Mixed variable, joint distribution, How do we know which one is continuous distribution, which one is discrete If we have one continuous r.v. $x$ and a discrete r.v. $y$ which takes one of the two values $y_1$ and $y_2$. Let's say we know the prior probabilities $P(y_1)$ and $P(y_2)$. 
From Bayes theorem we know
$p(y_i,x) = P(y_i|x)\cdot p(x)$
$P$ corresponds to prob. mass function, $p$ corresponds to prob. density function.
On LHS we have a probability value coming from a continuous distribution. On RHS we have product of two probabilities one coming from a discrete distribution, another coming from a continuous distribution.
My question is - Why is first term on RHS $P(y_i|x)$ and not $p(y_i|x)$ (Since it's possible that, given $x$, the probability that $y$ will take a value of $y_i$, might smoothly vary from $0$ to $1$)? Are there different ways of writing Bayes theorem that will have $p(y_i|x)$ instead of $P(y_i|x)$?
 A: 
(Since it's possible that, given x, the probability that y will take a value of yi, might smoothly vary from 0 to 1)

I believe you are misunderstanding what the distinction between a pdf and a pmf is based on. A pdf corresponds to the case when the argument of the probability function (for lack of a better alternative term) takes values on a continuum. Similarly, a pmf corresponds to the case when the argument of the probability function takes discrete values.
The argument of the probability function of $y|x$ is $y$, not $x$. You should think of $x$ as a constant (arbitrary, but fixed) in the probability function of $y|x$. Since $y$, the argument is discrete, the probability function of $y|x$ is a pmf, and indeed the notation $P(y_i|x)$ is correct.
Note: All my comments are based on the assumption that pdfs/pmfs exist when they need to. I am not talking about pathological cases here.
A: I think the way to understand this is to plot $P(y_i|x)$ Vs $y_i$. It's discrete distribution because X axis can only take discrete values. Thanks to @Glen_b for comment.
