If $X$ is a continuous random variable and $Y$ is a discrete random variable, is $P(X=Y) = 0$? I have a general question.
Lets say I have $2$ random variables

$X$ - Continuous Random Variable
$Y$ - Discrete Random Variable

For all $X$,$Y$ is $P(X=Y) = 0$ ?
 A: Let $n$ be all the values taken upon by $Y$, that is $P(Y=n) \neq 0$. Note that there can be at most a countable number of such $n$, so that we can use below the probability axioms.(On the third equality)
$$P(X = Y) = P(\{\omega: X(\omega) = Y(\omega)\}) = P\left(\bigcup_n \{\omega: X(\omega) = n, Y(\omega) = n\}\right) = $$$$=\sum_n P(\{\omega: X(\omega) = n, Y(\omega) = n\})$$
But $$\{\omega: X(\omega) = n, Y(\omega) = n\} \subset \{\omega: X(\omega) = n\} \implies$$$$ P(\{\omega: X(\omega) = n, Y(\omega) = n\})\le P(\{\omega: X(\omega) = n\}) = 0 $$
Hence $$P(X = Y) = \sum_n P(\{\omega: X(\omega) = n, Y(\omega) = n\}) = 0$$
A: Let me give a proof without involving explicit computations, when the two rv are independent. Let us define the new random variable Z=X-Y: we will prove that the law of Z has no atoms (and actually absolutely continuous). Indeed, the law of Z correponds to the convolution of the law of X with the law of -Y. Now, when you consider a convolution of two functions, one of which is absolutely continuous, then this convolution is absolutely continuous. QED 
