Noncentral Cochran Decompositions

From Central to Noncentral Decompositions

Cochran’s theorem describes how orthogonal projections of a centered spherical Gaussian vector divide its sum of squares into independent chi-square components. When the Gaussian vector has a nonzero mean, the same geometric division still gives independent components, but a component is generally noncentral. Its noncentrality parameter measures how much of the mean lies in the subspace selected by that projection.

The key is to distinguish the geometry of the projections from the location of the mean. Orthogonality determines independence and degrees of freedom; the projected mean determines noncentrality. When the projections partition the whole space, the component noncentralities add to the noncentrality of the total sum of squares.

Definitions and Setup

Let \(Z\sim N_n(\mu,\sigma^2 I_n)\), where \(\sigma^2>0\), and let \(P_1,\ldots,P_k\) be orthogonal projections satisfying \(P_iP_j=0\) for \(i\ne j\) and \(\sum_{i=1}^k P_i=I_n\). Write \(r_i=\operatorname{rank}(P_i)\). The corresponding scaled sums of squares and noncentrality parameters are

The parameter \(\lambda_i\) is nonnegative: since \(P_i=P_i^\top=P_i^2\), \(\mu^\top P_i\mu=\|P_i\mu\|^2\). Thus \(\lambda_i\) is the squared length of the mean component in the projected subspace, measured in units of the variance.

The Noncentral Cochran Decomposition

To prove the result, choose an orthonormal basis for the image of each \(P_i\). The images are mutually orthogonal because \(P_iP_j=0\), and together they span \(\mathbb R^n\) because the projections sum to \(I_n\). Put the basis vectors for each image into the columns of a matrix \(U_i\). The combined matrix \(U=[U_1\ \cdots\ U_k]\) is orthogonal, and \(P_i=U_iU_i^\top\).

The transformed vector \(U^\top Z\) is Gaussian with mean \(U^\top\mu\) and covariance \(\sigma^2 I_n\). Its coordinates are independent, since a jointly Gaussian vector with diagonal covariance has independent coordinates. In particular, its coordinate blocks \(U_i^\top Z\) are mutually independent. Also,

where the \(G_{ij}\) are independent \(N(0,1)\) variables. The squared shifts in block \(i\) sum to \(\|U_i^\top\mu\|^2/\sigma^2=\mu^\top U_iU_i^\top\mu/\sigma^2=\lambda_i\). The definition of the noncentral chi-square distribution therefore gives \(Q_i\sim\chi^2_{r_i}(\lambda_i)\). Each \(Q_i\) is a function only of its own independent coordinate block, so the \(Q_i\) are independent.

Finally, \(P_iP_j=0\) and \(\sum_iP_i=I_n\) imply \(\sum_iP_i^\top P_i=\sum_iP_i=I_n\). Hence \(\sum_i r_i=n\), and

Applying the same orthogonal-coordinate argument to the full space gives the stated distribution of the total. This proves the theorem. \(\square\)

The additivity also explains why grouping components preserves the form of the result. For any group \(A\) of indices, \(P_A=\sum_{i\in A}P_i\) is itself a projection: symmetry is preserved, and \(P_A^2=\sum_{i\in A}P_i=P_A\) because the cross-products vanish. Its rank is \(\sum_{i\in A}r_i\), its noncentrality is \(\sum_{i\in A}\lambda_i\), and its quadratic form is \(\sum_{i\in A}Q_i\). Thus the grouped sum has a noncentral chi-square distribution with those summed parameters.

Worked Decompositions

Interpretation and a Common Pitfall

A component is central precisely when \(P_i\mu=0\), because \(\lambda_i=\|P_i\mu\|^2/\sigma^2\). This does not require the full mean vector to be zero; it requires only that the mean have no component in that projection’s subspace. Conversely, a nonzero mean does not automatically make every component noncentral. The projection directions matter.

The independent components also provide a useful consistency check. For \(Q_i\sim\chi^2_{r_i}(\lambda_i)\), \(E[Q_i]=r_i+\lambda_i\). Thus the component means add to \(n+\mu^\top\mu/\sigma^2\), the mean of the total scaled squared norm. For the second worked example, the component means are \(1+1/4=5/4\) and \(3+3/4=15/4\), summing to \(5\), as does \(4+\mu^\top\mu/4=4+1\).

• Common mistake: treating every projected sum of squares as central because the projections are orthogonal. Orthogonality establishes independence; it does not make the projected means zero.
• Correction: compute \(\lambda_i=\mu^\top P_i\mu/\sigma^2\) for each projection. Use a central chi-square law only when this quantity is zero.
• Another check: if the projections partition the space, verify both \(\sum_i r_i=n\) and \(\sum_i\lambda_i=\mu^\top\mu/\sigma^2\). A failure usually signals a missing subspace or an incorrect projection calculation.