Exact F Reduction for a Rank-One Hypothesis

Why One Hypothesis Restriction Simplifies MANOVA

The four classical MANOVA criteria can look different because they summarize a spectrum of generalized roots in different ways. When the hypothesis has just one independent restriction, that spectrum collapses to a single nonzero root. Each criterion is then a simple transformation of the same number, and under the central null that root has an exact F calibration.

Here, “one restriction” means that the hypothesis has rank one, or equivalently one numerator hypothesis degree of freedom. This qualification matters: merely observing a rank-one hypothesis matrix in a particular data set does not give a higher-rank hypothesis the same null distribution.

Definition: Let $H$ and $E$ be the hypothesis and error sums-of-squares-and-cross-products matrices for a $p$-response MANOVA, with $E$ positive definite. The generalized roots are the eigenvalues of $E^{-1}H$. In the rank-one hypothesis case, write $\lambda$ for the unique nonzero generalized root.

For the exact central-null result, suppose the response matrix follows the multivariate normal linear model with a fixed, full-column-rank design and independent rows of errors having covariance matrix $\Sigma$, where $\Sigma$ is positive definite. Let $\nu$ be the error degrees of freedom, and suppose the hypothesis consists of one independent linear restriction. Under its null hypothesis, $H$ has a central Wishart distribution with one degree of freedom and scale $\Sigma$; independently, $E$ has a central Wishart distribution with $\nu$ degrees of freedom and scale $\Sigma$. Assume $\nu\geq p$, so that $E$ is positive definite with probability one.

The Root and the Four Criteria

A rank-one hypothesis matrix can be written $H=aa'$ for some $p$-vector $a$. The matrix $E^{-1}H$ has rank one, so it has at most one nonzero eigenvalue. Its trace is $a'E^{-1}a$; since the trace is the sum of its eigenvalues, its unique nonzero root is

\lambda=a'E^{-1}a.

This root is nonnegative, and is positive whenever $a\neq 0$, because $E^{-1}$ is positive definite. For a one-restriction hypothesis, $a\neq0$ with probability one under the stated model.

The four classical criteria are the product $\Lambda$ of $1/(1+\lambda_i)$, the sum $V$ of $\lambda_i/(1+\lambda_i)$, the sum $U$ of $\lambda_i$, and the largest root $\Theta$. With only one nonzero root, their definitions reduce to the following identities:

\Lambda=\frac{1}{1+\lambda},\qquad V=\frac{\lambda}{1+\lambda},\qquad U=\lambda,\qquad \Theta=\lambda.

Worked Example: Finding the Root From the Matrices

Take $p=3$, let $E=\operatorname{diag}(2,3,4)$, and let $a=(2,3,0)'$, so $H=aa'$. The error matrix is positive definite, and $H$ has rank one. The unique nonzero root is

\lambda=a'E^{-1}a =\frac{2^2}{2}+\frac{3^2}{3}+\frac{0^2}{4} =2+3+0=5.

Thus $\Lambda=1/(1+5)=1/6$, $V=5/(1+5)=5/6$, and $U=\Theta=5$. These four values are different summaries, but all are determined by the same root.

The Exact Central-Null F Reduction

Theorem: Under the multivariate normal linear model and central-null assumptions stated above, let $k=\nu-p+1$. For the unique nonzero root $\lambda$, $$ \frac{k}{p}\lambda \sim F_{p,k}. $$ Consequently, each criterion gives an exact central-null F statistic: $$ \frac{k}{p}\frac{1-\Lambda}{\Lambda} =\frac{k}{p}\frac{V}{1-V} =\frac{k}{p}U =\frac{k}{p}\Theta \sim F_{p,k}. $$

The proof uses the standard central Wishart representation of the hypothesis and error matrices under the null. Since the generalized roots are unchanged by applying the same nonsingular transformation to both matrices, we may whiten by $\Sigma^{-1/2}$. In the whitened coordinates, write $H=zz'$, where $z\sim N_p(0,I_p)$, and $E=X'X$, where $X$ is a $\nu\times p$ matrix of independent standard normal entries, independent of $z$. The root is $\lambda=z'E^{-1}z$.

Condition on $z\neq0$, an event of probability one. Choose an orthogonal matrix $R$ taking $z/\lVert z\rVert$ to the first coordinate vector. By rotational invariance, $RER'$ has the same distribution as $E$. Its conditional distribution does not depend on $z$, so it is independent of $z$. Therefore $\lambda$ has the distribution of $\lVert z\rVert^2(E^{-1})_{11}$, with the two factors independent. Since $z$ is standard normal, $\lVert z\rVert^2\sim\chi^2_p$.

To find the distribution of $(E^{-1})_{11}$, write the columns of $X$ as $x_1$ and $X_2$, with $X_2$ containing the other $p-1$ columns. The block inverse formula gives

(E^{-1})_{11} =\frac{1}{x_1'(I-P_{X_2})x_1},

where $P_{X_2}$ is the orthogonal projection onto the column space of $X_2$. Because $X_2$ has rank $p-1$ with probability one, $I-P_{X_2}$ has rank $\nu-p+1=k$. Conditional on $X_2$, the vector $x_1$ is independent standard normal, so its squared projection onto this $k$-dimensional subspace has a $\chi^2_k$ distribution. This conditional distribution does not depend on $X_2$. Hence $(E^{-1})_{11}$ has the distribution of $1/W$, where $W\sim\chi^2_k$, independently of $\lVert z\rVert^2$.

It follows that $\lambda$ has the distribution of $A/W$, where $A\sim\chi^2_p$ and $W\sim\chi^2_k$ are independent. By the definition of the F distribution, $A/W=(p/k)F$ for $F\sim F_{p,k}$. Therefore $(k/p)\lambda\sim F_{p,k}$. Finally, the criterion identities give $(1-\Lambda)/\Lambda=\lambda$, $V/(1-V)=\lambda$, and $U=\Theta=\lambda$. Substitution proves all four calibrations. $\square$

Worked Example: Calibrating a Root Under the Null

Suppose $p=3$, $\nu=12$, and an observed rank-one hypothesis produces $\lambda=5$, as in the matrix calculation above. Here $k=\nu-p+1=12-3+1=10$. The calibrated statistic is

F_{\mathrm{obs}}=\frac{10}{3}(5)=\frac{50}{3}\approx16.67, \qquad F_{\mathrm{null}}\sim F_{3,10}.

The equivalent criteria are $\Lambda=1/6$, $V=5/6$, and $U=\Theta=5$. For example, the Wilks conversion gives

\frac{10}{3}\frac{1-\Lambda}{\Lambda} =\frac{10}{3}\frac{1-1/6}{1/6} =\frac{10}{3}(5) =\frac{50}{3}.

Thus the root and each criterion lead to exactly the same F statistic and the same null reference distribution. This is a distributional calibration; the example does not claim that the observed value is typical under the null.

Worked Example: Recovering the F Statistic From Pillai’s Trace

Suppose $p=4$, $\nu=15$, and the one-root Pillai trace is $V=3/7$. The corresponding root is

\lambda=\frac{V}{1-V} =\frac{3/7}{1-3/7} =\frac{3/7}{4/7} =\frac{3}{4}.

The denominator degrees of freedom are $k=15-4+1=12$. Therefore,

F_{\mathrm{obs}}=\frac{12}{4}\lambda =3\left(\frac{3}{4}\right) =\frac{9}{4}=2.25, \qquad F_{\mathrm{null}}\sim F_{4,12}.

The same result follows directly from $V$: $(12/4)V/(1-V)=3(3/4)=2.25$. The one-root identity is what makes this inversion exact.

Scope, Use, and a Common Pitfall

The exact reduction is useful when a scientific question is represented by one independent multivariate restriction. The criterion values then contain no additional information about the root beyond their stated transformations, and the central-null reference distribution is determined by $p$ and $\nu$. The assumption $\nu\geq p$ is essential for this version: it ensures that the error matrix is nonsingular and that $k=\nu-p+1$ is positive.

Common mistake: treating any rank-one observed $H$ as proof that the hypothesis has one degree of freedom, and then assigning the $F_{p,\nu-p+1}$ law. The null distribution follows from the design and hypothesis rank, not just the numerical rank observed in a sample. With $q>1$ independent restrictions, the central null hypothesis matrix generally has Wishart degrees of freedom $q$; it need not have rank one. The one-root criterion identities and this exact F reduction cannot be inferred merely from a special observed matrix.

A second useful check is the degrees-of-freedom order: the numerator is $p$, not one, while the denominator is $\nu-p+1$. In the scalar-response case $p=1$, this becomes $F_{1,\nu}$, agreeing with the usual one-restriction scalar-response result. For more than one response, the $p$ numerator degrees of freedom account for the multivariate response dimension.

Key takeaway: For one independent hypothesis restriction, the hypothesis matrix has one nonzero generalized root $\lambda$. Wilks, Pillai, Hotelling–Lawley, and Roy reduce to functions of that root, and under the central null $(\nu-p+1)\lambda/p\sim F_{p,\nu-p+1}$, provided the stated normal-model and nonsingularity conditions hold.

Check Your Understanding

Use the one-restriction setting and assumptions stated above.

If the sole nonzero generalized root is $2$, what are $\Lambda$, $V$, $U$, and $\Theta$?
For $p=5$ and $\nu=14$, give the denominator degrees of freedom and the exact F statistic in terms of $\lambda$.
Explain why the nonzero root of $E^{-1}aa'$ equals $a'E^{-1}a$.
What distributional assumption about $H$ and $E$ under the central null supports the exact F reduction?
Why does an observed rank-one $H$ not, by itself, justify applying this one-restriction F law?

Exact F Reduction for a Rank-One Hypothesis

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