I am just getting started with trying to understand the theory behind Naive Bayes a bit.
$Y$ = boolean-valued rv $X_i$ = boolean-valued rv (part of random vector $\vec{X}$).
From what I understand, we want this:
$$ P(Y|X_1,X_2,X_3) $$
It can be rewritten as
$$ P(X_1,X_2,X_3|Y)P(Y) $$
The denominator can be ignored.
To get $P(X_1,X_2,X_3|Y)$ We have to do $(2^3-1)*2$ calculations.
And I think I get that. We have to this for every possible combination of values of $X_1,X_2,X_3$ (except one), once for $Y=1$ and once for $Y=0$.
However, if I understand correctly
$$ P(X_1,X_2,X_3|Y) = P(X_1|X_2,X_3,Y)*P(X_2|X_3,Y)*P(X_3|Y) $$
Doesn't this sum up to $2^4 + 2^3 + 2^2 $ calculations?
And then once we assume cond. independence, it looks like this:
$$ P(X_1,X_2,X_3|Y) = P(X_1|Y)*P(X_2|Y)*P(X_3|Y) $$
And we only have to do 2 + 2 + 2 calculations, right?
