7.6 Fishers Sample Linear Discriminate

Let \(\hat {\lambda }_1,\hat {\lambda }_2,\ldots , \hat {\lambda }_s>0\) denote the \(S\leq \min (g-1,p)\) non zero eigenvalues of \(\widehat {W}^{-1}\widehat {B}\) and \(\hat {e}_1,\hat {e}_2,\ldots , \hat {e}_s\) be the corresponding eigenvectors scaled so that \(\hat {e}'\) \[\hat {e}'S_p\hat {e}=1\]

Then the vector of coefficients \(\widehat {\underline {a}}= \widehat {L}\) that maximises the ratio \begin {equation} \tag {13} \frac {\widehat {L}'\widehat {B}\widehat {L}}{\widehat {L}'\widehat {W}\widehat {L}}=\frac {\widehat {L}'\Bigg (\sum \limits ^g_{i=1}\big (\overline {X}_i-\overline {X}\big )\big (\overline {X}_i-\overline {X}\big )'\Bigg )\widehat {L}}{\widehat {L}'\Bigg (\sum \limits ^g_{i=1}\sum \limits ^{n_i}_{j=1}\big (X_{ij}-\overline {X}_i\big )\big (X_{ij}-\overline {X}_i\big )'\Bigg )\widehat {L}} \end {equation}

is given by \(\widehat {L}_1=\widehat {e}_1\). The linear combination \(\widehat {L}_1X\) is called the sample first discriminate. The choice \(\widehat {L}_2 =\widehat {e}_2\) produces the sample second discriminate. Continuing we have that \[\widehat {L}'_kX =\widehat {e}'_kX\] is the sample \(k^{\text {th}}\) discriminate \(k\leq S\).

Sketch of proof of the Theorem on \(\displaystyle {\frac {\underline {a}'B\underline {a}}{\underline {a}'W\underline {a}}}\).

Let \(\displaystyle {f=\frac {\underline {a}'B\underline {a}}{\underline {a}'W\underline {a}}\in \mathbb {R}}\)

\(\implies \quad f\underline {a}'W\underline {a} = \underline {a}'B\underline {a}\) post multiplying by \(\underline {a}\)

\(\implies \quad \underline {a}\cdot \underline {a}' W \underline {a} = \underline {a}\cdot \underline {a}'B \underline {a}\)

Assume \(\underline {a}\, \underline {a}' >0\), pre-multiplying

\[f(\underline {a}\,\underline {a}')^{_1} \underline {a}\, \underline {a}' W \underline {a} = (\underline {a}\,\underline {a}')^{-1}\underline {a}\, \underline {a}'B\underline {a}\]

\[\implies \quad fW\underline {a} = B\underline {a}\]

\(W>0\) then \(fW^{-1} W \underline {a} = W^{-1}B\underline {a}\quad \implies \quad f\underline {a} = W^{-1}B\underline {a}\).

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