A Relationship Between the One-Way MANOVA Test Statistic and the Hotelling Lawley Trace Test Statistic

The multivariate analysis of variance (MANOVA) model yi = BT xi + ε i for i = 1, . . . , n has m ≥ 2 response variables Y1, . . . Ym and p predictor variables x1, x2, . . . , xp. The ith case is (xi , y T i ) = (xi1, ..., xip,Yi1, ..., Yim). If a constant xi1 = 1 is in the model, then xi1 could be omitted from the case. For the MANOVA model predictors are indicator variables. Sometimes the trivial predictor 1 is also in the model. The MANOVA model in matrix form is Z = XB + E and has E(εk) = 0 and Cov(εk) = Σε = (σi j) for k = 1, . . . , n. Also E(ei) = 0 while Cov(ei, e j) = σi jIn for i, j = 1, . . . ,m. Then B and Σε are unknown matrices to be estimated.


Introduction
We want to show that the One-Way MANOVA test statistic and the Hotelling Lawley trace test statistic are equivalent for a carefully chosen full rank design matrix.First we will describe the MANOVA model, and then the One-Way MANOVA model.The notation in this paper follows that used in Olive (2017) and closely follows Rupasinghe Arachchige Don (2017).

MANOVA
Multivariate analysis of variance (MANOVA) is analogous to an ANOVA, but there is more than one dependent variable.ANOVA tests for the difference in means between two or more groups, while MANOVA tests for the difference in two or more vectors of means.
The multivariate analysis of variance (MANOVA) model y i = B T x i + ϵ i for i = 1, . . ., n has m ≥ 2 response variables Y 1 , . . .Y m and p predictor variables x 1 , x 2 , . . ., x p .The ith case is (x T i , y T i ) = (x i1 , ..., x ip , Y i1 , ..., Y im ).If a constant x i1 = 1 is in the model, then x i1 could be omitted from the case.
For the MANOVA model predictors are indicator variables.Sometimes the trivial predictor 1 is also in the model.The MANOVA model in matrix form is Z = XB + E and has E(ϵ k ) = 0 and Cov(ϵ k ) = Σ ϵ = (σ i j ) for k = 1, . . ., n.Also E(e i ) = 0 while Cov(e i , e j ) = σ i j I n for i, j = 1, . . ., m.Then B and Σ ϵ are unknown matrices to be estimated.
The n × p matrix X is not necessarily of full rank p, and where often v 1 = 1.
The p × m coefficient matrix is The n × m error matrix is Each response variable in a MANOVA model follows an ANOVA model Y j = Xβ j + e j for j = 1, . . ., m, where it is assumed that E(e j ) = 0 and Cov(e j ) = σ j j I n .
MANOVA models are often fit by least squares.The least squares estimator where The predicted values or fitted values are

One Way MANOVA
Assume that there are independent random samples of size n i from p different populations, or n i cases are randomly assigned to p treatment groups.Let n = ∑ p i=1 n i be the total sample size.Also assume that m response variables y i j = (Y i j1 , . . ., Y i jm ) T are measured for the ith treatment group and the jth case.Assume E(y i j ) = µ i and Cov(y i j ) = Σ ϵ .
The one way MANOVA is used to test H 0 : Note that if m = 1 the one way MANOVA model becomes the one way ANOVA model.One might think that performing m ANOVA tests is sufficient to test the above hypotheses.But the separate ANOVA tests would not take the correlation between the m variables into account.On the other hand the MANOVA test will take the correlation into account.

Let ȳ = ∑ p i=1
∑ n i j=1 y i j /n be the overall mean.Let ȳi = ∑ n i j=1 y i j /n i .Several m × m matrices will be useful.Let S i be the sample covariance matrix corresponding to the ith treatment group.Then the within sum of squares and cross products matrix is W The treatment or between sum of squares and cross products matrix is The total corrected (for the mean) sum of squares and cross products matrix is is the usual sample covariance matrix of the y i j if it is assumed that all n of the y i j are iid so that the µ i ≡ µ for i = 1, ..., p.
The one way MANOVA model is y i j = µ i + ϵ i j where ϵ i j are iid with E(ϵ i j ) = 0 and Cov(ϵ i j ) = Σ ϵ .The summary one way MANOVA If the y i j − µ j are iid with common covariance matrix Σ ϵ , and if H 0 is true, then under regularity conditions Fujikoshi (2002) showed , and Note that the common covariance matrix assumption implies that each of the p treatment groups or populations has the same covariance matrix Σ i = Σ ϵ for i = 1, ..., p, an extremely strong assumption.Kakizawa (2009) and Olive et al. (2015) show that similar results hold for the multivariate linear model.The common covariance matrix assumption, Cov(ϵ k ) = Σ ϵ for k = 1, ..., n, is often reasonable for the multivariate linear regression model.

Hotelling Lawley Trace Test
Hotelling Lawley trace test statistic Hotelling (1951); Lawley (1938), and the asymptotic distribution by Fujikoshi (2002) are widely used.Olive et al. (2015) explains the large sample theory of the Wilks' Λ, Pillai's trace, and Hotelling Lawley trace test statistics and gives two theorems to show that the Hotelling Lawley test generalizes the usual partial F test for m = 1 response variable to m ≥ 1 response variables.

A Relationship Between the One-Way MANOVA Test and the Hotelling Lawley Trace Test
An alternative method for One-Way MANOVA is to use the model Z = XB + E where for i = 1, . . ., p and j = 1, . . ., n i Then X is a full rank where the ith column of X is an indicator for group i − 1 for i = 2, . . ., p. Then and Then the least squares estimator B of B, It can be shown that the inverse of the above matrix is

Then
Now consider the double sum in H.Note that n = n 1 p and Now consider the rest of H, Therefore by ( 5) and ( 6), it is clear that Now consider Therefore, B T becomes From ( 8) and ( 9) B T = H.
Proof.General case: First consider the double sum in H.
Now consider the rest of H, Therefore by ( 11) and ( 12) Now consider (13) and ( 14) proves that H = B T .

Cell Means Model
We can get the same result for the cell means model which is defined for X and B given below.) .
There are three commonly used test statistics to test the above hypotheses.Namely, 1. Hotelling Lawley trace statistic: U