Cochran's theorem

Cochran's theorem In statistics, Cochran's theorem, devised by William G. Cochran,[1] is a theorem used to justify results relating to the probability distributions of statistics that are used in the analysis of variance.[2] Inhalt 1 Aussage 1.1 Nachweisen 2 Beispiele 2.1 Sample mean and sample variance 2.2 Distributions 2.3 Estimation of variance 3 Alternative formulation 4 Siehe auch 5 References Statement Let U1, ..., UN be i.i.d. standard normally distributed random variables, und {Anzeigestil U=[U_{1},...,U_{N}]^{T}} . Lassen {Anzeigestil B^{(1)},B^{(2)},Punkte ,B^{(k)}} be symmetric matrices. Define ri to be the rank of {Anzeigestil B^{(ich)}} . Definieren {displaystyle Q_{ich}=U^{T}B^{(ich)}U} , so that the Qi are quadratic forms. Further assume {Anzeigestil Summe _{ich}Q_{ich}=U^{T}U} .

Cochran's theorem states that the following are equivalent: {Anzeigestil r_{1}+cdots +r_{k}=N} , the Qi are independent each Qi has a chi-squared distribution with ri degrees of freedom.[1][3] Often it's stated as {Anzeigestil Summe _{ich}EIN_{ich}=A} , wo {Anzeigestil A} is idempotent, und {Anzeigestil Summe _{ich}r_{ich}=N} is replaced by {Anzeigestil Summe _{ich}r_{ich}=rank(EIN)} . But after an orthogonal transform, {displaystyle A=diag(ICH_{M},0)} , and so we reduce to the above theorem.

Proof Claim: Lassen {Anzeigestil X} be a standard Gaussian in {Anzeigestil mathbb {R} ^{n}} , then for any symmetric matrices {Anzeigestil Q,Q'} , wenn {Anzeigestil X^{T}QX} und {Anzeigestil X^{T}Q'X} have the same distribution, dann {Anzeigestil Q,Q'} have the same eigenvalues (up to multiplicity).

Nachweisen: Let the eigenvalues of {Anzeigestil Q} be {Anzeigestil Lambda _{1},...,Lambda _{n}} , then calculate the characteristic function of {Anzeigestil X^{T}QX} . It comes out to be {Anzeigestil phi (t)=links(Produkt _{j}(1-2ilambda _{j}t)Rechts)^{-1/2}} (To calculate it, first diagonalize {Anzeigestil Q} , change into that frame, then use the fact that the characteristic function of the sum of independent variables is the product of their characteristic functions.) Zum {Anzeigestil X^{T}QX} und {Anzeigestil X^{T}Q'X} to be equal, their characteristic functions must be equal, Also {Anzeigestil Q,Q'} have the same eigenvalues (up to multiplicity).

Claim: {displaystyle I=sum _{ich}B_{ich}} .

Nachweisen: {Anzeigestil U^{T}(I-sum _{ich}B_{ich})U=0} . Seit {Anzeigestil (I-sum _{ich}B_{ich})} is symmetric, und {Anzeigestil U^{T}(I-sum _{ich}B_{ich})U=^{d}U^{T}0U} , by the previous claim, {Anzeigestil (I-sum _{ich}B_{ich})} has the same eigenvalues as 0.

Lemma: Wenn {Anzeigestil Summe _{ich}M_{ich}= Ich} , all {Anzeigestil M_{ich}} symmetric, and have eigenvalues 0, 1, then they are simultaneously diagonalizable.

Fix i, and consider the eigenvectors v of {Anzeigestil M_{ich}} so dass {Anzeigestil M_{ich}v=v} . Then we have {displaystyle v^{T}v=v^{T}Iv=v^{T}v+sum _{jneq i}v^{T}M_{j}v} , so all {displaystyle v^{T}M_{j}v=0} . Thus we obtain a split of {Anzeigestil mathbb {R} ^{N}} hinein {displaystyle Voplus V^{Täter }} , such that V is the 1-eigenspace of {Anzeigestil M_{ich}} , and in the 0-eigenspaces of all other {Anzeigestil M_{j}} . Now induct by moving into {Anzeigestil V^{Täter }} .

Fall: All {displaystyle Q_{ich}} are independent Fix some {Anzeigestil i} , definieren {Anzeigestil C_{ich}=I-B_{ich}= Summe _{jneq i}B_{j}} , and diagonalize {Anzeigestil B_{ich}} by an orthogonal transform {Anzeigestil O} . Then consider {displaystyle OC_{ich}O^{T}=I-OB_{ich}O^{T}} . It is diagonalized as well.

Lassen {displaystyle W=OU} , then it is also standard Gaussian. Then we have {displaystyle Q_{ich}=W^{T}(OB_{ich}O^{T})W;quad sum _{jneq i}Q_{j}=W^{T}(I-OB_{ich}O^{T})W} Inspect their diagonal entries, to see that {displaystyle Q_{ich}perp sum _{jneq i}Q_{j}} implies that their nonzero diagonal entries are disjoint.

Thus all eigenvalues of {Anzeigestil B_{ich}} sind 0, 1, Also {displaystyle Q_{ich}} ist ein {displaystyle chi ^{2}} dist with {Anzeigestil r_{ich}} degrees of freedom.

Fall: Jeder {displaystyle Q_{ich}} ist ein {displaystyle chi ^{2}(r_{ich})} distribution.

Fix any {Anzeigestil i} , diagonalize it by orthogonal transform {Anzeigestil O} , and reindex, so dass {displaystyle OB_{ich}O^{T}=diag(Lambda _{1},...,Lambda _{r_{ich}},0,...,0)} . Dann {displaystyle Q_{ich}= Summe _{j}Lambda _{j}{U'}_{j}^{2}} für einige {displaystyle U'_{j}} , a spherical rotation of {Anzeigestil U_{ich}} .

Seit {displaystyle Q_{ich}sim chi ^{2}(r_{ich})} , we get all {Anzeigestil Lambda _{j}=1} . So all {Anzeigestil B_{ich}succeq 0} , and have eigenvalues {Anzeigestil 0,1} .

So diagonalize them simultaneously, add them up, to find {Anzeigestil Summe _{ich}r_{ich}=N} .

Fall: {Anzeigestil r_{1}+cdots +r_{k}=N} We first show that the matrices B(ich) can be simultaneously diagonalized by an orthogonal matrix and that their non-zero eigenvalues are all equal to +1. Once that's shown, take this orthogonal transform to this simultaneous eigenbasis, in which the random vector {Anzeigestil [U_{1},...,U_{N}]^{T}} becomes {Anzeigestil [U'_{1},...,U'_{N}]^{T}} , but all {Anzeigestil U_{ich}'} are still independent and standard Gaussian. Then the result follows.

Each of the matrices B(ich) has rank ri and thus ri non-zero eigenvalues. For each i, the sum {Anzeigestil C^{(ich)}equiv sum _{jneq i}B^{(j)}} has at most rank {Anzeigestil Summe _{jneq i}r_{j}=N-r_{ich}} . Seit {Anzeigestil B^{(ich)}+C^{(ich)}=I_{Ntimes N}} , it follows that C(ich) has exactly rank N − ri.

Therefore B(ich) und C(ich) can be simultaneously diagonalized. This can be shown by first diagonalizing B(ich), by the spectral theorem. In this basis, it is of the form: {Anzeigestil {Start{bMatrix}Lambda _{1}&0&0&cdots &cdots &&0\0&lambda _{2}&0&cdots &cdots &&0\0&0&ddots &&&&vdots \vdots &vdots &&lambda _{r_{ich}}&&\vdots &vdots &&&0&\0&vdots &&&&ddots \0&0&ldots &&&&0end{bMatrix}}.} Thus the lower {Anzeigestil (N-r_{ich})} rows are zero. Seit {Anzeigestil C^{(ich)}=I-B^{(ich)}} , it follows that these rows in C(ich) in this basis contain a right block which is a {Anzeigestil (N-r_{ich})mal (N-r_{ich})} unit matrix, with zeros in the rest of these rows. But since C(ich) has rank N − ri, it must be zero elsewhere. Thus it is diagonal in this basis as well. It follows that all the non-zero eigenvalues of both B(ich) und C(ich) sind +1. This argument applies for all i, thus all B(ich) are positive semidefinite.

Darüber hinaus, the above analysis can be repeated in the diagonal basis for {Anzeigestil C^{(1)}=B^{(2)}+Summe _{j>2}B^{(j)}} . In this basis {Anzeigestil C^{(1)}} is the identity of an {Anzeigestil (N-r_{1})mal (N-r_{1})} vector space, so it follows that both B(2) und {Anzeigestil Summe _{j>2}B^{(j)}} are simultaneously diagonalizable in this vector space (and hence also together with B(1)). By iteration it follows that all B-s are simultaneously diagonalizable.

Thus there exists an orthogonal matrix {Anzeigestil S} so dass für alle {Anzeigestil i} , {Anzeigestil S^{Mathrm {T} }B^{(ich)}Sequiv B^{(ich)prim }} is diagonal, where any entry {Anzeigestil B_{x,j}^{(ich)prim }} with indices {Anzeigestil x=y} , {Anzeigestil Summe _{j=1}^{i-1}r_{j}

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