn p,, (x) = - Re J

e - -In` { 1 + K4 (t, n) } exp (- n2 itx) dt + 0

+BEn-Zb+1+E (14 4 20) Substituting ~ = tn2 ,

log n pn(x) = 1 Re I

e-22(1+K4(~n-2, n))e-'4xd~+BEn-Zb+1+E n

o

(14 4.21) and since

n 2

n-'/zlogn

f

b-1

t9 _ ~ Re

exp n - 2t2 + q gq

- un2 itx dt + 0

q=3

q +BEn-,b

(14 4 13)

(14.4 14)

for r<C1,

j logn J

e-Z4z~rd~ = B exp(-4 (log n)2)

(14 4 22) 0

Trang 10

14 5

INVESTIGATION OF THE AUXILIARY INTEGRALS

26 3

1

©©

Pn(x W =

n Re

e -442 (1+K4(cn -Z, n))e-` 4xdx+BEn -+6+i+E,

fo

± 5 Investigation of the auxiliary integrals

In this section we investigate more thoroughly the expression

00

E(x, r) = Re

J0 e- _J42 ~re-i4x

If r is even, then

E (x, r) = 2

00

e --1-42Zr

i4x

-00

is expressed in terms of e- Zx2Hr ©~ (x), where H,(©) is the r th Hermite poly-nomial We also remark that the assumption that g (x) be even is not essen-tial, though it simplifies the calculations

Ifr is odd, then E (x, r)does not fall off so sharply asx-+oo (for a discussion

of this function see [150]) For even r we have, for r bounded, E(x, r) = Bxr e-Zx2

,

(14 5 1) while for r odd,

E(x, r) = (-1)Z(r+1)r_

x-r- +Bx-r-2 .

(14 5.2) Let us now turn to the integral (14 4 25) The terms involving even powers of cn -Z are bounded by B(xn -Z)re -Zx2 and for x5 log n are therefore negligible compared with e- -L2X2 and for x > log n smaller than the remainder term in (14 2 25) Then (14.2.25) will have the form

Pn(x)=(27r) Ze Zx2 +r1(x,n')+BEn -Z6+1+E , where rl is a rational function of x

(14 4 24) 1

©©

_ (27r)-Ze -Ix2 + _ Re [7t

o

e -Z42 K4(~n -4 , n)e- `4xd~+

+BEn-Zb+i+E (14 4.25)

We therefore have to investigate

00

Re e-ZS2 re -i~x d (14 4.26)

J o0

Trang 11

264

INTEGRAL THEOREMS HOLDING ON THE WHOLE LINE

Chap 14

Now let

2+a -1 +E

1<x<xn = n (14 3 1) Then, from the last equation,

X n2 pn(x)dx= 1-O(x)+r2(x, n 2 )+BE n -2b+3+E

(14 5 3)

where r 2 (x, n2 ) is a rational function, since r 1 (x, n 2) can be represented,

up to accuracyB E n-2a+ 1 +E,

in the form of a power series in x -1 ,beginning with x-k (k >, 2), which may be integrated term by term to give r2 (x, 172) For y > x n ,

f

00

Pn(x) dx - nP (X 1 > yun 2 ) - nAa/o-a y a n 2a

Y

(14 5 4)

In particular, taking y = xn , nAa/faxan2a = BE n -2' 3+E ,

( 14.5 5) and so, combining (14 5 3) and (14 5.4),

f"o p

n (x) dx = P (Z,, > x) _

X

= 1-~(x)+r2(x, n2 )+BE n -26+3+E,

(14 5 6) for x < xn This formula is also true moreover for x n < X<, n4 , and conse-quently, for such values of x,

r 2 (x, n2 ) nA a /Q©xan2a

(14.5 7)

It is not difficult to see that (14 5 7) is also true forx > n , so that for x > 1,

CIO

P(Z n >x) - (2 )

X

e-2"2du+r(x, n 2 ) ,

(14.5 8)

X

where r is a rational function

We notice that the coefficients of the rational function are expressed in terms of a finite number of the derivatives at zero of the radial extension

y (t) of 0(t) These derivatives are called the pseudomoments of X,

If 0 (t) is differentiable h times at 0 (i.e if h < a - 1) then the first (h - 1) pseudomoments differ from the corresponding moments only by powers

Trang 12

14 5

INVESTIGATION OF THE AUXILIARY INTEGRALS

2 6 5

of i The pseudomoments play the role of the linear functionals a j , b;

described in Chapter 2

We remark that similar conclusions may be drawn when the densities have asymptotic expansions as x *co;

P (X, > x) = f g ( u)du =

6a dG (v)

+ 0 61 +

Jx

J a

X

x and similarly for x-* - oo, where G is of bounded variation (but not neces-sarily monotonic)

± 6 An example Suppose that

g (X) = 2

7r(1 +x2)2

so that a =1 and, for t > 0, q5(t) = e-'(t+ 1) Then

log 4(t) = -t+log (1+t),

so that, in 0 < t < l,

K(t) = - t -f- t - 2t 2 +3t 3- 4t4~

= -Zt 2 +K 3 (t) ,

where

K3(t)= -t+2t2+log (1+t), exp[nK 3 (t)] = e-nt+'Int 2(1 + t)n

Thus K4(t) will be a truncation of

e-5nlz+-z52(1 +fin ) _3 +

30 . Now

Re

f

00

e-z52 3e`5x

6

B

d ~ = - +

x5 ,

0

(14 6 1)

(14 6 2)

(14 6 3)

11

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