A Proof of the Herschel-Maxwell Theorem Using the Strong Law of Large Numbers

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by  Somabha Mukherjee

A Proof of the Herschel-Maxwell Theorem
Using the Strong Law of Large Numbers
Somabha Mukherjee
∗
Department of Statistics, Wharton School, University of Pennsylvania
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1 January10, 2017
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a Abstract
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In thisarticle, weuse the strong law of large numbers to give a proof of the Herschel-Maxwell theorem,
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which characterizes the normal distribution as the distribution of the components of a spherically symmetric

randomvector,providedtheyareindependent. Wepresentshorterproofsunderadditionalmomentassumptions,
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R andincludearemark,whichleadstoanotherstrikinglyshortproofofMaxwell’scharacterizationusingthecen-
trallimittheorem.
P
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h KEYWORDS: Sphericallysymmetric;Normaldistribution;Characteristicfunction;Stronglawoflargenum-
t
a bers;Centrallimittheorem.
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Contents
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2 1 Introduction 1
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2 2 SomeBasicPropertiesofaSphericallySymmetricDistribution 2
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1. 3 TheSphericalSymmetryCharacterizationanditsProof 2
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7 4 ShorterProofsUnderAdditionalMomentAssumptions 3
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: 5 Conclusion 4
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1 Introduction
r
a
TheHerschel-Maxwelltheoremisoneofthemanybeautifulcharacterizationsofthenormaldisribution. Itstates
that if the distribution of a random vector with independent components is invariant under rotations, then the
componentsmustbeidenticallydistributedasanormaldistribution.
Asmentionedin[2],J.C.Maxwelladdressedthefollowingquestion: Whatisthedistributionofvelocitiesof
the gas particles? The argumentbehind Maxwell’s claim that velocities are normally distributed, hinged upon
twoverynaturalassumptionsaboutthedistributionfunction,independenceandrotationinvariance. Evenbefore
Maxwell, astronomer J.F.W. Herschel addressed a similar issue while characterizing the errors in astronomical
measurements. Heassumedthatthecomponentsofthetwo-dimensionalerrorsinmeasurementareindependent,
andthatthedistributionoftheerrorisindependentofitsdirection.
Inthis paper, we givea proofof the Herschel-Maxwelltheoremusing the stronglaw of largenumbers, and
givea remarkaboutanotherunbelievablyshortproofofthe theoremusing the centrallimit theorem. The main
toolsofouranalysisarecharacteristicfunctionsandHaar’sTheoremforrotation-invariantmeasuresonthesurface
oftheunitsphereinEuclideanspaces.
∗Electronicaddress:[email protected], [email protected]
SomabhaMukherjee Herschel-MaxwellTheorem
2 Some Basic Properties of a Spherically Symmetric Distribution
Definition2.1. ArandomvectorXtakingvaluesinRnissaidtohaveasphericallysymmetricdistribution,ifX
andHXhavethesamedistributionforeveryn nreal,orthogonalmatrixH.
×
Inthefollowingtwotheorems,westatesomebasicpropertiesofasphericallysymmetricdistribution.
Theorem2.1. Theentriesofasphericallysymmetricrandomvectorhavethesamedistribution.Moreover,ifthat
distributionhasafinitemean,thenthemeanmustbe0,andifthatdistributionhasfinitesecondmoment,thenany
twodistinctentriesoftherandomvectorareuncorrelated.
Theorem2.2. TherandomvectorX = (X ,...,X )T hasasphericallysymmetricdistributionifandonlyifits
1 n
characteristicfunctionφsatisfiesφ(t)=Eei||t||X1 forallt Rn.
∈
ItfollowsimmediatelyfromTheorem2.2, thatif a randomvector(X ,...,X )T followsa sphericallysym-
1 n
metricdistribution,thensodoesallitssubvectors. Wenowstateandprovesomesortofa“converse”ofthisfact,
underanadditionalassumption,whichwillplayacrucialroleinourmainproof.
Theorem 2.3. Let F be a distribution on R with the propertythat if X and X are independentobservations
1 2
fromF,then(X ,X )Thasasphericallysymmetricdistribution.Thenforeveryn,ifX ,...,X areindependent
1 2 1 n
observationsfromF,(X ,...,X )T hasasphericallysymmetricdistribution.
1 n
Proof. In view of Theorem 2.2, it suffices to show that EeiPnj=1tjXj = Eei(√Pnj=1t2j)X1 for all n and for all
t ,...,t . By Theorem 2.2, this is true for n = 1 and 2. Assume that the proposition holds for some n. Let
1 n
X ,...,X ,X be(n+1)independentobservationsfromF. Then,byourinductionhypothesis,wehave:
1 n n+1
EeiPnj=+11tjXj = EeiPnj=1tjXj Eeitn+1Xn+1 = Eei(√Pnj=1t2j)X1 Eeitn+1Xn+1 =Eei(cid:16)qPnj=+11t2j(cid:17)X1
(cid:16) (cid:17)(cid:0) (cid:1) (cid:16) (cid:17)(cid:0) (cid:1)
forallt ,...,t . Wearedone.
1 n+1
Theorem 2.4. Let X and Y be two independent random variables. Suppose that (X,Y)T has a spherically
symmetricdistribution.Then,P(X =0)iseither0or1.
Proof. Suppose, towardsa contradiction, that 0 < P(X = 0) < 1. By Theorem2.1, X and Y have the same
distribution.SinceX =d Xcosθ+Y sinθforeveryθ R,wehaveforallθ R:
∈ ∈
P(Xcosθ+Y sinθ =0,(X,Y)=(0,0))
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= P(Xcosθ+Y sinθ =0) P(X =0,Y =0)
−
= P(X =0) P(X =0)P(Y =0)
−
= P(X =0)(1 P(X =0))>0.
−
However, the sets (x,y) = (0,0) : xcosθ+ysinθ = 0 0 θ π are pairwise disjoint, and they form
6 ≤ ≤ 2
an uncountablecol(cid:8)lection. This contradictsthe factthatfor(cid:9)an(cid:0)yset in this(cid:1)collection, the probabilityof (X,Y)
belongingtothatsetispositive.
3 The Spherical Symmetry Characterization and its Proof
We willrequirea simpleversionofHaar’sTheoremforrotation-invariantmeasureson n 1, thesurfaceofthe
−
unitsphereinRn. Itisstatedbelow. S
Theorem3.1. Let µ be a rotation-invariantBorelprobabilitymeasureon n 1 i.e. µ(HB) = µ(B) for every
−
BorelsetB n 1andeveryn northogonalmatrixH. Then,µistheuSniformmeasureon n 1.
− −
⊆S × S
ItfollowsfromTheorem3.1,thatifXisaunitnorm,sphericallysymmetricrandomvectorinRn,thenXhas
theuniformdistributionon n 1. Wearenowreadytostateandprovethemainresultofthispaper.
−
S
Theorem 3.2. Let X and Y be two independent random variables. Suppose that (X,Y)T has a spherically
symmetric distribution. Then, X and Y are identically distributed as a normal distribution with mean 0 (and
possibly0variance).
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SomabhaMukherjee Herschel-MaxwellTheorem
Proof. ByTheorem2.1,X andY havethesamedistribution,sayF. ByTheorem2.4,P(X =0)iseither0or1.
Inthelattercase,X hasthenormaldistributionwithmean0andvariance0. So,assumethatP(X =0)=0.
Generate a sequence X of independent random variables from the distribution F, and a sequence
Z ofindependentN{(0n,}1∞n)=ra1ndomvariables.Foreachn,callX =(X ,...,X )TandZ =(Z ,...,Z )T.
{Itfonl}lo∞nw=1sfromTheorem2.3thatX hasasphericallysymmetricdistrinbutioni.1e.forevneryn nnorthogo1nalmantrix
n
H,X andHX havethesamedistribution.So, ×
n n
X HX X
n =d n =H n
X HX X
n n n
|| || || || || ||
for every n and every n n orthogonalmatrix H. Thus, Xn is a unit norm spherically symmetric random
X
vector in Rn, and hence,×follows the uniformdistribution on|| nn|| 1. By the same argument, Zn also follows
− Z
S || n||
the uniformdistribution on n 1. Hence, Xn =d Zn for all n. This, in turn, implies that √nX1 =d √nZ1
− X Z X Z
foralln. BytheStrongLawSofLargeNumb||erns,||theri||ghnt||handsideconvergesalmostsurelytoZ||. On|b|serve||thna||t
1
EX2 < , since otherwise, by the Strong Law of Large Numbers for independentand identically distributed
1 ∞
random variables with expectation+ , it would follow that the left hand side convergesalmost surely to 0, a
∞
contradiction. So, by the Strong Law of Large Numbers for finite mean, the right hand side convergesalmost
surelyto X1 . Hence, X1 =d Z andwearedone.
√EX2 √EX2 1
1 1
Remark1. AslightmodificationoftheproofofTheorem3.2yieldsthefollowingresult:
Theorem3.3. Supposethat X isasequenceofrandomvariablessatisfyingthefollowingconditions:
1. P(X =0)=0, { n}∞n=1
1
2. EX4 < ,
1 ∞
3. (X ,...,X )T issphericallysymmetricforalln 1, and
1 n
≥
4. Cov(X2,X2)=0forall1 i<j.
i j ≤
Then,X ,X ,...areidenticallydistributedasanormaldistributionwithmean0andpositivevariance.
1 2
TheonlyobservationneededbeforereplicatingtheproofofTheorem3.2isthat,undertheaboveconditions,
||√Xnn|| −→P EX12. Theorem3.3isprobablyinterestingonlyfromtheanglethattheindependenceoftheXn’scan
berelaxedpinlieuofsomeadditionalassumptions,inordertoarriveatthesamenormalcharacterization.
4 Shorter Proofs Under Additional Moment Assumptions
Theorem3.2hasshorterproofsunderadditionalassumptionsoffinitenessofthefirstandsecondmomentsofX.
SupposethatweonlyhaveEX < . SinceXhasasymmetricdistributionaround0,thisconditionisequivalent
totheexistenceofEX. Inth|isc|ase,∞thecharacteristicfunctionφofX isdifferentiableonR.
ByanapplicationofTheorem2.2,wehaveφ(s)φ(t)=φ √s2+t2 foralls,t R. SincethedistributionofX
issymmetricaround0,φisarealvalued,evenfunction. W(cid:0)eclaimth(cid:1)atφ(t) > 0∈forallt R. Ifnot,thensince
∈
2
φ(0)=1,bytheintermediatevaluetheorem,thereisat R,suchthatφ(t )=0. Sinceφ(t)= φ t for
0 ∈ 0 h (cid:16)√2(cid:17)i
2n
allt R,aneasyinductiongivesφ(t)= φ t forallt Randalln 1. Thisimpliesthatφ t0 =0
n n
∈ h (cid:16)22 (cid:17)i ∈ ≥ (cid:16)22 (cid:17)
foralln 1,whichisnotpossible,sinceφiscontinuousat0andφ(0)=1. Thisprovesourclaim.
≥
If we denote logφ by ψ, then we have ψ(s) +ψ(t) = ψ √s2+t2 for all s,t R. Taking partial deriva-
∈
tivewithrespecttosonbothsidesoftheaboveidentity,we(cid:0)get: (cid:1)
s
ψ (s)=ψ s2+t2 forall (s,t)=(0,0).
′ ′
(cid:16)p (cid:17)(cid:18)√s2+t2(cid:19) 6
This implies that there is a constantc such that ψ′(s) = c for all s = 0. Solving this differentialequation and
s 6
rememberingthatψiscontinuousat0withψ(0)=0,wegetψ(s)= cs2 foralls R. Sinceψ(s) 0foralls,
2 ∈ ≤
wemusthavec 0. Now,φ(s)=ecs22 foralls RimpliesthatX N(0, c)andwearedone.
≤ ∈ ∼ −
Iffurther,weassumethatEX2 < ,theproofturnsouttobesurprisinglyshort,andisgivenbelow.
∞
3
SomabhaMukherjee Herschel-MaxwellTheorem
Lemma4.1. Let X beasequenceofindependentandidenticallydistributedrandomvariables,satisfying
that(X ,...,X )T{hans}a∞ns=p1hericallysymmetricdistributionforalln. Foreachn,denotethepartialsum n X
1 n i=1 i
bySn. Then, √Snn (n=1,2,...)areidenticallydistributedasX1. P
Proof. For each n, let H denote the orthogonal matrix whose first row is 1 , 1 ,..., 1 and let X =
n √n √n √n n
(cid:16) (cid:17)
(X ,...,X )T. SinceX andH X havethesamedistribution,theirfirstentrieshavethesamedistribution.
1 n n n n
Now,considerprovingTheorem3.2undertheassumptionEX2 < . ThecaseEX2 =0istrivial,soassume
thatEX2 > 0. AsintheproofofTheorem3.2,generateasequence ∞X ofindependentrandomvariables
fromthecommondistributionofX andY,andforeachn,letS = { nn}X∞n=1. ByTheorem2.3andLemma4.1,
n i=1 i
X =d Sn foralln. ByTheorem2.1, EX = 0. Hence,bytheCentrPalLimitTheorem, Sn d N(0,EX2). So,
1 √n √n −→
X N(0,EX2).
1
∼
Remark2. If{Xn}∞n=1 isani.i.d. sequenceofrandomvariableswithSn d=ef ni=1Xi,andif √Snn convergesin
distributionto a limit, then EX2 < (see exercise3.4.3 of [3]). The finitenPess of EX2 is now an immediate
1 ∞
consequenceofthisfactandLemma4.1,whichinturngivesasecondproofofTheorem3.2.
5 Conclusion
Theorem3.2 appearsin [2] (Theorem0.0.1) and an early proofof it appearsin [1]. The treatmentin [1] is
howevernotveryrigorousonprobabilisticgrounds.Corollary10of[4]isaconsequenceofTheorem3.2.
Ourfirst proofof Theorem3.2 can be dividedinto two broadideas. Thefirst idea is to derivethe spherical
symmetrypropertyofanynumberofindependentobservationsfromadistributionbasedontheknowledgeofthe
sphericalsymmetryoftwoindependentobservationsfromthatdistribution.Thesecondideaistousetheoutcome
of the first idea along with the Strong Law of Large Numbers, to conclude the result. In the process, the unit
normsphericalsymmetrycharacterizationoftheuniformdistributiononthesurfaceofanndimensionalsphere
was crucially used. The main advantageof this proofis that it is free of any calculation trickery, and is purely
conceptual.
It is possible to give a more “direct” proof of Theorem 3.2 by solving the functional equation φ(s)φ(t) =
φ √s2+t2 foralls,t R, forageneralcharacteristicfunctionφ. However,thisapproachreliesstronglyon
∈
th(cid:0)eindepend(cid:1)enceassumptionoftherandomvariables,andcannot,forexample,beusedtoproveTheorem3.3.
Acknowledgement
TheauthorthanksProfessorJ.MichaelSteeleforoneofhisassignmentproblemsintheSTAT930courseoffered
bytheUniversityofPennsylvania,whichwasthesourceoftheideabehindtheproofofTheorem3.2.
References
[1] Bartlett,M.S.(1934),Thevectorrepresentationofasample,Math.Proc.Cambr.Phil.Soc.30,pp.327-340.
[2] Bryc, W. (2005), NormalDistribution characterizationswith applications, Lecture Notes in Statistics 1995,
Vol100.
[3] Durrett,R.(2013),Probability:TheoryandExamples,CambridgeUniversityPress.
[4] Meckes,E.S.&Meckes,M.W.(2007),TheCentralLimitProblemForRandomVectorsWithSymmetries,J
Theoret.Probab.20,pp.697-720.
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