# 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] Contenu 1 Déclaration 1.1 Preuve 2 Exemples 2.1 Sample mean and sample variance 2.2 Distributions 2.3 Estimation of variance 3 Alternative formulation 4 Voir également 5 References Statement Let U1, ..., UN be i.i.d. standard normally distributed random variables, et {style d'affichage U=[U_{1},...,U_{N}]^{J}} . Laisser {style d'affichage B^{(1)},B^{(2)},ldots ,B^{(k)}} be symmetric matrices. Define ri to be the rank of {style d'affichage B^{(je)}} . Définir {style d'affichage Q_{je}=U^{J}B^{(je)}tu} , so that the Qi are quadratic forms. Further assume {somme de style d'affichage _{je}Q_{je}=U^{J}tu} .

Cochran's theorem states that the following are equivalent: {style d'affichage 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 {somme de style d'affichage _{je}UN_{je}=A} , où {style d'affichage A} is idempotent, et {somme de style d'affichage _{je}r_{je}=N} is replaced by {somme de style d'affichage _{je}r_{je}=rank(UN)} . But after an orthogonal transform, {displaystyle A=diag(JE_{M},0)} , and so we reduce to the above theorem.

Proof Claim: Laisser {style d'affichage X} be a standard Gaussian in {style d'affichage mathbb {R} ^{n}} , then for any symmetric matrices {style d'affichage Q,Q'} , si {style d'affichage X^{J}QX} et {style d'affichage X^{J}Q'X} have the same distribution, alors {style d'affichage Q,Q'} have the same eigenvalues (up to multiplicity).

Preuve: Let the eigenvalues of {style d'affichage Q} être {style d'affichage lambda _{1},...,lambda _{n}} , then calculate the characteristic function of {style d'affichage X^{J}QX} . It comes out to be {style d'affichage phi (t)=gauche(produit _{j}(1-2ilambda _{j}t)droit)^{-1/2}} (To calculate it, first diagonalize {style d'affichage 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.) Pour {style d'affichage X^{J}QX} et {style d'affichage X^{J}Q'X} to be equal, their characteristic functions must be equal, alors {style d'affichage Q,Q'} have the same eigenvalues (up to multiplicity).

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

Preuve: {style d'affichage U^{J}(I-sum _{je}B_{je})U=0} . Depuis {style d'affichage (I-sum _{je}B_{je})} is symmetric, et {style d'affichage U^{J}(I-sum _{je}B_{je})U=^{ré}U ^{J}0tu} , by the previous claim, {style d'affichage (I-sum _{je}B_{je})} has the same eigenvalues as 0.

Lemme: Si {somme de style d'affichage _{je}M_{je}=je} , all {style d'affichage M_{je}} symmetric, and have eigenvalues 0, 1, then they are simultaneously diagonalizable.

Fix i, and consider the eigenvectors v of {style d'affichage M_{je}} tel que {style d'affichage M_{je}v=v} . Then we have {displaystyle v^{J}v=v^{J}Iv=v^{J}v+sum _{jneq i}v^{J}M_{j}v} , so all {displaystyle v^{J}M_{j}v=0} . Thus we obtain a split of {style d'affichage mathbb {R} ^{N}} dans {displaystyle Voplus V^{perp }} , such that V is the 1-eigenspace of {style d'affichage M_{je}} , and in the 0-eigenspaces of all other {style d'affichage M_{j}} . Now induct by moving into {style d'affichage V^{perp }} .

Cas: All {style d'affichage Q_{je}} are independent Fix some {style d'affichage i} , définir {displaystyle C_{je}=I-B_{je}=somme _{jneq i}B_{j}} , and diagonalize {style d'affichage B_{je}} by an orthogonal transform {style d'affichage O} . Then consider {displaystyle OC_{je}O^{J}=I-OB_{je}O^{J}} . It is diagonalized as well.

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

Thus all eigenvalues of {style d'affichage B_{je}} sommes 0, 1, alors {style d'affichage Q_{je}} est un {displaystyle chi ^{2}} dist with {style d'affichage r_{je}} degrees of freedom.

Cas: Chaque {style d'affichage Q_{je}} est un {displaystyle chi ^{2}(r_{je})} distribution.

Fix any {style d'affichage i} , diagonalize it by orthogonal transform {style d'affichage O} , and reindex, pour que {displaystyle OB_{je}O^{J}=diag(lambda _{1},...,lambda _{r_{je}},0,...,0)} . Alors {style d'affichage Q_{je}=somme _{j}lambda _{j}{U'}_{j}^{2}} pour certains {displaystyle U'_{j}} , a spherical rotation of {style d'affichage U_{je}} .

Depuis {style d'affichage Q_{je}sim chi ^{2}(r_{je})} , we get all {style d'affichage lambda _{j}=1} . So all {style d'affichage B_{je}succeq 0} , and have eigenvalues {style d'affichage 0,1} .

So diagonalize them simultaneously, add them up, to find {somme de style d'affichage _{je}r_{je}=N} .

Cas: {style d'affichage r_{1}+cdots +r_{k}=N} We first show that the matrices B(je) 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 {style d'affichage [U_{1},...,U_{N}]^{J}} becomes {style d'affichage [U'_{1},...,U'_{N}]^{J}} , but all {style d'affichage U_{je}'} are still independent and standard Gaussian. Then the result follows.

Each of the matrices B(je) has rank ri and thus ri non-zero eigenvalues. For each i, la somme {displaystyle C^{(je)}equiv sum _{jneq i}B^{(j)}} has at most rank {somme de style d'affichage _{jneq i}r_{j}=N-r_{je}} . Depuis {style d'affichage B^{(je)}+C^{(je)}=I_{Ntimes N}} , it follows that C(je) has exactly rank N − ri.

Therefore B(je) et C(je) can be simultaneously diagonalized. This can be shown by first diagonalizing B(je), by the spectral theorem. In this basis, it is of the form: {style d'affichage {commencer{bmatrice}lambda _{1}&0&0&cdots &cdots &&0\0&lambda _{2}&0&cdots &cdots &&0\0&0&ddots &&&&vdots \vdots &vdots &&lambda _{r_{je}}&&\vdots &vdots &&&0&\0&vdots &&&&ddots \0&0&ldots &&&&0end{bmatrice}}.} Thus the lower {style d'affichage (N-r_{je})} rows are zero. Depuis {displaystyle C^{(je)}=I-B^{(je)}} , it follows that these rows in C(je) in this basis contain a right block which is a {style d'affichage (N-r_{je})fois (N-r_{je})} unit matrix, with zeros in the rest of these rows. But since C(je) 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(je) et C(je) sommes +1. This argument applies for all i, thus all B(je) are positive semidefinite.

En outre, the above analysis can be repeated in the diagonal basis for {displaystyle C^{(1)}=B^{(2)}+somme _{j>2}B^{(j)}} . In this basis {displaystyle C^{(1)}} is the identity of an {style d'affichage (N-r_{1})fois (N-r_{1})} vector space, so it follows that both B(2) et {somme de style d'affichage _{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 {style d'affichage S} telle que pour tout {style d'affichage i} , {style d'affichage S^{mathrm {J} }B^{(je)}Sequiv B^{(je)prime }} is diagonal, where any entry {style d'affichage B_{X,y}^{(je)prime }} with indices {style d'affichage x=y} , {somme de style d'affichage _{j=1}^{i-1}r_{j}

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