# How exactly does Wilks' Lambda distribution generalize the Hotelling distribution?

According to Wikipedia, Wilks' Lambda distribution generalizes Hotelling's distribution. I am having some problems seeing how this works. I can see how Hotelling's distribution generalizes Student's t-distribution (a RV distributed as Hotelling's law with $p=1$ is just the square of a RV distribution as Student's t), but cannot see how to get to Wilks' Lambda. Is there some setting of the parameters $p,m,n$ such that a RV distributed as Wilks' lambda is some transform of a Hotelling RV?

Multivariate tests in contrast to the overall F test, answer the question, "Is each effect significant?" or more specifically, "Is each effect significant for at least one of the dependent variables?" That is, where the F test focuses on the dependents, the multivariate tests focus on the independents and their interactions. These tests appear in the "Multivariate Tests" table of SPSS output. The multivariate formula for F is based not only on the sum of squares between and within groups, as in ANOVA, but also on the sum of crossproducts - that is, it takes covariance into account as well as group means....

Hotelling's T-Square is the most common, traditional test where there are two groups formed by the independent variables....

Wilks' lambda, U. This is the most common, traditional test where there are more than two groups formed by the independent variables.... The t-test, Hotelling's T, and the F test are special cases of Wilks's lambda....

So I presume that if you take Wilks' lambda, and reduce the number of groups formed by the independent variables to two, then you get something like Hotelling's T-Square.

• With 2 groups, all common multivariate test statistics (Wilks' $\Lambda$, Roy's largest root, Pillai-Bartlett-trace, Hotelling-Lawley-trace, Hotelling's $T^{2}$) are equivalent: they are functions of the largest eigenvalue of $A^{-1} B$, the only eigenvalue $\neq 0$ with 2 groups. They are strictly monotic functions of each other, and give the same $p$-value. Example: pastebin.com/Dq9jczE1 – caracal May 11 '11 at 9:37

My question was about the distributions, not the test, and I think I've figured out the answer: a t-distribution squared has an F(1,n) distribution, which is a Hotelling distribution (up to rescaling by a constant determined by the parameters). I believe one can say that an F(m,n) distribution is the same as a Wilks' $\Lambda(1,m,n)$ distribution, which is the kind of 'generalization' I was looking for. (I guess I'm now enough of an 'expert' to edit the wikipedia page ;)

• I think you're right. This seems to be along the lines of Kshirsagar's Multivariate Analysis - Chapter 8, pp.299--300. – idnavid May 24 '18 at 1:37