Bayes theorem in the context of generative classification models In the Introduction to Statistical Learning p. 142 in chapter 4.4 on generative models for classification the formula
$P(Y = k|X = x) = \frac{π_k \cdot f_k (x)} {\sum_{l=1}^{K}π_l f_l(x)}$ is given to determine the probability of class $k$ given the features $x$.
I understand that the Bayes theorem states $P(A|B) = \frac{P(A)\cdot P(B|A)}{P(B)} $. This implies $P(k) = π_k$, $P(x|k) = f_k(x)$, $P(x) = \sum_{l=1}^{K}π_l f_l(x)$.
I see that $P(k)$ is a prior probability, which can be found by looking at how often class $k$ occurs as opposed to the other class (calculating $n_k/n$). What is $P(x)$ though? How can I think about it in terms of a prior probability? How can there be a prior probability of $x$, if I can plug any kind of (unseen) data into $x$?
Note this question on why $P(x)  = \sum_{l=1}^{K}π_l f_l(x)$.
 A: The answer depends on the specific application. For example, the features/predictors $X$ (discrete, continuous, or mixed) could be results from a battery of clinical tests; say blood pressure, colesterol level, and so on. In this case $p(X)$ may represent the distribution (continuous/density, discrete/mass, or mixed) of such values in the population of interest, or our uncertainty about which particular values $X$ we'll observe in the next patient.
This distribution is important. Just as an example, consider the case where $X$ is binary with values ${+,-}$, the predictand $Y$ is also binary with values ${0,1}$, and we have the following conditional probabilities:
$$
\begin{aligned}p(0|+)&=0 & p(1|+)&=1,\\ p(0|-)&=1/2 & p(1|-)&=1/2\ .
\end{aligned}
$$
The value $X=\mathord{+}$ is a perfect predictor, whereas $X=\mathord{-}$ is useless. Is the predictor $X$ good or not, overall?
The answer depends on $p(X)$. If values $X=\mathord{-}$ are rarely seen in the population, i.e. $p(\mathord{+})/p(\mathord{-}) \gg 1$, then $X$ is overall a good predictor, because we'll mostly encounter perfect-prediction situations with $X=\mathord{+}$. If instead $p(\mathord{-})/p(\mathord{+}) \gg 1$, then $X$ is overall a poor predictor: most of the times we'll be clueless about $Y$, except few lucky cases where we can be completely certain about it. This is also one of the reasons why the mutual information between $X$ and $Y$ depends on the probability $p(X)$.
The distribution $p(X)$ thus also affects our choice between two predictors (assuming we can't use them jointly for some reason). Imagine there's a second predictor $Z$ with values "a", "b" and these conditional probabilities:
$$
\begin{aligned}p(0|\mathrm{a})&=1/4 & p(1|\mathrm{a})&=3/4,\\ p(0|\mathrm{b})&=1/2 &  p(1|\mathrm{b})&=1/2\ .
\end{aligned}
$$
Which to choose, $X$ or $Z$? At first $X$ would seem best, because $X=\mathord{-}$ and $Z=\mathrm{b}$ are equally useless, but $X=\mathord{+}$ predicts better than $Z=\mathrm{a}$.
Yet the answer again depends on $p(X)$ and $p(Z)$. If $p(X=\mathord{+})$ is very low (rare in the population), whereas $p(Z=\mathrm{a})$ is very high (frequent in the population), then on average we'll have better predictions by using $Z$.

In situations where you can arbitrarily set the value of $X$, then typically you don't use $p(X)$, which is formally just equal to 1 (you know which value you set). But also in such a context, you may be interested about some kind of future (or past but unknown) performance, and you may need to know how often any particular value $X=x$ will be set by you or whoever sets it. Then $p(X)$ enters again, and its assessment method differs wildly depending on the context.
I recommend MacKay's and Jaynes's books on these general matters, as well as a very illuminating paper by Lindley & Novick. You may also want to look up material about the base-rate fallacy.
