# Solving for the discriminant function in LDA

This is related to the question posted in:

The discriminant function in linear discriminant analysis

In the one dimensional case, where $$p_k(x) = \dfrac{f_k(x)\pi_k}{P(X = x)}$$, where $$f_k(x) = \dfrac{1}{\sqrt{2\pi}\sigma}\exp(- \frac{1}{2\sigma^2}(x-\mu_k)^2)$$.

After taking the logarithm, the discriminant function is given as:

$$\delta(x) = \log \pi_k + \dfrac{x \mu_k}{\sigma^2} - \dfrac{\mu_k^2}{2\sigma^2}$$.

Why is there no term involving $$-\dfrac{x^2}{2\sigma^2}$$? This is the first term in the expansion of $$(x-\mu_k)^2$$. Why is this neglected?

And now that I have a discriminant function, what is the next step?

It's ignored because you try to determine the class given $$X=x$$. So, $$-x^2/2\sigma^2$$ term is the same for all $$k$$, i.e. your classes. Therefore, it doesn't affect the decisions when two discriminant functions are compared, i.e. $$\delta_k(x)$$, $$\delta_l(x)$$. Once you have the discriminant functions, you put $$x$$ into them and choose the maximum one to get your class estimate. Note that this is also Bayes classifier with normal distribution.
Only the terms specifically dependent on the $$k^{th}$$ class are considered before writing the decision equation. Since $$\frac{-x^2}{2\sigma^2}$$ does not depend on $$k$$, the decision taken will not be affected by the numerical value of this quantity. Thus, it is neglected.