1927_Distribution of means.pdf

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Biometrika Trust

The Distribution of Means for Samples of Size N Drawn from a Population in which theVariate Takes Values Between 0 and 1, All Such Values Being Equally ProbableAuthor(s): Philip HallSource: Biometrika, Vol. 19, No. 3/4 (Dec., 1927), pp. 240-245Published by: Biometrika TrustStable URL: http://www.jstor.org/stable/2331961 .

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THE DISTRIBUTION OF MEANS FOR SAMPLES OF SIZE NDRAWN FROM A POPULATION IN WHICH THE VARIATETAKES VALUES BETWEEN 0 AND 1,ALL SUCH VALUESBEING EQUALLY PROBABLE.BY PHILIP HALL, B.A.

A PARTICULAR samplenwhichhevariate takes hevaluesx1, C2,.., X maybe representedy thepointP = (x1, x2, ..., xNy)ofN-dimensionalpace. Therepresentativeoints f a largenumber fsampleswilltend to be evenly is-tributedhroughouthe nteriorf heunithypercube0


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PHILIP HALL 241Let S be any point of Q, i.e. any point whose coordinates re all > 0, and let

justs of ts coordinatese > 1. Then will elong o ust(s)Ql's,2) 2'S,and (s) = 1Q. Now, fs > 0, we have

S ( 1)r r = ? ................ (7).Hence ifwhenever pointof 4) belongsto a Qrwe give it a " density" -1)r andthen sumoverall Q, the resultant ensitywill be equal to 1 or0 according s thepoint n questionbelongsto (1) or not. Let the segment of (4) lying n QOhavethe contentVN(p). Then the segmentof (4) lying n (5) will have. the contentVN(p r), (which s 0 whenr p). And the segment f 4) lying n (1) will havethecontent

E (-1)(r)VN(p-r)

................ (8),hereK= [p] is thegreatest nteger ess thanp.It remainsto findVN(p). Let VN-1p) be the content of the projectionofVN(p) perpendicularo one of the axes,so thatVNT(P) /NX VN-1P).Now VN(P) is the contentof the N-dimensional egionboundedby (4) and thecoordinatehyperplanes, regionwhosebase is thereforef contentVN(p). Theperpendicular rom on to this base is equal top/VlN.HenceVN(P)= NX = X VN ( P) ........................... (9).

Since V2p) = V/2p,we getVN(P)= N 1)ipN1 .............. (10).

Substituting n (8), we findfor the contentof the regioncommonto (1) and (4)thevaluef (p)= (fI) !O (_J)rN)(p r)N-l ........(11),

forvalues of p betweenK and K+ 1. To findthe actual frequency istributionof m, we note that ff(p) d (Pjf) - the volumeof the hypercube 1 and hencef:(Nrn) dn= . The distribution (m) ofrn s, therefore,iven by

(N-N1)! r=0 r ( N)}(12).K cK+l

Naturallyy= F (i) is symmetricalbout m= . It consistsof N arcs of degreeN- 1 havingNA- 1-point ontact t theiroins, viz.at thepointsB(icmrk2X....XN-1

Biometrika ix 16

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242 Distribution fMeans ofEqually Probable ValuesTheMoments.These are defined y Ma = jrM-J)a (m) dm ...................... (13)

Owingto the symmetryfF, we haveMf2a+=O, (a= O,1, ................ (14),

N~~~2NN112] Nrhile M21= IN[](_lr(f i(mi2 n- dm(NNN-l Io 1) (r(- N) B 2a +1, N),

orusingAN to denote the nthdifference,B(2a+1l N)( N 7Y N(N-1)NX r=O (r) (2 )(A(2a) IN N (XVN+2) ....................... (15)',(N+2a).! 22ax -or, n terms fthegeneralisedpolynomials f Bernoulli* defined yB(-~N) ( O=V . AN XcN+ ,.,............................... ,16))(x)( + v)!(1)

M2a 12) .........(17).These moments an,however, e obtained n a different orm nd by an entirelydifferent ethod.We consider n arbitrary requency istribution (x) of a variatex and writer+,saJ Af (x) dx .............................. (18).no I NWe may suppose a,,=1 and u,= 0. Then the distribution f meansm 7Ni 1xifor amplesofsize N will be given by

F (m)= A fif2..fNvd............................ (19),whereJ fxi) and dT is the N- 1-dimensionalolume-elementn thehyper-plane (4), theintegralbeing taken completely ver 4). Let us denote byMathemoments fthisdistribution.Thenr+0MO=J F(m) dm= 1,if A = VN, whichwe accordinglyuppose, nd

Ma = ...f mfi f2.. .fNdxl..dXN,_00 _ 001 a!i.e. M a ,! s2! ... S 2N .(20),* Notto be confusedwith he B of thecompleteBeta-functionn formula 15).



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PHILIP HALL 243the summation eingtaken over all integer olutions f

NiXl ss1a... ..................... (21).

Si>0, (i=1,2,...,N)}Since pu, 0,we mayalso add the condition j+-0. Now in ourcase0-= (s =1, 2,).

f(m - 1)28dm= 1(2)s+ .................. (22).Hence,we get(2a)! ~... (2)M24= (2N)2a X (2s, + 1) ! (2s,, + 1) ! . .. (2sN + 1) ! ...... 2)

takenas before verall integer olutions f 21)*.We returnnowto thegeneralformula20) truefor nyfrequencyistribution.To expressMa. s a polynomialn thep, let us take any partition f a intoN non-negative ummands,nd let therebe ao O's, , l's, a2 2's and so on, so thatYai=N, liaj=a, as>,0 ........................ (24).The number f terms f 20) correspondingo this partitionwillbe

N!a0! a,1!Hence we have Na !N (2a2 3! /-t. ?laopala2 * .......................25),a a ao! a,! ... (2 !)a2 (3 )a, ... L 2(2)thesummation eingnowextended oall possiblepartitions f 24). For example,the first ewcases maybe written own remembering1= 0) as follows:

m =0M = 2

3N2 1 ... .(26).M4N3[,u +3 (N- 1) 2] (2)M0=I[F +10 (N-1)u p2]Me= NJ so+ 15 (N-1) )/44/.L+10 (N-1) 33 15 (N-1) (N- 2) I]A* That (17) and (23) are consistentmaybe seen,e.g. byconsultingN6rlund'sDifferenzenrechnung

and comparinghe sixthequationof p. 139withthe ast equationofthesame section,on p. 140. Theactualvalues of the M2a forthe first ewvalues of a maybe readoff romN6rlund'stable 6, p. 460,D(-N) Tue.rememberinghat nhis notation, ?2 {2N- . Thus e.g.2=-2N as follows lso from26) and t22).1 , s olos lov26lad(2)16-2



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244 Distributionf Means ofEqually ProbableValuesIntroducing he/3's y the formulae

/328-2= - 3828S1 = /t28+2 .. (27),and similarly or heB's in termsof the M's, we have

B=NB2 3 2- 32--NB3 OB, -= - 10/3 ............ (28).B4-15B2-1OB, + 30= 4-1582-10,1 + 30

..................j

As N--oo, the dominant erm n (25) is that whichcorresponds o a partitionhaving the smallestpossiblevalue for o. This is the partition2 2 ...) if a is evenand (3 2 2 ...) if a is odd. Hence,we have as N--oo1 (2a)! N!N'if2 (2r !) (N -a)!

or Mr (2a-1)(2a-3). .53( N) H {(2r-1)M2} ...... (29),N/ r=1while M2l 1 (2a +1)! NN2a+l (2 !)a-13! (N - a)! (a - 1)!or M,0+ 'a (2a + 1) (2a-1) (2a-3)... 5. 3."fa+:3 (30).Whence B2'_2 ~ (2a-1) (2a-3)...7.5.3. ................3.. (31),

B2afi3 i(a -1) (2a -1) (2a -3) . . 7. 5.5 81= (a - 1)BB2a^_2 .. *(32).If we change our unit of length n the ratio1 VN*,we see at once fromheseequationsthatthe newmonmentsendto normal-curvealues as N-poo.

EDITORIAL NOTE.Proceedingby completelydifferentmethods, rwin and Hall have obtainedequations for the frequency istribution f means in samplesfrom populationfollowing "rectangular" aw ofdistributionp. 236, 19) and p. 241, (12)). Theirresultsare of considerable nterest, ot only n themselves, ut for he lighttheythrow ponthedistributionf momentsn samplesfrom opulationsn which hevariable ies within finite r limitedrange.Hall makesuse of a conceptionwhichhasbeen employednconnection iththesampling istributionfmeans nd standard eviations rom Normalpopulation ,* So as tomaintain constanttandardeviations N inereases.t R. A.Fisher, iometrika,Vol.x.p. 507.



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PHILIP HALL 245representinghe ample fN bya pointx1, 2, .., xN) in anN-dimensionalensityspace. For a populationof imited ainge,beying law y =f (x), we have, hoosingsuitable units,0 < x< 1, so that the appropriatedensity fieldwill be the unithyperculbe,ithpoint density t (xl, x2, ..* xN) off (cl) f (x2). f (xN). In thecaseof the"rectangular"populationforwhichf (x) is constant, hedensity s uniformthroughout he hypercube, ut for therforms f distribution,s for xample thatof the general Type I curve, his will not be so. The frequency istribution fmeans,mn,n samples of N, will still howeverbe obtained by integrating hedensity hroughout he N- 1-dimensioned egionwithin he hypercube ying atright ngles o the diagonal xisx1 x2 ... = xN. Just s forHall's simple ase,it would appear that the law representing he content f this regionwill changeits form t the N- 1 pointsm= Ic/N = 1, 2, .., N - 1) so that the resultingdistributionfmeans will consistof N connected rcs.

In the simple case of N= 3, the positioncan be graspedvisuallyfromHall'sdiagram (p. 240). The frequency f m is here proportional o the integral ofa density (x,) f (x2) f (x3) overtheplane perpendicular o thediagonal01. Thisplane is represented s cutting01 in M whereOM/OI= m. As M moves wayfrom the law of distributionemains f the sameform ntilOM I1 when thesectionof the plane by the cube changesfrom triangle o a hexagon. The newlaw holds until OM= 2 01, when the section changes back from hexagon toa triangle, nd the third aw holds until M reaches and m= 1. For thespecialcase of the "rectangular"population,rwin deduces these three laws from hegeneral equationsforN= 2, 3 and 4 on pp. 237-238.It seems probablethat the solution utlinedby Irwin at the end of his paper,forType II curves,will lead also to a system f separate arcs, s the considerationofthe hypercube uggests.

The case oftheType III curve s of interest s formaingtransitionype; onelimitofrange s removed o infinity,o thatonlyone corner f thehypercubeiesat a finite oint. As M movesawayfrom along the diagonal Hall's figure) heshape of theregion ying t right nglesto this ine, hroughout hich hedensityis to be integrated, oes not change as long as.m remainsfinite. Consequentlythere s a single aw ofdistributionfmeans, s givenfirst yChurch nd laterbyIrwin p. 229, (7)). This shouldalso be true forType VI curves limited in onedirection),s well as forType IV where he density ieldwillfill he whole of theN-dimensional pace,as for heNormal curve.A considerationfthehypercubelso suiggestshatthedistributionfstandarddeviationsn samplesfrom limited angepopulationwillalso consist fconnectedarcs. Withthe lettering f Hall's figure,he standard deviationof the sample sC-= MP/OI, and thefrequencyfa givenvalueof a-willbe obtainedby ntegratingthedensityhroughoutheregionforwhichMP is constant, hichmaybe describedas a "hypercylinder," ith axis 01, and cross-sectionn (N - 1)-dimensionedhypersphere.As MP increases he law representinghe form f this regionwillchange according o thenumber ffacesof thehypercubewhich t cuts. E. S. P.

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1927_Distribution of means.pdf