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I consider a standard probability space $(\Omega, \mathcal{F}, \mathbb{P})$.

In many sources, I found the definition which states that $X$ is a discrete random variable if:

  1. There exists a finite set $A\subset\mathbb{R}$ such that $P(A)=1$.

It is very intuitive for me.

However, in one source I found an additional condition, i.e.

  1. A probability measure $\mathbb{P}$ and a Lebesgue measure $\lambda$ are singular, which means $$\exists A\subset\mathbb{R} \quad \mathbb{P}(A) = 0 \quad\text{and}\quad \lambda(\Omega\backslash{A}) = 0$$

For me, the first condition (1.) implies the second one (2.) which is easy to prove.

I have two concerns:

  • Is condition 2. necessary when I want to define a discrete random variable?

  • I did not specify the set $\Omega$ in the above considerations, but regardless of whether it will be a finite or infinite set, the second condition will hold assuming that the first condition is satisfied.

The reason I posted this question was that I want to deeply understand the types of possible random variables and their formal definitions. For now, I know that there are 3 types: continuous, discrete, and singular random variables, but I cannot find a good source that explicitly states the conditions.

MMM
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1 Answers1

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The correct definition of a discrete random variable is there is a 'finite or countably infinite' set $A$ such that $P(A)=1$ where $P(A)=Pr (X \in A)$.

If $X$ is discrete then there is an induced measure $Q$ in the Borel sets of the real line defined by $Q(E)=Pr(X^{-1}(E))$. This measure is singular w.r.t. Lebesgue measure: $\lambda$: $Q(B)=0$ and $\lambda (\mathbb R \setminus B)=0$ are both satisfied when $B$ is the complement of the set $A$ in the definition of discrete measure.

The convesre is not true. Singularity of $Q$ does not guarantee that $X$ is discrete. $F_X$ can be a continuous function, at the same time being singular.

Kavi Rama Murthy
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