Minimizing the expected sample range

Journal

of Statistical

Planning

and Inference

347

8 (1983) 347-354

North-Holland

MINIMIZING THE EXPECTED SAMPLE RANGE Ester SAMUEL-CAHN Hebrew

University

Received

4 November

Recommended

Abstract: i=l,..., shown

of Jerusalem,

Israel

1982; revised

by Esther

manuscript

received

7 February

1983

Seiden

Let Xi be i.i.d.

random

variables

with finite expectations,

and 8i arbitrary

constants,

n. Yi=Xi+Bi.TheexpectedrangeoftheY’sisR,(B,,...,B,)=E(maxY;-minYi). that the expected

range

is minimized

Itis

if and only if 0, = ... = 8,. In the case where the X;

are independently and symmetrically distributed around the same constant, but not identically distributed, it is shown that f?t = ... = 8, are not necessarily the only (et, . . . , 0,) minimizing R,. Some lemmas AMS

which are applicable

I970 Subject

Key words:

1. Introduction

Classification:

Sample

range;

to more general

problems

of minimizing

R, are also given.

62G30.

Location

parameter;

Minimization.

and summary

Let Xi be independent random variables with finite expectations, where Bi are constants, --03
M, = max c, i=

Yj =Xi + Bi

I,....n

i=

I,....n

The sample range is W,, =M, - m,. We are interested in its expected value R,(B)=R,(B,,...,B,)=EgW,, as a function

of

8. Clearly,

for

any

8 and

constant

c, R, (~9,,. . . , e,) =

R,(e,+c,...,e,+c). A recent paper, Salama and Sen (1982), is devoted to proving: Theorem 1. Let the Xi be symmetric minimized if 8, = ... = 0,.

around the same constant.

Then R,(8)

is

Actually, the proof of Salama and Sen implies that the condition 13~= ... = 8, is both necessary and sufficient for minimization of R,(B). This is clearly false as the following example shows. 0378-3758/83/$3.00

0

1983, Elsevier

Science

Publishers

B.V. (North-Holland)

E. Samuel-Cahn / Minimizing

348

expected sample range

Example 1. Let n = 2 and let Xi equal -1 and 1, each with probability 3, and X, equal -2 and 2 each with probability +. Then it is easily checked directly that Rz(O, 0) = R,(O, 0) for all - 1 < t9< 1. A similar result is true if X1 is uniform on [-2, -11 and [1,2], and X, is uniform on [-4, -31 and [3,4] (in which case both Xi are absolutely continuous with respect to Lebesgue measure).

The proof of Theorem 1 of Salama and Sen (1982) is based on a series of inequalities, some of which are claimed to always be strict with positive Lebesgue measure. This is not correct. One could strengthen their Theorem 1 by checking for necessary and sufficient conditions for strict inequalities there. We have not attempted this. Let 0 be the empty set. We prove: Theorem

l*.

Let the Xi be symmetric around the same constant, and let At be the support of the distribution of Xi. (i) R,,(B) is minimized if 19,= --- = 8,. (ii) A necessary condition for 8, = --a = 8, to be the only 8’s minimizing R,(B) is #0

foraNj=l,...,n.

(1)

(iii) A sufficient condition for 8, = e-e= 8, to be the only B’s minimizing R,(B) is that the distributions of the Xi’s be continuous and AjflAjZ0

for all i+j.

(2)

Remark 1. It should be noticed that (1) is not a sufficient condition

sion of (ii), nor is (2) a necessary condition for the conclusion continuous case), as seen in Example 2 and 3, respectively. Example

for the concluof (iii) (even in the

2. Let XT and X,* have the distribution

of Xi in Example 1, and XT and X,* that of X, in that example. Then (1) is fulfilled, but it is easily verified that (e,e,e+t,e+E) for -1
Sufficient conditions other than (2) can also be given. Salama and Sen also say: “It would be of interest to show that when the Xi are independent and identically distributed then R,(O) may be minimized for 8r = -em= t9,, even without the symmetry condition”. The purpose of this note is to show that this

E. Samuel-Cahn / Minimizing expected sample range

is actually true for all distributions

(with finite expectation).

Theorem 2. Let Xi be independent and identically distributed. mized if and only if 8, = -.a = 0,.

349

We have: Then R,(8) is mini-

In Section 2 we prove the theorems: our proofs are based on several lemmas which can be utilized also in other problems of minimizing the range.

2. Proof of theorems Note that Theorem 2 is trivial if Xi is degenerate and Theorem l* is also trivial when all, or all but one of the variables are degenerate. If k < n - 2 of the variables are degenerate Theorem 1 again follows from the corresponding theorem where n is replaced by n-k. Thus we can, and shall, in the sequel assume that all random variables in our proofs are nondegenerate. Let t+ = t for t>O, 0 otherwise. The following simple lemma, is stated here in a somewhat greater generality than needed in our proofs. Lemma

1. Let U, V and X be any random variables with finite expectations, and let

h(e) =E(x+e- u)++E(v-x-e)+,

--oo
(3)

Then h(0) is minimised for 0= 8* if and o&y if 8* satisfies

P(x+e*>~)+aP(x-te*=~)=P(x+e*
(4)

forsomeO
mized for 0 being any median of the distribution. Proof of Lemma

1. If (X, U) and (X, V) have joint densities then (4) follows im-

E. Samuel-Cahn / Minimizing expected sample range

350

mediately (with a = p = 0) through differentiating h(B) =

m (x+&u)fx,,(x,u)dxdu 5 --oDc U-6 +

(u-x-@fx,,(x,u)dxdu

with respect to 8. In other cases h(B) may not be differentiable, and a slightly more detailed analysis is required to yield (4). We omit the details. Note that

e,_,)+E(x,+e,--bf,_,)+ +zqm,_ , -x, - en)+.

RIM 1, . . . . &)=R,-,(&...,

It therefore follows from Lemma 1 that, for given 8,, . . ..&_t. mizes R,(B) must satisfy

LetM~=max{~;j=l,..., Lemma

n, j#i}

and rn; =min(Yj;j=l,

(9

a 8, which mini-

. . ..n. j#i}.

We have:

2. A necessary and sufficient condition for (f9,, . . . ,t9,) to minimize R,(8)

is that P{Xi+8j>M~)+ariP{Xi+8j=M~) =P(Xi+Bi
i=l,...,n,

(6)

forsomeO
By Lemma 1 and the symmetric role of the Yi the necessity is immediate. By Remark 3 the sufficiency can also be shown to follow.

Proof of Theorem

2. For n = 2 the theorem

P{Yz>Yt}+aP{Yz=Y,}

is immediate since

=P{Y,
if and only if 0i = e2 (with properly chosen a and p). Clearly, when 8i = 1.0= B,, the Yi the i.i.d. and (6) holds (with all a’s and p’s equal). We shall show that 8, = . . . = 0, are the only solutions of (6). Suppose (6) holds for some 8 where not all components are equal, n > 3. Without loss of generality assume that 0, = max8i>8,_,=minBi. By (6),

E. Samuel-Cahn

/ Minimizing

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expected sample range

+a,_,P{X,_,+8,_,=max(M,-2,Xn+e,>> =P(X,_,+O,_,


=min(m,-2,Xn+&Jl

+&-,P{xn_,+8,_1

(7)

and also

P{x,+e,>max(M,_2rXn_1+en_1)} +~~~{x,+e,=max(M,_2,Xn-1+en-,)}

=P{x,+8,
(8)

We shall show below that

(9) Then, since the Xi are i.i.d., and by use of (7), we have P{x,+e,>max(M,_2,Xn-1+e,-1)> ~P{x,_,+en-,2-max(M,_2,Xn+0,)} ~P(X,_,+8,_,>max(M,_2rXn+8,)} +~r,_iP{x,_,+e,_,

=max(M,_,,X,+0,)}

=P{X,_I+e,_l
=min(m,_2,Xn+8,)}

2P{X,+t8,
(10)

But (10) implies that the left-hand side of (8) is greater than the right-hand side even if one takes (Y,,= 0 and /3,, = 1 in (8). This contradiction implies that there exists no

solution of (6) other than t9t= ... = 0,. We now prove (9). Clearly (9) holds with weak inequality, since if we denote by A and B the events in the left- and right-sides of (9) respectively, then A C B. We shall distinguish two possibilities concerning the distribution of the Xi and the value

8.

(i)

P{x,_,

+e,_,

aMn_2)

= 0;

(ii)

P{x,_,

+e,_,

aMn_z)

> 0.

In case (i) the value of the right hand side of (9) is also 0, whereas the value of the left hand side of (9) is always positive, so (9) holds. In case (ii) we have P(6>x,-x,_,>-6,x,_,+e”_,~MM,_~}>o, where 6 = 0, - 8, _ i > 0. Inequality (11) is true since zero is always the difference of two i.i.d. random variables.

(11) in the support

of

E. Samuel-Cahn / Minimizing expected sample range

352

Let the event in (11) be denoted by C. Then clearly CC A but C t-t B = 0, which established the strict inequality in (9). Remark 5. It is immediate that the result of Theorem 2 is valid if one assumes only

that the Xi are exchangeable

rather than independent.

Remark 6. In the setting of Theorem 2, let n = 2 and for simplicity assume that the Xi have a density f( *) (with respect to Lebesgue measure). It follows from the proof of Lemma 1 that R,(O, B) is strictly increasing in jB I. One may suspect that actually the stronger result, that W2 itself is stochastically nondecreasing, holds. This however, is true if and only if jf(x)f(x+r-O)dx>

jf(x)f(x+r+O)dx

for ally>O,

8>0,

as is easily verified. Though (12) is satisfied for many standard distributions, ample 4 shows that W2 need not be stochastically nondecreasing. Example

4. Let Xi be uniform

(12) Ex-

over [-2, -11 and [l ,2]. Then

(where the notation Pe should be obvious). It is easily seen that, for example, for e= 1, y=2, (12) fails. Remark 7. Let Xi, X, and X3 be as in Example 3. By Example 1, mine R2(B) = &(O, 1). However,

mm R3(0, 1, e,)

= Q > rnjn R3(@ = R,(O, 0,O) = Q.

This shows that stepwise minimization sional problem.

may not lead to a minimum for the n-dimen-

Proof of Theorem I*. Part (i) is just Theorem

1 and is proved (correctly) by Mama and Sen (1982). A simpler proof is obtained by using Lemma 2, since clearly 8i = .a. = 0, satisfies (6), with any 0 < oi = pi f 1, i = 1, . . . , n, by the symmetry of the Xi’s. To prove part (ii), assume (1) fails, and without loss of generality, assume it fails for j = n, i.e. n-1 A,n UAi =0. .( i=l > Then there exists an E > 0 such that P{Xn
=P{X,+&
for all i=l,...,n-1,

and because of the symmetry of the distributions this probability is also equal to P(X, >Xi} (=+). Using (5) and Lemma 1 it is immediate that R,(O,...,O)= R,(O, -.., 0, E). Therefore 8i = e-e= t$, are not the only solutions.

E. Samuel-Cahn

/ Minimizing

expected sample range

To prove (iii), notice that, with the continuity assumption, P(Xi>ML-08,)

=P{Xi
353

(6) can be.rewritten as

i=l,...,n*

(13)

Now suppose 8* minimizes R,(B) and e,* = max e,* > min ey, i=l,...,n i=l.....n

and that the point of symmetry of the Xi’s is 0. Then setting i = n and -Xi instead of Xi in the left hand side of (13), we must have P{X,<-M,_r+f?,*}

=P(X,.

(14)

We shall show that (14) cannot hold. Let m=

min Xi *=1,....n-I

and

M =

max

i=l,....n-1

Xi,

and note that because of the symmetry m and -M have the same distribution. -M,,_,+8,*=-max(Xi+tJ,*)+O,*>-A4

and

Now

m,_,-l?,*
i$n-I and thus the left hand side of (14) is at least as large as the right hand side. We shall show that

P{m,_,-e,*GX,
>O,

which shows that (14) fails. Without loss of generality assume minign 0;= 19:. Then it suffices to show P{x,-(e,*-e,*)
~0,

since X, + ~9;2 m, _ 1. Notice that P{X,
since the distributions

,...) X,_,20}>0,

are symmetric around 0. Thus it suffices to show that

P{x,-6~x,o,

where 6 = e,* - 0: > 0. But by (2) and symmetry we have A 1fl A, rl R- # 0, where The desired result now follows from the continuity of the distributions.

R- = (x:xcO}.

Though not needed for the cases of Theorems l* and 2, it may be of interest to have a general necessary condition for 0 to minimize R,(B). We have: Theorem 3. Let Xi be any independent random variables with finite expectation, Yi = Xi + 0,. A necessary condition for 8 to be such that it minimizes R,(B) is that, for aN i=l,..., n, P(Yi=M,) >O and P(Yi=m,)>O. Proof.

Suppose false, and suppose e.g. P{ q =M,}

= 0. Then there exists a j such

E. Samuel-Cahn / Minimizing expected sample range

354

that P{ Yj > Y;} = 1, and also P( Yj > Mi} > 0. But then P{Yj&}


=o,

so B cannot satisfy (6). Similarly if P( Yj = RZ,} = 0.

Reference Salama, 1.A. and P.K. Sen (1982). On expected sample range from heterogeneous tributions. J. Statist. Plann. Inference 6, 235-239.

symmetric dis-