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MANOVA Criteria and Distributions · Tutorial 353 of 975

Local Alternatives and Asymptotic MANOVA Power

Learn how local alternatives produce limiting MANOVA power, and distinguish sensitivity to total noncentrality from sensitivity to its distribution across directions.

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What You'll Learn

  • Define a local-alternative MANOVA sequence with fixed hypothesis dimensions and noncentrality.
  • Derive the limiting distributions of Wilks, Pillai, Hotelling–Lawley, and Roy criteria.
  • Obtain the trace criteria’s noncentral chi-square power formula.
  • Identify the rank condition required for a Gaussian representation of a noncentral Wishart matrix.
  • Compare criterion sensitivity to total noncentrality and to its distribution across directions.
  • Distinguish changes in a criterion’s value from changes in its rejection probability.

Local Alternatives and the Scale of MANOVA Roots

A useful way to compare MANOVA criteria is to study alternatives that become weaker as the sample size grows, but remain statistically visible. In the classical fixed-dimension regime, this means that the noncentrality matrix stays fixed while the residual degrees of freedom increase. The generalized roots then shrink, so comparing unscaled criterion values can conceal their common leading behavior.

Let \(p\) be the response dimension, \(m\) the numerator degrees of freedom, and \(q\) the residual degrees of freedom. Assume the hypothesis and error sum-of-squares-and-cross-products matrices are independent, with \(H\sim W_p(m,I_p,\Omega)\) and \(E\sim W_p(q,I_p)\). Here \(I_p\) is the \(p\times p\) identity matrix and \(\Omega\) is a fixed positive-semidefinite noncentrality matrix. As \(q\) increases, \(p\), \(m\), and \(\Omega\) are fixed. In a sequence of models, fixed \(\Omega\) represents effects shrinking at the rate that keeps noncentrality bounded.

Definition: The generalized roots \(\lambda_1\geq\cdots\geq\lambda_s\geq0\), where \(s=\min(p,m)\), are the nonzero eigenvalues (including multiplicities) of \(E^{-1}H\). The four criteria considered here are Wilks’ lambda \(\Lambda=\prod_{i=1}^s(1+\lambda_i)^{-1}\), Pillai’s trace \(V=\sum_{i=1}^s\lambda_i/(1+\lambda_i)\), the Hotelling–Lawley trace \(U=\sum_{i=1}^s\lambda_i\), and Roy’s largest root \(R=\lambda_1\).

For a Gaussian representation of \(H\), take a \(p\times m\) matrix \(Z\) with independent unit-variance normal entries and mean matrix \(M\), so \(H=ZZ'\) and \(MM'=\Omega\). This requires \(\operatorname{rank}(\Omega)\leq m\): a matrix \(M\) with \(m\) columns cannot have \(MM'\) of rank greater than \(m\). The condition holds for the usual MANOVA construction, in which the noncentrality is generated by a mean matrix with \(m\) columns. We impose it whenever using this representation.

Limiting Criteria and Power

The residual matrix grows with \(q\), while the hypothesis matrix remains on a fixed scale. Since \(E/q\) converges in probability to \(I_p\), the rescaled generalized roots converge to the eigenvalues of \(H\). This yields a common first-order limit for the three trace criteria, but not for Roy’s criterion.

Theorem: Under the assumptions above, let \(\mu_1\geq\cdots\geq\mu_s\geq0\) be the nonzero eigenvalues of \(H\), with zeros included if needed. Then, jointly, \(qU\xrightarrow{d}\operatorname{tr}(H)\), \(qV\xrightarrow{d}\operatorname{tr}(H)\), \(q(-\log\Lambda)\xrightarrow{d}\operatorname{tr}(H)\), and \(qR\xrightarrow{d}\mu_1\). Under the null, \(\Omega=0\); under a fixed-noncentrality local alternative, \(\Omega\) remains fixed.

Proof. For fixed \(p\), the entries of \(E\) are sums of \(q\) independent contributions with finite variances, so the law of large numbers gives \(E/q\xrightarrow{p}I_p\). The matrix \(E^{-1}H\) is similar to the symmetric matrix \(E^{-1/2}HE^{-1/2}\). Thus its roots are the eigenvalues of \((E/q)^{-1/2}H(E/q)^{-1/2}\). By continuity of the inverse square root at \(I_p\), this symmetric matrix differs from \(H\) by a quantity converging to zero in probability. The eigenvalues of symmetric matrices are continuous functions of their entries; consequently, the ordered rescaled roots \(q\lambda_i\) converge jointly in distribution to the ordered eigenvalues \(\mu_i\) of \(H\).

Write \(x_i=q\lambda_i\). Then \(qU=\sum_i x_i\), \(qV=\sum_i x_i/(1+x_i/q)\), and \(q(-\log\Lambda)=\sum_i q\log(1+x_i/q)\). For every bounded \(x\geq0\), \(x/(1+x/q)\to x\) and \(q\log(1+x/q)\to x\). The limiting random vector \((\mu_1,\ldots,\mu_s)\) is finite almost surely, so applying these continuous limits to the jointly converging roots gives the first three claims. Also \(qR=\max_i x_i\), which converges to \(\max_i\mu_i\). These arguments establish the joint limits. \(\square\)

The trace limit has a simple distribution. With \(H=ZZ'\), \(\operatorname{tr}(H)\) is the sum of the squares of all \(pm\) entries of \(Z\). A sum of squares of independent unit-variance normal variables with means whose squared sum is \(\operatorname{tr}(\Omega)\) has a noncentral chi-square distribution. Therefore \(\operatorname{tr}(H)\sim\chi^2_{pm}(\operatorname{tr}\Omega)\). For a level-\(\alpha\) upper-tail test based on any of the three scaled trace criteria, the limiting power is \(P\{\chi^2_{pm}(\operatorname{tr}\Omega)>c_\alpha\}\), where \(c_\alpha\) is the \(1-\alpha\) quantile of the central \(\chi^2_{pm}\) distribution.

To justify this power statement, under the null the scaled statistics converge to the central chi-square variable, whose distribution is continuous. Their level-\(\alpha\) upper-tail critical values therefore converge to \(c_\alpha\). Under the alternative, the same statistics converge to the stated noncentral chi-square variable, which also has a continuous distribution. The probability of exceeding the converging critical value consequently converges to the displayed tail probability. Roy’s limiting power instead uses a central-Wishart largest-root critical value and the alternative probability \(P\{\mu_1>r_\alpha\}\). It does not reduce to the trace chi-square calculation.

Worked Example: Limiting Power for a Trace Criterion

Take \(p=2\), \(m=2\), and \(\Omega=\begin{pmatrix}1&0\\0&0\end{pmatrix}\). Its rank is 1, so the Gaussian representation with two columns is possible. The noncentrality is \(\operatorname{tr}(\Omega)=1\), and the limiting trace statistic has distribution \(\chi^2_4(1)\). At a level-\(\alpha\) test, the limiting power of the scaled Hotelling–Lawley, Pillai, or negative-log-Wilks statistic is therefore \(P\{\chi^2_4(1)>\chi^2_{4,\,1-\alpha}\}\).

For a numerical check on the alternative distribution, its mean is \(4+1=5\) and its variance is \(2(4+2\cdot1)=12\). These follow from the noncentral chi-square formulas \(E[X]=\nu+\delta\) and \(\operatorname{Var}(X)=2(\nu+2\delta)\), with \(\nu=4\) and \(\delta=1\). The formula specifies power at any chosen \(\alpha\); it does not require treating the random roots as fixed values.

What the Noncentrality Spectrum Changes

The trace criteria’s limiting power depends on \(\Omega\) only through its trace. In particular, the eigenvalues of \(\Omega\)—the noncentrality spectrum—can be redistributed while keeping their sum fixed without changing this limiting trace power. Roy’s limit is different: it is the largest eigenvalue of a noncentral Wishart matrix, whose distribution reflects the directional structure of the alternative. It does not have a general reduction to a noncentral chi-square distribution indexed only by \(\operatorname{tr}(\Omega)\). Thus Roy retains spectral information that the limiting trace power discards.

Worked Example: Equal Total Noncentrality, Different Spectra

For \(p=m=2\), compare \(\Omega_A=\begin{pmatrix}2&0\\0&0\end{pmatrix}\) and \(\Omega_B=\begin{pmatrix}1&0\\0&1\end{pmatrix}\). Both are positive semidefinite, have rank at most \(m=2\), and have trace 2. The trace limits under both alternatives are therefore \(\chi^2_4(2)\), with mean \(4+2=6\) and variance \(2(4+4)=16\). The three trace criteria have the same limiting power under these two alternatives.

The spectra differ: \(\Omega_A\) concentrates all noncentrality in one direction, whereas \(\Omega_B\) distributes it equally. Roy’s limiting statistic is the largest eigenvalue of \(H\), not \(\operatorname{tr}(H)\). Its alternative distribution is not determined by the chi-square trace calculation, so the equality of trace-test powers does not establish equality of Roy powers or a universal ordering between them.

The criteria also have different sensitivities as functions of a given root vector. If one root changes by a small amount while the others are fixed, the derivatives with respect to \(\lambda_i\) are \(1\) for \(U\), \(1/(1+\lambda_i)^2\) for \(V\), and \(1/(1+\lambda_i)\) for \(-\log\Lambda\). Roy responds to the largest root; when that root is unique, its derivative with respect to it is 1 and with respect to the other roots is 0. Under local alternatives the roots are of order \(1/q\), so both trace derivatives approach 1. This is the differential version of their shared first-order trace limit. At non-small roots, their weightings differ.

Worked Example: First-Order Sensitivity to a Root Shift

Suppose the roots at a particular point are \((0.5,0.2)\), and the first root increases by \(0.01\) while the second is unchanged. The first-order changes are:

For \(U\), the derivative is \(1\), giving change \(0.01\). For \(V\), it is \(1/(1+0.5)^2=4/9\), giving approximately \(0.00444\). For \(-\log\Lambda\), it is \(1/(1+0.5)=2/3\), giving approximately \(0.00667\). Since the first root is uniquely largest, Roy’s root changes by exactly \(0.01\).

These are changes in criterion values for a specified root perturbation, not changes in rejection probabilities. Power also depends on the null critical value and the full sampling distribution under the alternative. It is not valid to infer a universal power ranking from these derivatives alone.

Rank Conditions and Common Pitfalls

The Gaussian representation places a concrete restriction on examples and derivations. If \(H=ZZ'\) and \(Z\) has \(m\) columns, then \(\operatorname{rank}(H)\leq m\), and likewise \(\operatorname{rank}(MM')\leq m\). A proposed noncentrality matrix with larger rank cannot arise from that representation. This matters especially when describing the distribution of the limiting Roy root, since that description is based on the noncentral Wishart matrix \(H\).

Worked Example: Checking a Noncentrality Rank

Let \(p=3\) and \(m=2\). The matrix \(\Omega=\operatorname{diag}(1,0,0)\) is admissible: choose \(M\) to have first column \((1,0,0)'\) and second column zero. Then \(MM'=\Omega\).

By contrast, \(\widetilde{\Omega}=\operatorname{diag}(1,1,1)\) has rank 3. If a \(3\times2\) matrix \(M\) satisfied \(MM'=\widetilde{\Omega}\), then \(\operatorname{rank}(MM')=\operatorname{rank}(M)\leq2\), contradicting \(\operatorname{rank}(\widetilde{\Omega})=3\). It cannot be used with this two-column Gaussian construction.

  • Comparing unscaled statistics. Under these local alternatives, the roots shrink with \(q\). The nondegenerate comparisons use scaled criteria such as \(qU\), not \(U\) alone.
  • Claiming all four criteria have the same limit. The three trace criteria share the trace limit; Roy converges to a largest eigenvalue.
  • Equating equal trace with equal spectral sensitivity. Equal \(\operatorname{tr}(\Omega)\) gives equal limiting trace power, but does not establish equal Roy power.
  • Ignoring the rank condition. A \(p\times m\) mean matrix cannot generate a noncentrality matrix of rank greater than \(m\).
Key takeaway: With fixed MANOVA dimensions and fixed noncentrality as residual degrees of freedom grow, Wilks, Pillai, and Hotelling–Lawley have a common scaled limit governed by total noncentrality. Roy’s limit retains sensitivity to the largest random root and hence to directional structure, but this does not imply a universal power advantage.

Check Your Understanding

Use the assumptions and results above to answer the following.

  1. Why are the MANOVA roots multiplied by \(q\) when studying this local-alternative regime?
  2. For \(p=3\), \(m=2\), and \(\operatorname{tr}(\Omega)=0.5\), what is the limiting distribution of \(qU\)?
  3. Two admissible noncentrality matrices have equal trace but different eigenvalues. What can be concluded about limiting trace-test power, and what cannot be concluded about Roy’s power?
  4. Why can a \(3\times2\) mean matrix not generate a noncentrality matrix of rank 3?
  5. At a tie for the largest root, why should Roy’s response to a perturbation be described using the maximum among the tied-root changes?
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