Exact F Reduction for a Scalar Response

Why the Four Criteria Coincide with One Response

In a multivariate test, Wilks' lambda, Pillai's trace, the Hotelling–Lawley trace, and Roy's largest root summarize the hypothesis and error sums of squares in different ways. With one response variable, those summaries cannot describe different directions in response space: there is only one response direction. Each criterion is therefore a transformation of the same scalar ratio. Under the central null hypothesis, that ratio has an exact $F$ distribution, provided the normal linear model assumptions hold.

The distinction between a scalar response and a hypothesis with one restriction matters. The response count is one here, but the hypothesis may impose $h$ independent restrictions, with $h$ greater than one. The $F$ numerator degrees of freedom reflect $h$; they do not have to equal one.

Model, Hypothesis, and Criteria

Definition: Consider the normal linear model $y=X\beta+\varepsilon$, where $y$ is an $n$-vector, $X$ is a fixed $n\times k$ matrix of full column rank, $\beta$ is a $k$-vector, and $\varepsilon\sim N_n(0,\sigma^2 I_n)$, with $\sigma^2>0$. Test $H_0:L\beta=0$, where $L$ has $h$ independent, estimable rows. Let $P_X$ be the orthogonal projection onto the column space of $X$, and let $P_0$ project onto the fitted-mean space under $H_0$. Define the hypothesis and error sums of squares by $H=y'(P_X-P_0)y$ and $E=y'(I_n-P_X)y$. The residual degrees of freedom are $\nu=n-k$.

Assume $h\geq1$ and $\nu>0$. Under the null, the constrained fitted-mean space has dimension $k-h$, so $P_X-P_0$ projects onto an $h$-dimensional space. The error projection $I_n-P_X$ has rank $\nu$. The criteria below require $E>0$, which holds with probability one in this model when $\nu>0$.

Definition: For one response, there is one generalized root, $\lambda=H/E$. The four MANOVA criteria are Wilks' lambda $\Lambda=E/(E+H)$, Pillai's trace $V=H/(H+E)$, the Hotelling–Lawley trace $U=H/E$, and Roy's largest root $\lambda_{\max}=H/E$. In this scalar setting, the Hotelling–Lawley trace and Roy's root are identical.

Worked Example: Calculate All Four Criteria

Suppose a test has $h=3$, $\nu=20$, and observed sums of squares $H=12$ and $E=30$. The hypothesis-to-error ratio is $H/E=12/30=0.4$, and the conventional test statistic is

F=\frac{H/h}{E/\nu} =\frac{12/3}{30/20} =\frac{4}{1.5} =\frac{8}{3}.

The criteria are $\Lambda=30/(30+12)=5/7$, $V=12/(30+12)=2/7$, and $U=\lambda_{\max}=12/30=0.4$. The ratio recovered from the $F$ statistic is $hF/\nu=3(8/3)/20=0.4$, as required. Thus the four reported numbers differ in scale and direction, but they encode the same observed ratio.

Exact Central-Null F Calibration

The exact calibration follows from the distributions of the two sums of squares. Under $H_0$, the mean $X\beta$ lies in the constrained fitted-mean space. The hypothesis and error projections are onto orthogonal spaces, of dimensions $h$ and $\nu$, respectively. In orthonormal coordinates adapted to these spaces, the corresponding components of $y$ are independent normal variables with mean zero and variance $\sigma^2$.

Theorem: Under the model and hypotheses above, under $H_0$, $H/\sigma^2\sim\chi^2_h$ and $E/\sigma^2\sim\chi^2_\nu$, independently. Consequently, $$ F=\frac{H/h}{E/\nu}\sim F_{h,\nu}. $$

Proof. Choose an orthonormal basis for the column space of $P_X-P_0$, another for the column space of $I_n-P_X$, and complete them to an orthonormal basis of $\mathbb{R}^n$. In these coordinates, the first $h$ components of $y$ are independent $N(0,\sigma^2)$ variables under $H_0$, because the null mean has zero projection onto the hypothesis space. Their squared sum is $H$, so $H/\sigma^2\sim\chi^2_h$. The $\nu$ error-space components are also independent $N(0,\sigma^2)$ variables; their squared sum is $E$, so $E/\sigma^2\sim\chi^2_\nu$. The two sets of coordinates are disjoint components of a jointly normal vector in an orthonormal basis, hence independent. Dividing each independent chi-square variable by its degrees of freedom and taking their ratio gives the stated $F$ distribution. $\square$

Set $T=H/E$. Since $F=(\nu/h)T$, the criteria and the $F$ statistic are related by

T=\frac{h}{\nu}F,\qquad \Lambda=\frac{1}{1+T}=\frac{\nu}{\nu+hF},\qquad V=\frac{T}{1+T}=\frac{hF}{\nu+hF},\qquad U=\lambda_{\max}=T=\frac{hF}{\nu}.

These are one-to-one transformations on their possible ranges. Wilks' lambda decreases as $F$ increases, while Pillai's trace, the Hotelling–Lawley trace, and Roy's root increase. Hence the exact central-null rejection rule can be expressed using any one of the four criteria, with the inequality direction chosen accordingly.

Beta Distributions and Equivalent Tail Areas

The bounded criteria have especially simple null laws. If independent random variables $A\sim\chi^2_h$ and $B\sim\chi^2_\nu$, then $A/(A+B)\sim\operatorname{Beta}(h/2,\nu/2)$. To see why, transform their joint density using $S=A+B$ and $Z=A/(A+B)$. The inverse transformation is $A=SZ$, $B=S(1-Z)$, with Jacobian $S$. The joint density, up to a constant, becomes

s^{(h+\nu)/2-1}e^{-s/2}\, z^{h/2-1}(1-z)^{\nu/2-1}, \qquad s>0,\quad 0<z<1.

The expression factors into a function of $s$ and a function of $z$. Normalizing the latter gives the beta density, proving the stated ratio law.

Result: Under the central null, $$ V=\frac{H}{H+E}\sim\operatorname{Beta}\!\left(\frac h2,\frac\nu2\right), \qquad \Lambda=\frac{E}{H+E}\sim\operatorname{Beta}\!\left(\frac\nu2,\frac h2\right). $$ Also $U=\lambda_{\max}=hF/\nu$, where $F\sim F_{h,\nu}$. The four exact null calibrations are equivalent: $V=1-\Lambda$, and each criterion is a one-to-one function of $F$.

For an observed $F=f$, the upper-tail p-value is $P(F_{h,\nu}\geq f)$. The same p-value is obtained by taking the lower tail for Wilks' lambda, $P(\Lambda\leq\Lambda_{\mathrm{obs}})$, or the upper tail for the other three criteria. A critical value $f_\alpha$ for an upper-tail test translates into $\Lambda\leq\nu/(\nu+h f_\alpha)$, $V\geq h f_\alpha/(\nu+h f_\alpha)$, and $U,\lambda_{\max}\geq h f_\alpha/\nu$.

Worked Example: Obtain an Exact P-Value from Pillai's Trace

Let $h=2$, $\nu=6$, and suppose $H=6$ and $E=6$. Then $F=(6/2)/(6/6)=3$, while $V=6/(6+6)=1/2$ and $\Lambda=1/2$. The null distribution of $V$ is $\operatorname{Beta}(1,3)$, whose cumulative distribution function is $P(V\leq v)=1-(1-v)^3$ for $0\leq v\leq1$. Therefore the upper-tail p-value for the observed Pillai trace is

P(V\geq 1/2)=(1-1/2)^3=\frac18.

The corresponding lower-tail probability for Wilks' lambda is also $P(\Lambda\leq1/2)=1/8$. Equivalently, the $F_{2,6}$ upper tail at 3 is $(1+2(3)/6)^{-3}=1/8$. All three calculations agree.

Worked Example: Recover F from Each Criterion

Suppose $h=4$, $\nu=10$, $H=20$, and $E=25$. Then $F=(20/4)/(25/10)=5/2.5=2$. The observed criteria are $\Lambda=25/45=5/9$, $V=20/45=4/9$, and $U=\lambda_{\max}=20/25=4/5$.

Each statistic recovers the same $F$. From the Hotelling–Lawley trace, $F=(\nu/h)U=(10/4)(4/5)=2$. From Pillai's trace, $F=(\nu/h)V/(1-V)=(10/4)(4/5)=2$. From Wilks' lambda, $F=(\nu/h)(1-\Lambda)/\Lambda=(10/4)(4/5)=2$. Thus using any of these criteria with its correctly transformed reference distribution gives the same test result.

Why the Reduction Matters—and a Common Pitfall

With one response, selecting among the four criteria cannot change the exact central-null evidence against the hypothesis. Their numbers may look different, and their conventional rejection directions differ, but there is no additional information in one criterion beyond the ratio $H/E$. This gives a useful check on both calculations and software output: convert each reported criterion to $F$, and the results should agree up to rounding.

Common mistake: treating a scalar response as a one-restriction test. One might set $h=1$ merely because there is only one response, then use $F_{1,\nu}$. That confuses the response dimension with the hypothesis degrees of freedom. For example, a scalar response with four independent restrictions has $h=4$, and its null statistic is $F_{4,\nu}$. Use the rank of the tested restrictions for $h$; use the number of response variables to decide whether the MANOVA criteria reduce to a single root.

The result is exact only under the stated normal, homoscedastic linear model and the central null. The chi-square and independence argument is not a general justification for using the same calibration under a noncentral alternative or under a different error structure.

Check Your Understanding

Use the definitions and transformations above to answer these questions.

For $h=3$, $\nu=18$, and $F=2$, calculate $H/E$, Wilks' lambda, and Pillai's trace.
Why can a scalar-response test have numerator degrees of freedom greater than one?
State the exact central-null distribution of Pillai's trace and of Wilks' lambda.
If an upper-tail $F$ test rejects for $F\geq f_\alpha$, which direction of inequality gives the equivalent rejection region for Wilks' lambda?
Explain why the hypothesis and error sums of squares are independent under the central null in the stated model.