Common Errors in Applying Cochran's Theorem

Why Careful Checks Matter

Cochran’s theorem turns a suitable decomposition of a Gaussian squared norm into a powerful distributional statement: the components are independent chi-square variables, with degrees of freedom determined by matrix ranks. But a sum-of-squares identity by itself does not justify that conclusion. The matrices must have the right structure, the decomposition must include every component, and the mean determines whether each chi-square distribution is central.

The safest approach is to check each condition separately. In particular, verify that the matrices are symmetric idempotent projections, that they sum to the identity when a full decomposition is claimed, and that the ranks add correctly. Then check the Gaussian mean before calling a component central.

Definition: A real matrix \(A\) is a symmetric idempotent matrix if \(A'=A\) and \(A^2=A\). Such a matrix is an orthogonal projection. Its quadratic form \(Y'AY\) is the squared length of the projection of \(Y\) onto the subspace onto which \(A\) projects.

The Conditions Behind the Conclusion

Here is a useful form of Cochran’s theorem. Let \(Y\sim N_n(\mu,\sigma^2 I_n)\), where \(\sigma^2>0\), and let \(A_1,\ldots,A_k\) be fixed, symmetric idempotent \(n\times n\) matrices. If they form a full decomposition, \( \sum_{j=1}^k A_j=I_n \), then their ranges are mutually orthogonal. Consequently, the quadratic forms \(Q_j=Y'A_jY\) are mutually independent, and \(Q_j/\sigma^2\) has a noncentral chi-square distribution with degrees of freedom \(\operatorname{rank}(A_j)\) and noncentrality parameter \(\mu'A_j\mu/\sigma^2\). It is central precisely when \(A_j\mu=0\).

This statement is deliberately explicit about both the matrices and the distribution of \(Y\). A symmetric matrix need not be idempotent; a list of projections need not be complete; and a nonzero mean can make a component noncentral. Each issue can invalidate a different part of an attempted application.

Theorem: Under the conditions just stated, \(A_iA_j=0\) whenever \(i\ne j\), the ranks satisfy \(\sum_{j=1}^k \operatorname{rank}(A_j)=n\), and the quadratic forms have the independent noncentral chi-square distributions described above.

Proof. Fix an index \(i\). Using \(\sum_j A_j=I_n\) and \(A_i^2=A_i\),

A_i=A_i I_n A_i =A_i\left(\sum_{j=1}^k A_j\right)A_i =A_i+\sum_{j\ne i}A_iA_jA_i.

Thus \(\sum_{j\ne i}A_iA_jA_i=0\). For each \(j\), symmetry and idempotence give \(A_iA_jA_i=(A_jA_i)'(A_jA_i)\), a positive semidefinite matrix. A sum of positive semidefinite matrices is zero only if each summand is zero: for every vector \(x\), the sum of the nonnegative quantities \(\lVert A_jA_i x\rVert^2\) is zero. Hence \(A_jA_i=0\), and by transposition \(A_iA_j=0\).

The ranges are therefore mutually orthogonal. Choose an orthonormal basis adapted to these ranges. In that basis, the matrices are diagonal blocks of zeros and ones, with a block of ones of size \(\operatorname{rank}(A_j)\) for \(A_j\). Since their sum is \(I_n\), these blocks cover all \(n\) coordinates, so their ranks sum to \(n\). The coordinates of \(Y\) in this basis are independent normal variables with variance \(\sigma^2\). Each \(Q_j/\sigma^2\) is a sum of squares of \(\operatorname{rank}(A_j)\) independent unit-variance normal variables, whose means are the coordinates of the projected mean. Disjoint coordinate blocks are independent. This gives the stated noncentral chi-square distributions and independence. The distribution is central exactly when the projected mean is zero, equivalently \(A_j\mu=0\). \(\square\)

Worked Checks for Common Failure Modes

Worked Example: A Symmetric Matrix That Is Not a Projection

Let \(Y\sim N_2(0,I_2)\), and consider \(A=\begin{pmatrix}1&0\\0&1/2\end{pmatrix}\). It is symmetric, but

A^2=\begin{pmatrix}1&0\\0&1/4\end{pmatrix}\ne A.

Writing \(Y=(Y_1,Y_2)'\), the quadratic form is \(Q=Y_1^2+\tfrac12Y_2^2\). Since \(Y_1,Y_2\) are independent standard normal variables,

E[Q]=1+\tfrac12=\tfrac32,\qquad \operatorname{Var}(Q)=2\left(1^2+\left(\tfrac12\right)^2\right)=\tfrac52.

The rank of \(A\) is two, but a central chi-square variable with two degrees of freedom has mean two and variance four. So \(Q\) is not \(\chi^2_2\). The failure is not a small bookkeeping issue: the matrix is not idempotent, and the theorem’s projection-based conclusion does not apply.

Worked Example: An Omitted Component Changes the Decomposition

Take six observations \(y=(2,4,3,5,7,6)'\), arranged as two groups of size three. Let \(J\) project onto the constant vector, \(H\) project onto the group-contrast subspace, and \(M\) project onto the within-group residual subspace. Their ranks are \(1,1,4\), respectively, and they form the full decomposition \(J+H+M=I_6\).

The grand mean is \(4.5\), and the group means are \(3\) and \(6\). The squared length of the grand-mean component is

$$ y'Jy=6(4.5)^2=121.5. $$

The between-group component and within-group residual component are, respectively,

y'Hy=3(3-4.5)^2+3(6-4.5)^2=13.5, \qquad y'My=(1^2+(-1)^2+0^2)+( (-1)^2+1^2+0^2)=4.

The full sum is \(121.5+13.5+4=139\), which agrees with \(y'y=2^2+4^2+3^2+5^2+7^2+6^2=139\). If the grand-mean component is omitted, the remaining components sum to \(17\), not \(139\); their ranks sum to \(1+4=5\), not \(6\). They describe the centered variation, not the full squared norm. An omitted component may be appropriate for a centered analysis, but it must not be silently treated as a complete decomposition of \(I_6\).

Worked Example: Orthogonal Components Can Be Noncentral

Let \(Y\sim N_2(\mu,I_2)\), where \(\mu=(1,0)'\), and use the orthogonal projections \(A_1=\operatorname{diag}(1,0)\) and \(A_2=\operatorname{diag}(0,1)\). They are symmetric and idempotent, sum to \(I_2\), and both have rank one.

Here \(Q_1=Y_1^2\) and \(Q_2=Y_2^2\). Since \(Y_1\sim N(1,1)\), \(Q_1\) has a noncentral chi-square distribution with one degree of freedom and noncentrality \(1\). Since \(Y_2\sim N(0,1)\), \(Q_2\sim\chi^2_1\) is central. The components are independent because they depend on distinct independent Gaussian coordinates. Their means check the distinction: \(E[Q_1]=1+1=2\), while \(E[Q_2]=1\). Orthogonality gives independence here, but it does not make every component central.

Worked Example: An Algebraic Identity Does Not Guarantee Independence

Let \(Z\sim N(0,1)\), let \(S\) be independent of \(Z\) with \(P(S=1)=P(S=-1)=1/2\), and set \(Y=(Z,SZ)'\). Then \(E[Y]=0\) and

\operatorname{Cov}(Y)= \begin{pmatrix} E[Z^2]&E[SZ^2]\\ E[SZ^2]&E[S^2Z^2] \end{pmatrix} =I_2.

Nevertheless, \(Y\) is not jointly Gaussian. With \(A_1=\operatorname{diag}(1,0)\) and \(A_2=\operatorname{diag}(0,1)\), the matrices are orthogonal projections summing to \(I_2\), but the quadratic forms are \(Q_1=Z^2\) and \(Q_2=(SZ)^2=Z^2\). They are equal with probability one and are not independent: for instance, \(P(Q_1\le 1,Q_2\le 1)=P(Z^2\le 1)\), which is strictly greater than \(P(Z^2\le 1)^2\). This shows why the Gaussian assumption matters. Even identity covariance and an exact algebraic decomposition do not suffice.

Rank Totals and Independence: What to Check

When a complete decomposition into symmetric idempotents is valid, the rank total follows from the geometry: mutually orthogonal ranges span \(\mathbb R^n\), so their dimensions add to \(n\). If reported ranks do not add to \(n\), then at least one of the claims about completeness, idempotence, or the rank calculation is wrong. Conversely, merely obtaining a rank total of \(n\) does not prove that the matrices are projections or that their ranges are orthogonal.

Independence also needs more than the scalar equation \(Y'Y=\sum_j Y'A_jY\). That equation records an equality of quadratic forms; it does not establish the joint distribution of the terms. In the theorem, independence comes from orthogonal Gaussian coordinate blocks. For Gaussian vectors with a different covariance structure, or for non-Gaussian vectors, that argument cannot simply be reused. Do not infer independence from an algebraic sum alone.

Check each matrix.
Verify \(A_j'=A_j\) and \(A_j^2=A_j\). Symmetry alone is not enough.

Check what is being decomposed.
For a full decomposition of \(Y'Y\), verify \(\sum_j A_j=I_n\). If the target is centered variation, identify the omitted subspace explicitly.

Check the rank accounting.
For a complete orthogonal projection decomposition, the ranks must sum to \(n\).

Check the distribution and mean.
Confirm the spherical Gaussian assumption \(Y\sim N_n(\mu,\sigma^2I_n)\). A component is central only if \(A_j\mu=0\).

State independence only after these checks.
The theorem’s independence conclusion follows from the orthogonal projections and Gaussian coordinates, not from the sum-of-squares identity by itself.

Common Mistake

Wrong reasoning: “The quadratic forms add to the total sum of squares, so they must be independent chi-square variables with degrees of freedom equal to the matrix ranks.” This skips the conditions that make the conclusion true. The matrices might not be idempotent; a component might have been left out; the ranks might be inconsistent; the Gaussian mean might yield noncentral components; or the vector might not have spherical Gaussian distribution.

Correction: Treat the conclusion as conditional, not automatic. Verify the projection structure and completeness, check that the rank total matches the dimension for a full partition, and use the mean to determine centrality. Then use the Gaussian assumption and orthogonal subspaces to justify independence. If one of these checks fails, the particular Cochran-theorem conclusion has not been established; further analysis is needed rather than a guess based on the identity.

Key takeaway: A valid application requires more than a sum-of-squares equation. Check projection matrices, completeness, ranks, Gaussian spherical errors, and projected means separately; each controls a different part of the conclusion.

Check Your Understanding

For each question, identify which condition or conclusion is at issue.

A symmetric matrix has rank three but does not satisfy \(A^2=A\). Can its quadratic form be called chi-square with three degrees of freedom by Cochran’s theorem? Why or why not?
In a four-dimensional full decomposition, three orthogonal projection matrices have ranks \(1,1,1\). What should you check before accepting the claim that they sum to the identity?
For a projection \(A_j\), what condition on \(\mu\) makes \(Y'A_jY/\sigma^2\) central rather than noncentral?
Why does an identity \(Y'Y=Q_1+Q_2\) not, by itself, establish that \(Q_1\) and \(Q_2\) are independent?
Can a complete orthogonal projection decomposition have a rank total greater than \(n\)? Explain the role of the projection assumptions.