Consider the following problem, from Tijms's Understanding Probability:
A receiver gets as input a random signal that is represented by a discrete random variable $X$, where $X$ takes on the value +1 with probability $p$ and the value -1 with probability $1- p$. The output $Y$ is a continuous random variable which is equal to the input $X$ plus random noise, where the random noise has an $N (0, \sigma^2 )$ distribution. You can only observe the output. What is the conditional probability of $X = 1$ given the observed value $Y = y$?
This problem is interesting because it seems a mix of discrete and continuous random variables.
My attempt.
For the probability that $Y\le y$, we have two possibilities: either $X=1$ or $X=-1$, and $P(Y\le y) = P(Y\le y|X=1)P(X=1) + P(Y\le y|X=-1)P(X=-1)$, so
$$
P(Y\le y)=p\int_{-\infty}^y \frac{1}{\sqrt{2\pi}}e^{-\frac12 (s-1)^2}ds + (1-p)\int_{-\infty}^y \frac{1}{\sqrt{2\pi}}e^{-\frac12 (s+1)^2}ds.
$$
Here I used the fact that:
$$
P(Y\le y|X=1) = \int_{-\infty}^y \frac{1}{\sqrt{2\pi}}e^{-\frac12(s-1)^2}ds.
$$
Then, we might use:
$$
P(X=1|Y=y) = P(Y=y|X=1)\frac{P(Y= y)}{P(X=1)},
$$
however if I replace the terms $P(Y\le y)$ and $P(Y\le y|X=1)$ I derived earlier, first I get a complicated simple expression, and then I am puzzled by having something like $P(Y=y)$ in the context of a continuous random variable.
Best Answer
I know this sounds like nonsense, but for conditioning on a continuous variable $Y$ having a particular value $Y=y$, you can simply substitute the pdf where you would normally have $P(Y=y)$ (whose value is technically $0$).
Indeed, your situation is exactly Bayes theorem with one discrete variable and one continuous variable.
I don't know enough of the underlying theory (presumably measure theory?) to explain why this works. Sorry!