A calculation with a finite sample space can look harmless until the sample space stops being finite. If a coin is tossed until the first head appears, the possible values are $1,2,3,\dots$, and no finite table contains the whole distribution. Yet the random variable is still described by a list of point probabilities, and the total probability is still obtained by adding them. Discrete random variables are the class for which this pointwise bookkeeping remains exact, even when the list is infinite.

The central issue is how a measurable map $X: (\Omega, \mathcal F) \to (\mathbb R, \mathcal B(\mathbb R))$ can be studied through the probabilities of its individual values. For a general real-valued random variable, singletons may carry no probability at all. A continuous uniform random variable on $(0,1)$ satisfies $\mathbb P(X=x)=0$ for every $x$, so the values $\mathbb P(X=x)$ do not reconstruct its law. In the discrete case, the opposite happens: the law is concentrated on a [countable set](/page/Countable%20Set), and the masses at the atoms contain all information about probabilities, expectations, and generating functions.

[example: Waiting Time for the First Head] Let a fair coin be tossed independently until the first head appears, and let $X$ be the toss number on which the first head occurs. For $k \in \mathbb N$, the event $\{X=k\}$ is the event that the first $k-1$ tosses are tails and the $k$-th toss is heads. By independence of the tosses and fairness of the coin, \begin{align*} \mathbb P(X=k)=\left(\frac{1}{2}\right)^{k-1}\frac{1}{2}=\left(\frac{1}{2}\right)^k. \end{align*} Thus the possible values of $X$ are $1,2,3,\dots$, so the range is countable but not finite. To check that the listed point probabilities carry total mass $1$, compute the partial sums: \begin{align*} \sum_{k=1}^{n}\mathbb P(X=k)=\sum_{k=1}^{n}\left(\frac{1}{2}\right)^k=\frac{1}{2}\frac{1-\left(\frac{1}{2}\right)^n}{1-\frac{1}{2}}=1-\left(\frac{1}{2}\right)^n. \end{align*} Since $\left(\frac{1}{2}\right)^n \to 0$ as $n\to\infty$, it follows that \begin{align*} \sum_{k=1}^{\infty}\mathbb P(X=k)=\lim_{n\to\infty}\left(1-\left(\frac{1}{2}\right)^n\right)=1. \end{align*} This example shows why discreteness should not mean finite-valued: what matters is that the possible values can be listed and the probabilities assigned to those listed points exhaust the total mass. [/example]

The example also shows the main advantage of the discrete theory. Events involving $X$ can be reduced to sums over values: the event $\{X \text{ is even}\}$ has probability

\begin{align*} \mathbb P(X \text{ is even}) = \sum_{j=1}^{\infty} \mathbb P(X=2j) = \sum_{j=1}^{\infty} \left(\frac{1}{2}\right)^{2j} = \frac{1}{3}. \end{align*}

Instead of integrating over a continuum, we add over a countable set.

## Definition

The parent concept is a [Random Variable](/page/Random%20Variable): a measurable map from a [probability space](/page/Probability%20Space) into a measurable state space. The discrete version answers a more specific question: when is the random variable's law supported on a list of possible real values? This is the exact setting in which probability mass functions replace densities, sums replace integrals, and distributional questions become questions about weighted countable sets.

[definition: Discrete Random Variable] Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. A real-valued random variable $X: (\Omega, \mathcal F) \to (\mathbb R, \mathcal B(\mathbb R))$ is a discrete random variable if there exists a countable set $S \subseteq \mathbb R$ such that \begin{align*} \mathbb P(X \in S) = 1. \end{align*} [/definition]

The set $S$ is not part of the data of $X$; it is a countable support on which the law is concentrated. It may contain values that occur with probability $0$, so computations need a sharper object than an arbitrary countable carrier. To calculate probabilities, we need a function that records how much probability is placed at each possible value.

[definition: Probability Mass Function] Let $X$ be a discrete random variable. The probability mass function of $X$ is the function \begin{align*} p_X: \mathbb R \to [0,1] \end{align*} defined by \begin{align*} p_X(x)=\mathbb P(X=x). \end{align*} [/definition]

Although $p_X$ is defined on all of $\mathbb R$, it is nonzero at at most countably many points. A table of values is often easier to read if the zero-probability points have been removed. The next definition names the exact set of atoms that carry positive mass.

[definition: Support of a Discrete Random Variable] Let $X$ be a discrete random variable with probability mass function $p_X$. The support of $X$ is the set \begin{align*} \operatorname{supp}(X) = \{x \in \mathbb R : p_X(x) > 0\}. \end{align*} [/definition]

To ask probabilities of intervals, inverse images, or Borel events, a list of atoms must be interpreted as a probability measure on the real line. This is the same object attached to every real-valued random variable, but the discrete case lets us later reconstruct it from point masses. We therefore name the induced measure before proving the reconstruction formula.

[definition: Law of a Discrete Random Variable] Let $X$ be a discrete random variable. The law of $X$ is the probability measure $\mu_X$ on $(\mathbb R, \mathcal B(\mathbb R))$ defined by \begin{align*} \mu_X(A) = \mathbb P(X \in A) \end{align*} for every $A \in \mathcal B(\mathbb R)$. [/definition]

For discrete $X$, the law is concentrated on $\operatorname{supp}(X)$, but the law still assigns probabilities to all Borel sets. The point-mass table is useful only if it determines those probabilities without ambiguity. The following result gives exactly that reconstruction principle: on a countable state space, the probability of a set is obtained by adding the masses of the points it contains.

[quotetheorem:9348]

This theorem is the reason discrete random variables can be treated by summation. It also separates two questions that are often conflated: whether a random variable is discrete, and whether a proposed list of numbers really is a probability distribution. It is a statement about countable additivity of point masses, not about distribution functions or jump sizes.

A compact example helps fix the difference between the random variable, its support, and its mass function.

[example: A Non-Injective Discrete Random Variable] Let $\Omega=\{1,2,3,4\}$, let $\mathcal F$ be the power set of $\Omega$, and suppose each outcome has probability $1/4$. Define $X:\Omega\to\mathbb R$ by $X(1)=0$, $X(2)=0$, $X(3)=1$, and $X(4)=3$. The range of $X$ is $\{0,1,3\}$, which is finite and therefore countable, so $X$ is a discrete random variable. The mass at $0$ is the probability of the inverse image of $\{0\}$: \begin{align*} \{X=0\}=\{\omega\in\Omega:X(\omega)=0\}=\{1,2\}. \end{align*} Since the two outcomes $1$ and $2$ are disjoint and each has probability $1/4$, \begin{align*} p_X(0)=\mathbb P(X=0)=\mathbb P(\{1,2\})=\mathbb P(\{1\})+\mathbb P(\{2\})=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}. \end{align*} Similarly, \begin{align*} \{X=1\}=\{3\}, \end{align*} so \begin{align*} p_X(1)=\mathbb P(X=1)=\mathbb P(\{3\})=\frac{1}{4}. \end{align*} Also, \begin{align*} \{X=3\}=\{4\}, \end{align*} so \begin{align*} p_X(3)=\mathbb P(X=3)=\mathbb P(\{4\})=\frac{1}{4}. \end{align*} If $x\notin\{0,1,3\}$, then no outcome is mapped to $x$, hence \begin{align*} \{X=x\}=\varnothing \end{align*} and therefore \begin{align*} p_X(x)=\mathbb P(X=x)=\mathbb P(\varnothing)=0. \end{align*} Thus $\operatorname{supp}(X)=\{0,1,3\}$. The value $0$ has larger mass than $1$ or $3$ because two distinct sample outcomes are merged into the same value of the random variable. [/example]