Worked Comparison of Four Criteria From One Root Spectrum

One Spectrum, Four Summaries

The four classical MANOVA criteria summarize the same set of roots in different ways. A worked comparison is useful because a value such as $2.04$ for one criterion is not directly comparable with a value such as $1.00$ for another: the formulas, scales, and directions of evidence differ. The roots are the common input; each criterion is a distinct summary of them.

Let $H$ be the hypothesis sum-of-squares-and-cross-products matrix and $E$ the error sum-of-squares-and-cross-products matrix for a MANOVA hypothesis. Assume $H$ is positive semidefinite and $E$ is positive definite. We use bounded roots $\lambda_i$, which lie between zero and one. This convention makes the product form of Wilks’ lambda especially direct.

Definition: The bounded MANOVA roots $\lambda_1,\ldots,\lambda_p$ are the eigenvalues of $(H+E)^{-1}H$, counted with multiplicity. Under the stated assumptions, $0\leq \lambda_i<1$. The corresponding generalized roots are $r_i=\lambda_i/(1-\lambda_i)$, the eigenvalues of $E^{-1}H$. The four criteria are $$ \Lambda=\prod_{i=1}^p(1-\lambda_i),\qquad V=\sum_{i=1}^p\lambda_i,\qquad U=\sum_{i=1}^p\frac{\lambda_i}{1-\lambda_i},\qquad \Theta=\max_i\frac{\lambda_i}{1-\lambda_i}. $$

Here $\Lambda$ is Wilks’ lambda, $V$ is Pillai’s trace, $U$ is the Hotelling–Lawley trace, and $\Theta$ is Roy’s largest root on the conventional generalized-root scale. Some presentations call the largest bounded root itself Roy’s root. Under the convention above, that bounded root is $\lambda_{\max}$, while the conventional generalized root is $\lambda_{\max}/(1-\lambda_{\max})$. State the scale whenever reporting Roy’s value.

Why These Formulas Describe the Same Roots

The bounded and generalized roots are linked by a simple transformation. To see this, form the symmetric positive semidefinite matrix $A=E^{-1/2}HE^{-1/2}$. Its eigenvalues $r_i$ are the generalized roots of the pair $(H,E)$. The matrix $(H+E)^{-1}H$ is similar to $(I+A)^{-1}A$, whose eigenvalues are $r_i/(1+r_i)$. Thus $\lambda_i=r_i/(1+r_i)$, or equivalently $r_i=\lambda_i/(1-\lambda_i)$. Since each $r_i\geq0$ is finite, each $\lambda_i$ lies in $[0,1)$.

Theorem: When $H$ is positive semidefinite and $E$ is positive definite, the criteria computed from the bounded roots satisfy $$ \Lambda=\frac{|E|}{|H+E|},\qquad V=\operatorname{tr}\{H(H+E)^{-1}\},\qquad U=\operatorname{tr}(E^{-1}H),\qquad \Theta=\lambda_{\max}(E^{-1}H). $$ In the last expression, $\lambda_{\max}(E^{-1}H)$ denotes the largest eigenvalue of $E^{-1}H$, not the largest bounded root.

Proof. With $A=E^{-1/2}HE^{-1/2}$, the determinant factorizes as $$ |H+E|=|E^{1/2}(I+A)E^{1/2}|=|E||I+A|=|E|\prod_{i=1}^p(1+r_i). $$ Therefore $|E|/|H+E|=\prod_i(1+r_i)^{-1}=\prod_i(1-\lambda_i)=\Lambda$. The matrices $H(H+E)^{-1}$ and $(H+E)^{-1}H$ have the same trace, so the trace defining $V$ is the sum of the bounded roots. Also, $E^{-1}H$ is similar to $A$, so its trace is $\sum_i r_i=U$, and its largest eigenvalue is $\max_i r_i=\Theta$. This proves all four identities. $\square$

Worked Comparisons

Worked Example: Calculating All Four Criteria

Suppose the bounded roots are $0.60,\ 0.30,\ 0.10$. Calculate each criterion, retaining enough precision to make the comparison clear.

Wilks’ lambda. Multiply the complements: $$ \Lambda=(1-0.60)(1-0.30)(1-0.10) =0.40(0.70)(0.90)=0.252. $$

Pillai’s trace. Add the bounded roots: $$ V=0.60+0.30+0.10=1.00. $$

Hotelling–Lawley trace. Convert each root to its generalized scale, then add: $$ U=\frac{0.60}{0.40}+\frac{0.30}{0.70}+\frac{0.10}{0.90} =\frac{3}{2}+\frac{3}{7}+\frac{1}{9} =\frac{189+54+14}{126} =\frac{257}{126}\approx2.0397. $$

Roy’s largest root. The largest bounded root is $0.60$. On the generalized-root scale it is $$ \Theta=\frac{0.60}{1-0.60}=\frac{3}{2}=1.50. $$

The four reported summaries are therefore $\Lambda=0.252$, $V=1.00$, $U\approx2.0397$, and $\Theta=1.50$ on the generalized-root scale. The numbers are not four estimates of one quantity: each is a different function of the same spectrum.

Worked Example: Equal Pillai Traces, Different Summaries

Compare spectra A, $(0.50,0.25,0.25)$, and B, $(0.40,0.40,0.20)$. Both have Pillai’s trace equal to one, but the remaining criteria need not match.

For spectrum A, $$ V_A=0.50+0.25+0.25=1, $$ $$ \Lambda_A=(0.50)(0.75)(0.75)=0.28125, $$ $$ U_A=\frac{0.50}{0.50}+\frac{0.25}{0.75}+\frac{0.25}{0.75} =1+\frac{1}{3}+\frac{1}{3}=\frac{5}{3}\approx1.6667, $$ and $$ \Theta_A=\frac{0.50}{0.50}=1. $$

For spectrum B, $$ V_B=0.40+0.40+0.20=1, $$ $$ \Lambda_B=(0.60)(0.60)(0.80)=0.288, $$ $$ U_B=\frac{0.40}{0.60}+\frac{0.40}{0.60}+\frac{0.20}{0.80} =\frac{2}{3}+\frac{2}{3}+\frac{1}{4} =\frac{19}{12}\approx1.5833, $$ and $$ \Theta_B=\frac{0.40}{0.60}=\frac{2}{3}\approx0.6667. $$

The Pillai traces tie. Spectrum A has the smaller Wilks’ lambda, larger Hotelling–Lawley trace, and larger Roy generalized root. Those directions all indicate a more extreme value for A under the usual criterion directions. The example shows why a fixed sum of bounded roots does not fix the other summaries: the roots’ distribution across coordinates matters.

Worked Example: Zero Roots and the Effective Contribution

Take the four-root spectrum $(0.70,0.20,0,0)$. The zero roots are included in the calculations; they simply contribute zero to each sum and a factor of one to the product.

The Wilks product is $$ \Lambda=(0.30)(0.80)(1)(1)=0.24. $$ Pillai’s trace is $$ V=0.70+0.20+0+0=0.90. $$ The Hotelling–Lawley trace is $$ U=\frac{0.70}{0.30}+\frac{0.20}{0.80}+0+0 =\frac{7}{3}+\frac{1}{4} =\frac{28+3}{12}=\frac{31}{12}\approx2.5833. $$ Finally, the largest generalized root is $$ \Theta=\frac{0.70}{0.30}=\frac{7}{3}\approx2.3333. $$

Thus the four values are $0.24,\ 0.90,\ 31/12,$ and $7/3$, respectively. Adding zero roots does not alter any criterion, although keeping the full spectrum makes the response dimension explicit.

How Criterion Values Move with the Roots

A useful comparison is possible when two spectra can be paired so that every root in one is at least as large as its partner in the other. This is a coordinatewise comparison, not a claim that every pair of spectra can be ordered in this way.

Theorem: Let $a_i$ and $b_i$ be two bounded-root spectra of the same length, with $0\leq a_i\leq b_i<1$ for every $i$. Then Wilks’ lambda for the $b_i$ spectrum is no greater than that for the $a_i$ spectrum; Pillai’s trace, the Hotelling–Lawley trace, and Roy’s generalized root for the $b_i$ spectrum are no smaller.

Proof. Since $1-b_i\leq1-a_i$ and both are nonnegative, multiplying these inequalities gives $$ \prod_i(1-b_i)\leq\prod_i(1-a_i). $$ This is the stated direction for Wilks’ lambda. Summing $b_i\geq a_i$ gives the direction for Pillai’s trace. For the transformed roots, for $0\leq a\leq b<1$, $$ \frac{b}{1-b}-\frac{a}{1-a} =\frac{b-a}{(1-a)(1-b)}\geq0. $$ Summing this inequality gives the Hotelling–Lawley direction. Taking maxima of the coordinatewise ordered transformed roots gives the direction for Roy’s generalized root. $\square$

Interpretation Without Mixing Scales

For a fixed test and its usual upper-tail interpretation, a smaller Wilks’ lambda and larger values of Pillai’s trace, the Hotelling–Lawley trace, or Roy’s root correspond to more extreme criterion values. But comparing raw magnitudes across criteria is not meaningful. In the first worked example, the Hotelling–Lawley value exceeds the Pillai value numerically; that does not establish that it provides stronger evidence. Each statistic has its own scale and calibration, which also depends on the MANOVA design dimensions.

A second distinction is between a criterion’s descriptive value and a significance conclusion. The root spectrum gives the four observed criterion values, but by itself it does not give a p-value or establish statistical significance. Calibration requires the appropriate null distribution or approximation for the design. The focus here is the arithmetic and interpretation of the four summaries, not that separate calibration step.

Common Mistake

Wrong reasoning: “The Hotelling–Lawley trace is $2.04$, while Pillai’s trace is $1.00$, so Hotelling–Lawley shows a larger effect.”

Correction: The values are on different scales, so their numerical sizes cannot be compared that way. First identify which criterion is being reported and its direction: Wilks’ lambda is interpreted toward smaller values, while the other three are interpreted toward larger values. For formal evidence, use the calibration appropriate to that criterion and design. Also state whether Roy’s root is reported as a bounded root $\lambda_{\max}$ or as the generalized root $\lambda_{\max}/(1-\lambda_{\max})$.

Key takeaway: Start with one clearly defined root spectrum, apply each criterion’s own formula, and report the scale—especially for Roy’s root. Compare a criterion with its own calibrated reference, not with the raw numerical value of another criterion.

Check Your Understanding

Use the bounded-root convention throughout unless a question asks for the generalized-root scale.

For roots $0.50$ and $0.20$, calculate Wilks’ lambda and Pillai’s trace.
For a bounded root of $0.75$, what is the corresponding generalized root?
Why does adding a zero root leave all four criteria unchanged?
Two spectra have equal Pillai’s traces. Does that imply equal Wilks’ lambdas? Explain.
Why is it invalid to conclude that one criterion gives stronger evidence just because its raw value is numerically larger than another criterion’s?