 Solution

Let $n={\sum }_{\nu =1}^g{n}_{\nu }$, ${\bar{x}}_{\nu }=n_{\nu }^{-1}{\sum }_j{x}_j^{(\nu )}$ and $\bar{x}=n^{-1}{\sum }_{\nu }{n}_{\nu }{\bar{x}}_{\nu }$. Define the within-group and between-group scatter matrices byThe within-group and between-group scatter decomposition gives total scatter $T=W+B$: in each group the residuals about ${\bar{x}}_{\nu }$ sum to zero, so all cross terms vanish. Under unrestricted group means, the maximum-likelihood estimators are ${\hat{\mu }}_{\nu }={\bar{x}}_{\nu }$ and $\hat{V}=W/n$. Under the common-mean null hypothesis, they are $\hat{\mu }=\bar{x}$ and ${\hat{V}}_0=(W+B)/n$.

For either fitted scatter matrix $A$, substitution into the log-likelihood givessince $\mathrm{tr}[(A/n{)}^{-1}A]=np$. Thus the maximized likelihood ratio isA generalized likelihood-ratio test rejects for small ${\Lambda }_{\mathrm{LR}}$, equivalentlyThe cutoff $c$ is chosen for the desired significance level. The determinant ratio $|W|/|W+B|$ itself is the Wilks lambda statistic. The derivation assumes $W$ is positive-definite, which for continuous normal samples requires enough within-group degrees of freedom, $n-g\ge p$; singular $W$ does not support this ordinary determinant-based test without a modified model.