On the definition - Probability distribution over a set

Question

I am watching some video lectures in Crypto, where there is a crash course in Discrete Probabilities.

The instructor, even before defining what a random variable is, he introduces the following definition:

0. Def: A probability distribution over a finite set $U$ is a function $P:U\longrightarrow [0,1]$ such that $\sum_{x\in U}P(x)=1$.

As an example, he explains that the function $P(x)=\frac{1}{|U|}$ for any $x\in U$ is a probability distribution and is called Uniform Distribution.

As far as I know, to define distributions on a discrete samplespace, we think as follows:

1. A random variable $X:U\longrightarrow \Bbb R$ is called discrete random variable, if it takes values in an at most countable set $\{x_1,x_2,\dots\}\subset \Bbb R$ with $\mathbb P(X\in\{x_1,x_2,\dots\})=1$.

2. Now, if $U$ is our samplespace and $X:U\longrightarrow \Bbb R$ is a discrete random variable, the probability mass function of $X$ is the real function $$f:\Bbb R\longrightarrow [0,1],\quad x\longmapsto f(x):=\mathbb P(X=x).$$

and there is a theorem which is often used as an equivalent definition:

3. The function $f:\Bbb R\longrightarrow [0,1]$ is a pmf of a random variable $X:U\longrightarrow \Bbb R$ if and only if $f(x)\geq 0$ and $\sum_{x\in X(U)}f(x)=1$.

So the author talks about probability distribution.

Does he mean the probability in 1.? Does he implicitly mean in 0. that we have a random variable $X:U\longrightarrow \Bbb R$ and $P$ is in fact the probability distribution?

I am absolutely confused. Any help please?

Answer 1

The distribution associated to a random variable is a special case of a more general notion of probability distribution, which in your course is first introduced on any finite space. You will see gradually more and more general definitions of a probability distribution.

There is nothing unusual about that, you can read more about that there.

Answer 2

So, the definition of a probability distribution as given is just a more general definition of the probability mass function.

Namely, suppose that $P:U\longrightarrow[0,1]$ is a probability distribution and let $X:U\longrightarrow \Bbb R$ be a discrete random variable, where $U=\{\omega_1,\dots,\omega_n\}$ is a finite set. Then, $$X(U)=R_U=\{X(\omega_1),\dots,X(\omega_n)\}\subset \Bbb R.$$ Say $X(\omega_1)=x_1,\dots,X(\omega_n)=x_n$. Then, \begin{gather*} P(\omega_1)+\dotsb+P(\omega_n)=1 \iff \\ P(\{\omega\in U:X(\omega)=x_1\})+\dotsb+P(\{\omega\in U:X(\omega)=x_n\})=1 \iff \\ P(X=x_1)+\dotsb+P(X=x_n)=1 \iff \\ p_X(x_1)+\dotsb+p_X(x_n)=1 \end{gather*} where $p_X:\Bbb R\longrightarrow [0,1]$ is the probability mass function of $X$.

So, from a probability distribution we can construct using a discrete random variable the probability mass function and vice versa. Is that correct?