Does the joint pdf $f_{x, y} (x, y)$ equal to the conditional $p_{y | x} (y | x)$ for all random variables? So I have this question where you are given two random variables, $X$ and $Y$. $X$ is a continuous random variable (represented as a mean) with a distribution of $Exp(1)$ (exponential with $\lambda = 1$) and $Y$ is a discrete random variable (represented as the actual probability) with the distribution of $Pois(\lambda)$ and the parameter is $X$. I am asked to find the joint probability distribution. 
The solution suggests that $f_{x, y} (x, y) = p_{y | x} (y | x)$ but I am not sure how to got to this or why. I do not even know why they use the discrete formula for this or why they even used a conditional. After, they multiple the two distribution formulae for each variable. I am not sure why this is either. 
 A: Let me clear the notation. $X\sim\exp(\lambda=1)$, and given $X$, $Y$ is Poisson distributed with $\lambda=X$. By the definition of conditional probability, we have:$$f_{X,Y}(x,y)=f_{Y|X}(y|x)f_X(x)$$
where $$f_{Y|X}(y|x)=e^{-x}\frac{x^y}{y!},f_X(x)=e^{-x} $$
when $x> 0, y\in \mathbb{Z^+}$. 
So, your answer is correct but $f_{x,y}(x,y)\neq p_{y|x}(y|x)$
A: The $Y$ with parameter $X$ just means: $Y | X = x$ (where $x$ is constant). $Y$ is distributed as $Pois(X = x)$. 
Now, to find the joint probability density function of the two variables, you can use $p_{y | x} (y | x)$ since $Y$ is dependent on a constant $X = x$ value and the joint conditional probability has its second variable as a constant. Therefore, $f_{x, y} (x, y) = p_{y | x} (y | x)$. Now, since $x$ is constant, to find the probability for some $y$ given $x$, it is simply the probability of the $y$ times the probability of $x$ or mathematically: $= \frac{e^{-x}x^y}{y!}\cdot e^{-x}$ for $x > 0$ and $y = 0, 1, 2, \dots$ (all from the pdfs of the respective distributions). 
The other way i.e. $p_{x | y} (x | y)$ is harder to get (or not possible) since you would need a random variable where $y$ is constant, but for this question, there is none. 
