I have a question on diagonal LDA in the book "Elements of statistical learning" by Hastie, Tibshirani et al. On page 652, it says
The discriminant score of the diagonal-covariance LDA for class $k$ is $$ \delta_{k}(x)= - \sum_{j=1}^{p} \dfrac{(x_{j}-\bar{x}_{kj})^{2}}{s_{j}^{2}} + 2 \log \pi_{k}, $$ [see (4.12) on page 110].
First thing I'm wondering about is the fact that (4.12) is the discriminant function of QDA and not of LDA, I thought diagonal LDA is based on LDA?
Nevertheless, this is (4.12):
$$ \delta_{k}(x) = - \frac{1}{2} \log |\mathbf{\Sigma}_k| - \frac{1}{2}(x-\mu_k)^T\mathbf{\Sigma}_k^{-1}(x-\mu_k)+\log \pi_k $$
$\mathbf{\Sigma}_k$ is the covariance matrix of each class $k$. For the transformation below I used $\mathbf{\Sigma}_k = \mathrm{diag}(\sigma_j^2) \; \forall k$.
So now I would like to justify why dLDA has the above discriminant function. With what I know, I was able to transform (4.12) into
$$\delta_{k}(x) = - \frac{1}{2} \log |\mathrm{diag}(\sigma_j^2)| - \frac{1}{2} \sum\limits_{j=1}^p\frac{(x_j-\mu_k)^2}{s_j^2}+\log \pi_k. $$
Explanation of the variables:
- $s_j$ is the Standard Deviation of the $j$th gene
- $\bar{x}_{kj}=\sum_{i \in C_{k}}\frac{x_{ij}}{N_{k}} $ mean of the $N_{k}$ values for gene $j$ in class $k$
- $C_{k}$ Indexset for class $k$
- $\pi_k$ Prior probability of class $k$, $\sum_{k=1}^{K}\pi_{k}=1$
- I think $\mu_k$ = $\bar{x}_{kj}$
So now I would really appreciate some hints how to proceed in order to transform some more so I end up with the discriminant function of dLDA.