Can you have a "PMF-PDF" Together? This is a question I have always struggled to understand:

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*For a discrete random variable, you can have a "Probability Mass Function" (PMF) : For several discrete random variables, you can model their joint probabilities using a "Joint Probability Mass Function".


*For a continuous random variable, you can have a "Probability Distribution Function" (PDF) : For several continuous random variables, you can model their joint probabilities using a "Joint Probability Density Function".
This brings me to my question: Suppose you have three discrete random variables X1, X2 and X3 - and suppose you also have three continuous random variables X4, X5, X6 : Is it possible to create a joint probability function for these 6 variables such as P(X1, X2, X3, X4, X5, X6)? Could you also have a conditional probability function P(X1 | X2 = x2, X3 = x3, X4 = x4, X5 = x5, X6 = x6)? Or is this by definition impossible?
Thanks!

This is possible. It is standard measure theory and is done the same way as if all the random variables are continuous or all are discrete: You have a probability space $(\Omega, F, p)$ and your random variables $X_i$ are all, by definition, measurable maps from $\Omega$ to the measurable space $(\mathbb{R}, B)$ of real numbers $\mathbb{R}$, where $B$ is the sigma field generated by the open sets. The difference between continuous and discrete random variables is simply that the image of discrete random variables is discrete.
If you want to, you can now use your random variables $(X_1, \ldots, X_n)$ to push forward the measure $p$ from $\Omega$ to $\mathbb{R}^n$. This will define a distribution on $\mathbb{R}^n$, i.e. a probability measure on $\mathbb{R}^n$. It is just an ordinary measure. It might "look" a little different compared to the "pure" cases. E.g. if you have one binary random variable with image just the two points $0$ and $1$, i.e. $X_1: \Omega\to\mathbb{R}, Im(X_1) = \{0, 1\}$, and a second variable $X_2$ which is e.g. the standard normal, you could choose to visualize this as two weighted one-dimensional standard normals, one along the line $x_1=0$ and the other along the line $x_1=1$. All the members $b\in B$ of the sigma field $B$ that do not intersect with those two lines have measure zero.
You could even go one step further and define random elements, which don't even require anymore that the range of the $X_i$ be $\mathbb{R}$. It is all standard measure theory.
Conditional probabilities can be defined as usual, with the same possible problems if densities become zero.
