NATIONAL BUREAU OF STANDARDS REPORT

2267

ON INDEPENDENT SAMPLES FROM NORMAL POPULATIONS

by

A* A. Zinger

U. S. DEPARTMENT OF COMMERCE

NATIONAL BUREAU OF STANDARDS
U. S. DEPARTMENT OF COMMERCE

Sinclair Weeks, Secretary

NATIONAL BUREAU OF STANDARDS

A. V. Astin, Director THE NATIONAL BUREAU OF STANDARDS

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NATIONAL BUREAU OF STANDARDS REPORT

NBS PROJECT NBS REPORT

1103 - 10-1107 11 February 1953 226?

ON INDEPENDENT SAMPLES FROM NORMAL POPULATIONS

by

A- Ao Zinger

Published in Uspehi Math. Nauk* (N,S s )&,

No . 5(45), PP* 172-175 (1951

)

Translated from the Russian

by

L Rhodes

Approved for public release by the

Director of the National Institute of

Standards and Technology (NIST) on October 9, 2015

The publication

unless permlssU 25.D.C. Such

eally prepared

lor In part, Is prohibited

r Standards, Washington

1 report has been speclfir report for Its own use,

ON INDEPENDENT SAMPLES FROM NORMAL POPULATIONS

by

A. A. Zinger

It is well known that for a sample from a normal pop-

. „ 2 illation, the sample mean x and the sample variance s are

independent. This theorem was proven by R.A. Fisher,

The question arises: To what extent does this property (the

independence of x amd s^) characterize a general population?

Here we must point out the work of Lukacs where it is proven

that if the general population has a moment of the second

order, and if the sample mean and sample variance are independ

ent in samples of si&e n > 1, then the general population is

distributed normally. However, this theorem does not answer

completely the stated question, since the condition demanding

the existence of a general second moment imposes a considerabl

limitation., There remains the question of what the general

distribution might be if we do not demand a priori the existence of a general second moment,

THEOREM » Let us consider n (n>l) mutually independent

random variables x, « x_, .... x , each of which is distributed

1 2 n

according to the law F(x). Let us w- x = (x^+x^+.o* +x

n )/n

P 2 2

and s d = [ (x^-x) +.„« + (x^-xf 1 /n. If x and s^ are independent

then the distribution F(x) is normal.
£

Proof* We note first the obvious algebraic identity

• ^ 2 2 p

2 x . = nx + ns^ (1)

j=i -

and introduce the function f(t):

f(t) = E(e ltx ~x

) (2)

The symbol E here, and elsewhere, denotes the mathematical

expectation of the random variable within the parentheses.

Let us note, that due to the finiteness of the integrals

+00 2

J

x

n e“x dp(x) (3)

-00

the function f(t) is differentiable any number of times. To

prove the theorem, it is necessary that f(t) be differentiable

only twice.

Examine the expression

E f ns 2 exp [ 2 (itx.-x^)]} * (II)

j=l J 3

p Making use of the independence of x and s and the x- -• ntity

(1), let us write

E{ns 2 exp[ 2 (itx.-x?)]} = E(ns 2e“ns )E[exp(it2Xi -nx'

1

' ) ] J=1 J J 1

Let us transform separately both sides of (5>) :

o ^c, 2 n _ 2 -ns 2 ,, n

E(ns‘-e“ )E [ exp ( it Si x .-nx2

) ] = Ejris— 'E[exp(it 2xr

j~l 3 E(e“ns2 ) j=l J nx2

) ]E (

( 5 )

2

e~ns

)

( 6 )
3

Again using the fact that x and s

r~ are independent, we write

.2 n o n

E(ns 2

e LS

) E[exp(it 2 x r nx

L

)] = aE{exp[ 2 (itx . - x*T ]} (7 ) J=1

J 3=1 J J where we define

a „ / 2 -ns \ Elns^e ) E(e"nS

)

( 8 )

Furthermore, since all x^ are by definition mutually independent

and since each is distributed by the law of F(x),

n

Efexpt 2 (itx*-x2

3 = fn

(t)

3=1 J J (9)

Let us consider the transformation of the left side of

« n

p ^2

equation (5). Let us substitute here from (IK ns ^ ~ 2 xf ~ nx

y i i-l

n

E{ns 2 exp[ 2 (itx.-x.)]} =

j-1 J J

n

= E { 2 x

2 exp[ 2 (itx lr ~ x

2

)] } — E Inx

2 exp[ 2 ( itx , - x

2

) ]

- 1 •’ ' K

j=l J J

n n

J-1 J k=3

Me make fur t he r the following t r an s forma t i o n

n

Efns 2 exp [ 2 (itx. - x

2

) ] } =

j=i 3 3

1 ? p n n o n n2 = nr”1

(t )E[x exp (itx - x

2

)] - — e{( 2 x.) exp [it 2 x.- 2 x.]}

n j j=i J j=i j

n n n 2

Let us evaluate E [x

2

exp ( itx-x2

) ] and E{( 2 xj^exp[it 2 x . » 2 x'Tjl

3=1 j=i 3 j=i
h

E(x 2

e itx ”x2

) = »f”(t)

n 9 n

e{( 21 x ) '"'exp t it 21 x . - x'; j j-1 J j=l J J=1 J ~[rn (t)V

It follows that the identity (5) may be rewritten as follows

-nf

n ~ 1

(t)f"(t) + fn_1

(t)f"(t) + (n-l)f

n ' 2 (t)[f>(t)] 2 = a fn

(t)

or

fn_1

(t)f ,, (t)-f

n "2

(t)tf'(t )]

2 =—

xr

ri (t) , (a

x

= ~I) (11)

p Clearly f(t) is a continuous function and f(0) = J

e"x dF) > Os

therefore, in some neighborhood of zero f(t) J- 0,

Let us solve equation (11) in this region*. We divide both

n

sides of the equation by f (t), so as to obtain

f"(t)

TTtT

_rf 5 (t)

L nrr ( 12 )

orflog f(t)] w “ -a_ „ The solution of this equation is

Here

f (t) = k exp[itt- t 2

]

k = f(0) = Je“x dF(x) *1

2

1 = j

,f » (0) = Jxe“x dF(x)

(13)
5

It 5 s easy to show that this solution may be extended to the

entire line. We shall make use of the fact that the set of zeros

of f(t) is closed. Assuming that f(t) does have zeros, then there

exists a value t^ such that f(t^) = 0, while for all values of

t for which [tj<|t j, f(t) ^ 0, Since f(t) is the solution of

° a l a

the equation it may be written in the form k exp[i/t- -*r*t ] „ Let, for instance t^>0. We can choose an e sufficiently small,

for which

f(t

Q

-e) f 0,

f(t

Q

-e) = k exp[i^(t - c)--^-(t

Q

- e) 1

From this it follows that f(t~) = lim f(t - e) cannot be equal to

0

0 3^ 2

gero. Wehave thus shown that the solution f(t) = k exp[i/t~ 2*” t ]

is true over the entire line*

Let us now investigate the function Q(x), defined as follows 2

x 2 • Q(x) = J

e”

X dF (x ) , ( lip

)

“00

Then by definition f(t) is the Fourier transform of the function

fi(x)* But f(t) is the characteristic function of the Gaussian law,,

and therefore S2(x) may be represented as follows

( x-r

k

f V '2*1 V dx . </2Tca^ J

-oo" 1

£ (x) (15)
6

In other words, we may define the function F(x) by the equation

_ (x-?Q

P e"x2dF(x) = ~~rr P e 2al dx • -co Aai -od

( 16 )

To solve this equation, let us note that for arbitrary x^ and x

x 0 « Xo - (x -/)

2

k , — - (i 7 ) r

: 2 2

J

e x

dP(x) = , r

J

Xi

e 2al dx .

We shall prove that F(x) is differentiable at every point, and

we shall evaluate its derivative.

Let x

Q

be some point, for example x^>0, and let h>0, then

xD+h 2

j e“

X dF (x

)

k

x^+h

/2-jia f

e

1 x.

(x-r)

2

a

l dx

and therefore

. ,2. XQ+h

(x°+h)

j dF (x

)

‘o

xQ+h 2 t

xo+h

< J e X

dF(x) < e X

° J dF(x). (18)

From this it follows that

>/2n;ai

-2 Xsb - (x-y)

e 2al dx <

h

F(x0+h)-F(x0) E

‘

k e < -^===

““"TT

( x„+h) 2

/2ita5 J

£ 19 )
7

An analogous inequality nay be obtained for h < 0, and we thus obtain

(x o

} ~ exp[xQ - UQ -r ) 2

2a-j_

(20)

The &bove theorem may be somewhat generalized as follows*

Let x^, x^j x

n

be mutually independent random variables,

identically distributed according to the law F(x), Let us form the

statistics x ~

°*zx

2 + ‘ ° ,<yn

x

n ^ we assu^e that all are nonnegative

,

and that

n

2 oL k=l

2

k

1)

2 n 2 2

and s = Z xv +, t ,oi x ) , k=l

K 1 1 n n

»>v 2

If x and s are independent, then the general population is

normally distributed.

Proof » As in the previous theorem, let us introduce the function

f(t) = |e ltx -x2dF(x)

Then in the manner utilized in the previous theorem, we obtain the

equation

2 gnat*! . L2 flhth - * o* v *.

jc=i f(oikt) k=X 1 J f(oijt)

'The solution of this equation yields the desired answer.

References

E, Lukacs, A characterization of the normal distribution* Annals of

Mathematical Statistics, 13 (1942), 91 *

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