Criterion Performance for Diffuse and Concentrated Effects

Why the Shape of an Effect Spectrum Matters

A multivariate alternative can distribute its signal across several response directions or concentrate it in one direction. The four standard MANOVA criteria summarize the roots of the hypothesis relative to the error matrix in different ways. Consequently, two alternatives with the same total root mass can look quite different to the different criteria.

The comparison here is algebraic: it asks how the criteria change when a fixed nonnegative spectrum is redistributed. This is useful for understanding their sensitivity, but it is not a substitute for a power calculation. Under an alternative, the sample roots are random, and each criterion has its own null calibration and rejection threshold.

Definition: Let \(\lambda_1,\ldots,\lambda_m\) be the nonnegative MANOVA roots for a hypothesis, with zero roots included if needed. The four criteria are Wilks’ Lambda, \(\Lambda=\prod_{i=1}^m(1+\lambda_i)^{-1}\); Pillai’s trace, \(V=\sum_{i=1}^m\lambda_i/(1+\lambda_i)\); the Hotelling–Lawley trace, \(T=\sum_{i=1}^m\lambda_i\); and Roy’s largest root, \(\Theta=\max_i\lambda_i\). Smaller \(\Lambda\), or equivalently larger \(W=-\log\Lambda=\sum_i\log(1+\lambda_i)\), represents stronger evidence in the direction of the alternative. Larger \(V\), \(T\), and \(\Theta\) represent stronger evidence.

A spectrum is diffuse when its total is spread relatively evenly across roots; it is concentrated when one or a few roots account for most of that total. These are comparative descriptions, not extra assumptions about the MANOVA model. In the examples below, the total \(T\) is held fixed so that changes in the other summaries reflect allocation rather than a change in total root mass.

Worked Comparisons at Fixed Total Root Mass

Worked Example: Four Moderate Roots Versus One Dominant Root

Compare the spectra \((1.5,1.5,1.5,1.5)\) and \((5.4,0.2,0.2,0.2)\). Both have total root mass \(6\).

For the diffuse spectrum, \(T=4(1.5)=6\) and \(\Theta=1.5\). Its Wilks product is \(\prod_i(1+\lambda_i)=2.5^4=39.0625\), so \(\Lambda=1/39.0625=0.0256\) and \(W=\log(39.0625)\approx3.6652\). Its Pillai trace is \(V=4(1.5/2.5)=4(0.6)=2.4\).

For the concentrated spectrum, \(T=5.4+3(0.2)=6\) and \(\Theta=5.4\). The Wilks product is \(6.4(1.2)^3=6.4(1.728)=11.0592\), so \(\Lambda=1/11.0592\approx0.09042\) and \(W=\log(11.0592)\approx2.4031\). The Pillai trace is \(V=5.4/6.4+3(0.2/1.2)=0.84375+0.5=1.34375\).

Thus the diffuse spectrum has larger \(W\) and \(V\), while the concentrated spectrum has the larger Roy root. The Hotelling–Lawley trace is \(6\) for both. These are differences in the criteria’s summaries, not a proof that one calibrated test has greater power.

Worked Example: A Single Equalization Step

Start with \((4,0)\), then transfer one unit from the larger root to the smaller, obtaining \((3,1)\). The transfer is a mean-preserving equalization: the gap changes from \(4\) to \(2\), and the transfer of \(1\) does not exceed half the original gap, \(2\).

Both spectra have \(T=4\). For \((4,0)\), \(W=\log 5\approx1.6094\), \(V=4/5+0=0.8\), and \(\Theta=4\). For \((3,1)\), \(W=\log 4+\log 2=\log 8\approx2.0794\), \(V=3/4+1/2=1.25\), and \(\Theta=3\). Thus this equalization raises the transformed Wilks summary and Pillai’s trace, leaves \(T\) unchanged, and lowers Roy’s root.

The size restriction on the transfer matters. A transfer that overshoots equality can make the allocation more concentrated rather than more diffuse; it cannot be treated as an equalization step.

Worked Example: Equal Roots and the Bounds on Roy’s Root

Take \(m=3\) roots with total \(T=6\). For the equal spectrum \((2,2,2)\), Roy’s root is \(2\). For the concentrated spectrum \((5,0.5,0.5)\), the total is also \(5+0.5+0.5=6\), and Roy’s root is \(5\).

For any nonnegative roots with this total, the largest root is at least their average: \(\Theta\ge T/m=2\). It is also at most their sum: \(\Theta\le T=6\). The equal spectrum attains the lower bound. A spectrum such as \((6,0,0)\) attains the upper bound. The second spectrum in this example lies between these extremes.

More generally, for \(m\) nonnegative roots summing to \(T\), each root is at most the maximum, so \(T\le m\Theta\), giving \(\Theta\ge T/m\). Since all roots are nonnegative, the maximum is no larger than their sum, giving \(\Theta\le T\). Equality at the lower bound requires all roots to equal \(T/m\); equality at the upper bound requires all other roots to be zero. \(\square\)

Why Equalization Favors Pillai and Transformed Wilks

The behavior in the second example follows from the curvature of the functions applied to each root. Define \(g_P(x)=x/(1+x)\) and \(g_W(x)=\log(1+x)\) for \(x\ge0\). Then \(V=\sum_i g_P(\lambda_i)\) and \(W=\sum_i g_W(\lambda_i)\). Both functions are strictly increasing and strictly concave: \(g_P'(x)=1/(1+x)^2\), \(g_P''(x)=-2/(1+x)^3<0\); and \(g_W'(x)=1/(1+x)\), \(g_W''(x)=-1/(1+x)^2<0\).

Theorem: Suppose two roots satisfy \(x\ge y\ge0\), and replace them by \(x-\delta\) and \(y+\delta\), where \(0\le\delta\le(x-y)/2\). The total is unchanged. For either \(g=g_P\) or \(g=g_W\), the sum \(g(x-\delta)+g(y+\delta)\) is at least \(g(x)+g(y)\). It is strictly larger when \(x>y\) and \(0<\delta\le(x-y)/2\).

Proof. Define \(h(t)=g(x-t)+g(y+t)\) for \(0\le t\le(x-y)/2\). Throughout this interval, \(x-t\ge y+t\). Since \(g\) is concave, its derivative is decreasing, so \(g'(y+t)\ge g'(x-t)\). Therefore

h'(t)=-g'(x-t)+g'(y+t)\ge0.

Thus \(h(\delta)\ge h(0)\), which is the claimed inequality. If \(x>y\), strict concavity makes \(g'\) strictly decreasing; for \(0\le t<(x-y)/2\), the inequality between the derivatives is strict. Integrating \(h'(t)\) over any interval of positive length from \(0\) to \(\delta\) gives \(h(\delta)>h(0)\). This proves the strict case as well. \(\square\)

Applying the result to one pair of roots at a time shows that a sequence of such mean-preserving equalization steps increases \(V\) and \(W\), while preserving \(T\). Since \(\Lambda=\exp(-W)\), an increase in \(W\) corresponds to a decrease in Wilks’ Lambda. The result does not say that every arbitrary transfer from a larger root to a smaller one is an equalization: the transfer must not exceed half their difference.

How the Four Criteria Emphasize Different Features

Criterion	What its formula emphasizes	Fixed-total comparison
Wilks’ Lambda	The product of the factors \(1+\lambda_i\); equivalently, \(W\) sums their logarithms.	Equalization raises \(W\), so it lowers \(\Lambda\).
Pillai’s trace	A bounded contribution \(\lambda_i/(1+\lambda_i)\) from each root.	Equalization raises \(V\).
Hotelling–Lawley trace	The total root mass.	It is unchanged when the total is fixed.
Roy’s largest root	The strongest single root.	It is small for equal roots and can be large when signal is concentrated.

Pillai’s contribution from any one root is bounded above by \(1\), and a root’s additional contribution diminishes as that root grows. The logarithmic contribution to \(W\) also grows at a diminishing rate. These features help explain why both summaries respond to signal spread across directions. The Hotelling–Lawley trace instead counts root mass linearly, without regard to how it is divided. Roy’s root asks whether there is a strong direction and therefore can respond sharply to a single dominant root.

Interpreting These Comparisons as Power Intuition

If a scientific alternative is expected to produce several moderate effects, the algebraic behavior of Pillai’s trace and Wilks’ Lambda can make them natural summaries to examine: at a fixed total, both favor a more even allocation than a single-root summary does. If the substantive alternative is specifically a strong effect in one direction, Roy’s root directly targets the largest root. The Hotelling–Lawley trace is sensitive to the sum, but by itself cannot distinguish a diffuse spectrum from a concentrated one with the same total.

These observations are about the criterion values for a fixed spectrum. Actual power is the probability that the statistic exceeds its calibrated rejection threshold (or, for Wilks’ Lambda, falls below its threshold) under a specified alternative. That probability also depends on the distribution of the random roots, the design dimensions, the significance level, and the calibration used. In particular, a larger value of one criterion’s numerical summary cannot be compared directly with a larger value of another criterion’s summary, whose scale and null distribution differ.

Common Pitfall

Wrong reasoning: “Moving any positive amount from a larger root to a smaller root makes the spectrum more diffuse, so Pillai’s trace and the transformed Wilks statistic must increase.”

Correction: The transfer must be a mean-preserving equalization. For a pair \(x\ge y\), its size must satisfy \(0\le\delta\le(x-y)/2\). A larger transfer crosses the equal point and makes the pair more unequal in the opposite direction; the equalization theorem does not apply. Also, even a valid equalization result compares criterion values, not the power of differently calibrated tests.

Key takeaway: At a fixed total root mass, Pillai’s trace and \(-\log\Lambda\) increase under mean-preserving equalization, the Hotelling–Lawley trace stays fixed, and Roy’s root emphasizes concentration. Treat these as descriptions of criterion sensitivity, not as a universal ranking of power.

Check Your Understanding

Use the root-based definitions and distinguish algebraic comparisons from calibrated power claims.

For spectra \((2,2)\) and \((4,0)\), which has the larger Hotelling–Lawley trace, and which has the larger Roy root?
For \(x=7\) and \(y=3\), what is the largest transfer \(\delta\) that qualifies as a mean-preserving equalization under the theorem?
Explain why an equalization step raises both Pillai’s trace and \(-\log\Lambda\).
Can two spectra with the same Hotelling–Lawley trace have different Roy roots? Give a pair of examples.
Why does a larger value of Pillai’s trace than of Roy’s root not, by itself, show that Pillai’s test has greater power?