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MANOVA Criteria and Distributions · Tutorial 350 of 975

Choosing an Approximation by Design Dimensions

Learn how MANOVA design dimensions determine the available null calibrations, and how to distinguish a defensible approximation from an exact special case.

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What You'll Learn

  • Identify the design dimensions that enter Bartlett and F calibrations.
  • Distinguish nominal hypothesis rank from the number of nonzero MANOVA roots.
  • Calculate the standard chi-square and F calibration parameters.
  • Recognize scalar-response and rank-one designs where exact F reductions are available.
  • Choose a calibration with appropriate caution when dimensions are large or error degrees of freedom are limited.

Why the Design Dimensions Matter

MANOVA criteria summarize the same set of hypothesis roots in different ways, but their null calibrations do not respond identically to the design. A chi-square approximation for Wilks' lambda, for example, uses a degrees-of-freedom count tied to both the response dimension and the hypothesis rank. F calibrations also depend on the error degrees of freedom, and their formulas can involve the number of available roots. Choosing a calibration therefore starts with the design, not with whichever test statistic happens to look most familiar.

This tutorial compares the main calibration dimensions and gives a practical way to assess them. The comparison does not produce a universal ranking: an approximation can be more or less credible depending on the design and the null distribution it is intended to represent. In particular, an exact reduction in a special design should take precedence over a general approximation.

Design Dimensions and Root Count

Let \(p\) be the response dimension, \(q\) the rank of the tested hypothesis, and \(v\) the error degrees of freedom. Under the central MANOVA null, assume the usual multivariate normal linear model, with independent hypothesis and error sums-of-squares-and-cross-products matrices and a nonsingular error matrix. The MANOVA roots are the \(s\) eigenvalues of \(E^{-1}H\), counted with zeros when necessary, where \(s=\min(p,q)\). At most \(s\) of these roots can be nonzero.

Definition: The design root count is \(s=\min(p,q)\), the maximum number of nonzero MANOVA roots allowed by the response dimension and hypothesis rank. The realized nonzero-root count is \(r=\operatorname{rank}(H)\), with \(r\leq s\). Standard calibration formulas use the design dimensions, including \(s\); do not substitute an observed \(r\) into those formulas unless a separate result justifies doing so.

The distinction matters because the root count in a particular data set can be smaller than its design maximum. For example, if \(H=0\), every root is zero, even though \(s\) may be positive. Conversely, when the hypothesis matrix has its generic rank, the design may permit all \(s\) roots to be nonzero. The calibration formulas are defined by the model dimensions, not by counting the roots that happen to appear numerically in one sample.

Result: If \(E\) is positive definite and \(H\) is positive semidefinite, then \(E^{-1}H\) has nonnegative real eigenvalues, and the number of its nonzero eigenvalues, counted with multiplicity, is \(\operatorname{rank}(H)\). In the MANOVA design, \(\operatorname{rank}(H)\leq\min(p,q)\).

To verify this, write \(E^{-1}H=E^{-1/2}(E^{-1/2}HE^{-1/2})E^{1/2}\). Thus \(E^{-1}H\) is similar to the symmetric positive semidefinite matrix \(E^{-1/2}HE^{-1/2}\); similarity preserves eigenvalues. The latter matrix has nonnegative eigenvalues. Multiplication on either side by the invertible matrix \(E^{-1/2}\) preserves rank, so its rank is \(\operatorname{rank}(H)\). A positive semidefinite matrix has exactly its rank many nonzero eigenvalues. Finally, \(H\) is formed from a hypothesis of rank \(q\) in a \(p\)-dimensional response space, so \(\operatorname{rank}(H)\leq\min(p,q)\). \(\square\)

What the Standard Calibrations Use

Write the roots in decreasing order as \(\theta_1\geq\cdots\geq\theta_s\geq0\), including zeros when needed. The four criteria are Wilks' lambda, Pillai's trace, the Hotelling–Lawley trace, and Roy's largest root. Their standard null calibrations use different summaries of these roots. The following parameterizations make the design dependence explicit.

Calibration formulas: With \(s=\min(p,q)\), a common Bartlett chi-square calibration for Wilks' lambda is $$ -b\log\Lambda \ \approx\ \chi^2_{pq},\qquad b=v-\frac{p-q+1}{2}. $$ For the F calibrations, put \(U=\sum_{i=1}^s\theta_i\) (Hotelling–Lawley trace) and \(V=\sum_{i=1}^s\frac{\theta_i}{1+\theta_i}\) (Pillai trace). The standard Pillai approximation uses $$ F_P=\frac{s(v-p+s)}{pq}\frac{V}{s-V}, \qquad (df_1,df_2)=\bigl(pq,\ s(v-p+s)\bigr). $$ The standard Hotelling–Lawley approximation uses $$ F_{HL}=\frac{2(sn+1)}{s^2(2m+s+1)}U, \qquad (df_1,df_2)=\bigl(s(2m+s+1),\ 2(sn+1)\bigr). $$ Roy's upper-bound F calibration uses $$ F_R=\frac{v-p+s}{\max(p,q)}\theta_1, \qquad (df_1,df_2)=\bigl(\max(p,q),\ v-p+s\bigr). $$ Here “\(\approx\)” indicates a calibration, not an exact distribution in general. The Bartlett factor must be positive for this form to be used.

The degrees of freedom for Pillai and Hotelling–Lawley are not generally common. In the usual parameterization, let \(m=(|p-q|-1)/2\) and \(n=(v-p-1)/2\). Pillai's F degrees of freedom are \(s(2m+s+1)\) and \(s(2n+s+1)\). Since \(2m+s+1=|p-q|+s=\max(p,q)\), the first is \(s\max(p,q)=pq\); since \(2n+s+1=v-p+s\), the second is \(s(v-p+s)\). Hotelling–Lawley's F degrees of freedom are \(s(2m+s+1)\) and \(2(sn+1)\). These formulas apply provided the design has the positive error degrees of freedom required by the underlying MANOVA calibration.

These formulas also show why a single dimension is not enough to choose a method. The Bartlett degrees of freedom are \(pq\), while its multiplier depends on \(v,p,q\). The Pillai and Hotelling–Lawley F calibrations generally use different degrees of freedom and transform different criteria. Roy's calibration has numerator degrees of freedom \(\max(p,q)\), not \(pq\), and focuses on one root. A large response dimension can therefore affect calibrations differently from a large hypothesis rank.

Worked Example: A Scalar Response With Several Hypothesis Degrees of Freedom

Suppose \(p=1\), \(q=3\), and \(v=40\). Then \(s=1\), so there is one possible nonzero root, denoted \(\theta_1\). The Bartlett factor is \(b=40-(1-3+1)/2=40.5\), and its approximate reference distribution has \(pq=3\) degrees of freedom.

For Pillai's trace, \(V=\theta_1/(1+\theta_1)\). The F degrees of freedom are \(df_1=3\) and \(df_2=1(40-1+1)=40\). Its transformed statistic is $$ F_P=\frac{40}{3}\frac{V}{1-V} =\frac{40}{3}\frac{\theta_1/(1+\theta_1)}{1/(1+\theta_1)} =\frac{40}{3}\theta_1. $$

For Hotelling–Lawley, \(U=\theta_1\), giving \(F_{HL}=(40/3)\theta_1\) with the same degrees of freedom. Roy's calibration also has \(df_1=3\), \(df_2=40\), and \(F_R=(40/3)\theta_1\). In this scalar-response design, however, the classical univariate test provides an exact F reduction. It is not necessary to treat the general approximation as the best available choice merely because its formula can be evaluated.

Worked Example: Comparing Calibrations for a Multivariate Design

Take \(p=4\), \(q=2\), and \(v=30\), and suppose the roots are \(\theta_1=0.8\) and \(\theta_2=0.2\). Here \(s=2\), and \(pq=8\). The Bartlett factor is \(b=30-(4-2+1)/2=28.5\). Wilks' lambda is $$ \Lambda=\frac{1}{(1+0.8)(1+0.2)} =\frac{1}{2.16}\approx0.46296. $$ Thus the Bartlett statistic is \(28.5[-\log(0.46296)]\approx28.5(0.77011)=21.95\), compared with a \(\chi^2_8\) reference distribution.

The Pillai trace is \(V=0.8/1.8+0.2/1.2=4/9+1/6=11/18\). Its F degrees of freedom are \(8\) and \(2(30-4+2)=56\), and $$ F_P=\frac{56}{8}\frac{11/18}{2-11/18} =7\frac{11/18}{25/18} =3.08. $$

The Hotelling–Lawley trace is \(U=0.8+0.2=1\), so \(F_{HL}=(52/16)(1)=3.25\) with degrees of freedom \((8,52)\). Roy's statistic uses only \(\theta_1=0.8\); its degrees of freedom are \((4,28)\), and \(F_R=(28/4)(0.8)=5.6\). These numerical values are on different reference scales and should not be compared as if they were interchangeable test statistics. Each must be assessed against its own specified null calibration.

Worked Example: A Rank-One Hypothesis

Suppose \(p=6\), \(q=1\), and \(v=20\). Then \(s=1\), \(df_1=pq=6\), and \(df_2=s(v-p+s)=15\). Let the single root be \(\theta_1=0.5\). The Pillai trace is \(V=0.5/1.5=1/3\), giving $$ F_P=\frac{15}{6}\frac{1/3}{1-1/3} =\frac{15}{6}\frac{1}{2} =1.25. $$ The Hotelling–Lawley calibration gives \(F_{HL}=(15/6)(0.5)=1.25\). Roy's upper-bound calibration also gives \(F_R=(15/6)(0.5)=1.25\), with degrees of freedom \((6,15)\). In this rank-one case the classical exact F reduction is available, so these formulas coincide with an exact calibration under the standard model assumptions. The example illustrates that exactness follows from the special design, not from the numerical agreement alone.

A Design-Based Choice

A defensible choice can be made in a fixed order. First use the actual rank of the tested hypothesis, after removing redundant hypothesis directions; a nominal count of contrasts is not necessarily the rank \(q\). Next record \(p\) and \(v\), checking that the error matrix is nonsingular and that the correction or F degrees of freedom are admissible. Then compute \(s=\min(p,q)\), the maximum possible number of nonzero roots. Finally ask whether the design has a known exact reduction before relying on an approximation.

When \(p=1\), the problem is univariate and an exact F test is available under the classical model. When \(q=1\), the rank-one exact reduction applies. Outside these cases, Bartlett's chi-square and the named F calibrations are approximations or, for Roy, an upper-bound calibration. The formulas alone do not establish that one will be more accurate for every choice of dimensions.

The error degrees of freedom \(v\) deserve particular attention. In the Bartlett expression they affect the scale \(b\); in the F calibrations they affect the denominator degrees of freedom and scaling. Limited \(v\) means the calibration is being used with less error information, so interpreting a reference tail probability as exact is especially unwarranted. Meanwhile \(p\) and \(q\) determine the scale of the hypothesis: their product appears in the Wilks chi-square degrees of freedom and in the Pillai and Hotelling–Lawley transformations.

Common mistake: Counting only the observed nonzero roots and replacing \(s\) by that count in the standard F degrees of freedom. This confuses realized rank with design root count. If \(p=4\) and \(q=2\), then \(s=2\), even if one observed root is numerically zero. The standard formula still uses \(s=2\); a reduced-root calibration requires its own justification.

A second pitfall is to call the Bartlett chi-square or a general F approximation “exact” because the formula is explicit. Exactness is a distributional claim, not a consequence of having a closed-form statistic. Use an exact reduction when the design supplies one; otherwise state which approximation is being used and retain its assumptions and degrees of freedom. This makes the chosen calibration auditable and prevents a design feature such as small \(v\) or a high response dimension from being hidden behind a single reported p-value.

Check Your Understanding

Use the design dimensions and calibration formulas to answer these questions.

  1. If \(p=5\) and \(q=2\), what is the design root count \(s\), and what is the largest possible number of nonzero roots?
  2. For \(p=3\), \(q=4\), and \(v=25\), calculate the Bartlett factor and its chi-square degrees of freedom.
  3. Why should an observed nonzero-root count not automatically replace \(s\) in a standard Pillai F calibration?
  4. For \(p=1\), explain why an exact univariate F reduction should be considered before a general MANOVA approximation.
  5. Which design dimension enters the denominator degrees of freedom \(s(v-p+s)\) for the Pillai and Hotelling–Lawley calibrations, and why does it matter?
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