# Is there a generalization of Pillai trace and the Hotelling-Lawley trace?

In the setting of multivariate multiple regression (vector regressor and regressand), the four major tests for the general hypothesis (Wilk's Lambda, Pillai-Bartlett, Hotelling-Lawley, and Roy's Largest Root) all depend on the eigenvalues of the matrix $H E^{-1}$, where $H$ and $E$ are the 'explained' and 'total' variation matrices.

I had noticed that the Pillai and Hotelling-Lawley statistics could both be expressed as $$\psi_{\kappa} = \mbox{Tr}\left(H\left[\kappa H + E\right]^{-1}\right),$$ for, respectively, $\kappa = 1, 0$. I am looking at an application where the distribution of this trace, defined for the population analogues of $H$ and $E$, is of interest for the $\kappa = 2$ case. (modulo errors in my work.) I am curious if there is some known unification of the sample statistics for general $\kappa$, or some other generalization that captures two or more of the four classical tests. I realize that for $\kappa$ not equal to $0$ or $1$, the numerator no longer looks like a Chi-square under the null, and so a central F approximation seems questionable, so maybe this is a dead end.

I am hoping that there has been some research on the distribution of $\psi_{\kappa}$ under the null (i.e. the true matrix of regression coefficients is all zero), and under the alternative. I am interested particularly in the $\kappa = 2$ case, but if there is work on the general $\kappa$ case, I could, of course, use that.

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Wait, $H$ is the 'E'xplained variation and $E$ is the 'T'otal variation? Just checking my mnemonics. –  cardinal Jun 19 '12 at 13:58
@cardinal, that is correct. When $\hat{B}$ is the multivariate least squares fits to the correlation coefficients, we have $H=\hat{B}^{\top} \left(X^{\top}X\right)\hat{B}$ and $E=\left(Y - X\hat{B}\right)^{\top}\left(Y - X\hat{B}\right).$ A (literally) big picture overview from Michael Friendly has been pretty useful for me: psych.yorku.ca/lab/psy6140/lectures/… –  shabbychef Jun 19 '12 at 20:07
Thanks! I'll have a look. (By the way, I was kind of just teasing based on the choice of letters, 'h' for 'explained' and 'e' for 'total'.) Interesting question, by the way; (+1) from me. –  cardinal Jun 19 '12 at 20:18
@cardinal I was insufficiently caffeinated to notice the joke. Yes, the mnemonics are bad, but the choice of $H$ and $E$ (and $T=H+E$) is rather standard. –  shabbychef Jun 19 '12 at 20:39
The joke was sufficiently bad that it would have required lots of caffeine to notice. –  cardinal Jun 19 '12 at 21:09
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## 1 Answer

I imagine that productive generalizations would come out from observations that

1. some of these tests are norms of the vector ${\rm spec}[HE^{-1}]=\{\lambda_1, \ldots, \lambda_p\}$, so Hotelling-Lawley's trace is the $l_1$ norm, $\| \{\lambda_1, \ldots, \lambda_p\} \|_1$, and Roy's largest root is the $l_\infty$ norm, $\| \{\lambda_1, \ldots, \lambda_p\} \|_\infty$.
2. some of these tests may be a norm of the matrix $HE^{-1}$, e.g., Roy's largest root is the spectral, or $l_2$, norm $\| H E^{-1} \|_2$.
3. some of the tests may be of the generalized entropy form, e.g., Hotelling-Lawley's trace is GE(1), Roy's largest root is GE($\infty$), and Wilks' $\Lambda$ is GE(-1) on $\{1+\lambda_1, \ldots, 1+\lambda_p\}$, up to a monotone transformation each.

When other norms or other generalized entropy parameters are entertained, other statistics may be arrived at that might be meaningful. I doubt that any of them would produce your $\psi_2$, though.

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I believe we have $\psi_{\kappa} = \sum_i \frac{\lambda_i}{1 + \kappa\lambda_i}$, where $\lambda_i$ are the eigenvalues of $HE^{-1}$. But that doesn't seem to get me anywhere. I guess I don't know enough about the distribution of sums of the eigenvalues ... –  shabbychef Jun 25 '12 at 18:01
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