in the context of our lecture a $\textbf{random variable}$ is a map $X\colon \Omega\to S$ between a probability space $(\Omega, \mathcal{F}, P)$ a measurable space $(S,\mathcal{S})$ with $X^{-1}(A)\in\mathcal{F}$ for $A\in\mathcal{S}$, and the random variable is called $\textbf{discrete}$ iff $X(\Omega)$ is countable. $X$' $\textbf{distribution}$ is $P_X\colon\mathcal{S}\to [0,1],\, A\mapsto P(X\in A)$.

At this point our lecture differentiates the notion of discrete and non-discrete ($\mathbb{R}^n$-valued) random variables, but I see some similarities in the following definitions:

Discrete case: Given a probability measure $\mu$ on $(S,\mathcal{P}(S))$, $S$ being countable, the map $\rho\colon S\to [0,1],\, x\mapsto\mu(x)$ is its $\textbf{counting density}$. The counting density $\rho_X$ of a discrete random variable $X\colon\Omega\to S$ is the counting density of its distribution $P_X$.

Non-discrete case: Given a probability measure $\mu$ on $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))$ the map $f\colon \mathbb{R}^n\to [0,\infty[$ is its $\textbf{density}$ according to the Lebesgue measure, iff $f$ is existent with $\mu(A)=\int_A f(x) dx$ for $A\in\mathcal{B}(\mathbb{R}^n)$. The density according to the Lebesgue measure $f_X$ of a random variable $X\colon\Omega\to\mathbb{R}^n$ is the density according to the Lebesgue-measure of its distribution $P_X$.

Now, as in the situation of the first definition it is $P_X(A)=\sum_{x\in A} \rho_X(x)$, I wondered if the above definitions cannot be covered in only one? Unfortunately, I am not into measure theory yet, but I can imagine a measure-dependent definition that results in the above sum being the integral according to the counting measure. However, further research led me to the distinction of the discrete and non-discrete case, always, so, here I am now, asking this question.

Thanks alot in advantage

  • $\begingroup$ It's actually something that's done pretty often - it's very handy. See e.g. "Unified Signal Theory" by G. Cariolaro. $\endgroup$ – Anonymous Jan 13 '17 at 14:45
  • $\begingroup$ Thanks, I guessed so. Is a large amount of work involved? There seems to be a strict distinction in undergrad material. $\endgroup$ – Ramen Jan 13 '17 at 15:00
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    $\begingroup$ Hmm, I don't think you need a lot of work per se. It's just a bit tricky, and you need distributions - things like functions, but that can assume "infinite values", and yet are still integrable - the most basic example would be Dirac's delta (en.wikipedia.org/wiki/Dirac_delta_function). $\endgroup$ – Anonymous Jan 13 '17 at 15:06
  • $\begingroup$ I decided to make an answer of the comment above, so it's easier to find! $\endgroup$ – Anonymous Jan 13 '17 at 15:15

This is an excellent question! One can "unify" discrete and continuous probabilities using generalized functions like Dirac's delta - the wikipedia page has all the information necessary to get started.


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