Discrete Random Variables and PMFs

How do you assign a number to every possible outcome of a coin toss? Then how do you measure its likelihood? A discrete random variable answers the first question. A probability mass function (PMF) answers the second, precisely. This pairing turns vague talk of "chance" into arithmetic. It involves sums over a countable set of values. These values must add to exactly one. It is the single most-used object in elementary probability. It underlies the Bernoulli and Binomial Distributions, the Poisson and geometric families. It also underlies every discrete simulation ever coded. The idea matters because it converts probability questions into counting and summation problems. Once a PMF is written down, every event about the random variable becomes simple. Examples include "at least 3 successes," "an even number," or "value in some set A." Each becomes a sum of PMF values. This is the bridge between Combinatorial Probability (counting outcomes) and analytic tools. Those tools include Expectation and generating functions (manipulating sums). The big picture: a random variable is a function from a sample space to the real numbers. It is built on top of Random Variables as an abstract concept. When the range of that function is finite or countably infinite, the variable is discrete. Its entire probabilistic behavior is captured by one function of one real variable — the PMF. This PMF is subject to two non-negotiable axioms: nonnegativity and total mass one. This article gives the rigorous definition of a discrete random variable and its PMF. It proves the fundamental normalization theorem from the axioms of probability. It works through two nontrivial asymmetric examples. It shows exactly what fails if the axioms are relaxed. It closes the loop toward Cumulative Distribution Functions and expectation. These are built directly on the PMF machinery developed here.

Key takeaways
  • A discrete random variable XX takes values in a finite or countably infinite set, called its support.
  • The PMF pX(x)=P(X=x)p_X(x) = P(X=x) assigns a probability to each possible value.
  • Two axioms characterize any valid PMF: pX(x)≥0p_X(x) \geq 0 for all xx, and ∑xpX(x)=1\sum_x p_X(x) = 1.
  • Any event about XX has probability equal to a sum of PMF values: P(X∈A)=∑x∈ApX(x)P(X \in A) = \sum_{x \in A} p_X(x).
  • The normalization condition is a theorem derived from Kolmogorov's axioms, not an independent assumption.
  • Key idea: a discrete random variable's entire probabilistic content is a countable list of (value, probability) pairs summing to 1.

Where this lives today

Discrete PMFs are the computational backbone of applied probability and its neighboring fields. In statistics, maximum likelihood estimation for discrete models (Poisson regression, logistic regression's Bernoulli likelihood, multinomial models for categorical data) is built entirely on products and sums of PMF values. In computer science, randomized algorithms and probabilistic analysis of hashing, load balancing, and random graphs rely on discrete PMFs — for instance, the number of collisions in a hash table follows a PMF derivable from combinatorial counting. In information theory, the entropy of a discrete source, H(X)=−∑xpX(x)log⁡pX(x)H(X) = -\sum_x p_X(x)\log p_X(x), is defined directly from the PMF, and Shannon's source coding theorem is a statement about how well a PMF can be compressed. In cryptography, the security of certain schemes (e.g., analyzing guessing attacks) depends on bounding the maximum value of a PMF (the "min-entropy"). In physics, discrete PMFs describe quantum measurement outcomes over discrete eigenvalues (Born rule probabilities sum to 1 by the same normalization logic) and particle counting statistics. In engineering, reliability analysis models the number of failures in a batch using discrete PMFs like the binomial or hypergeometric, both of which will be built from this same normalization principle in the treatment of The Bernoulli and Binomial Distributions.

Ask the tutor: How do I verify normalization for a PMF whose formula involves a parameter, like p(k)=c rkp(k) = c\,r^k, and solve for the correct constant cc?

DefinitionDiscrete Random Variable and Probability Mass Function

Let (Ω,F,P)(\Omega, \mathcal{F}, P) be a probability space. A random variable X:Ω→RX: \Omega \to \mathbb{R} is discrete if its range

RX=X(Ω)={X(ω):ω∈Ω}R_X = X(\Omega) = \{X(\omega) : \omega \in \Omega\}

is a finite or countably infinite subset of R\mathbb{R}. We call RXR_X (or any countable superset containing all values of positive probability) the support of XX, sometimes denoted supp⁡(X)\operatorname{supp}(X). For a discrete random variable XX with range RX={x1,x2,x3,… }R_X = \{x_1, x_2, x_3, \dots\}, the probability mass function (PMF) of XX is the function pX:R→[0,1]p_X : \mathbb{R} \to [0,1] defined by

pX(x)=P(X=x)=P({ω∈Ω:X(ω)=x}).p_X(x) = P(X = x) = P(\{\omega \in \Omega : X(\omega) = x\}).

By convention pX(x)=0p_X(x) = 0 for any x∉RXx \notin R_X. A function p:R→Rp : \mathbb{R} \to \mathbb{R} is called a valid PMF if there is a countable set R⊂RR \subset \mathbb{R} such that: 1. (Nonnegativity) p(x)≥0p(x) \geq 0 for all x∈Rx \in R, and p(x)=0p(x) = 0 for x∉Rx \notin R; 2. (Normalization) ∑x∈Rp(x)=1\displaystyle\sum_{x \in R} p(x) = 1.

Intuition

Think of a discrete random variable as a labeled spinner with a countable — possibly infinite — number of stopping positions, each engraved with a real number and a weight. The PMF is the list of weights. Since the spinner must stop somewhere, the weights must add up to the whole of probability, namely 1. Since a weight represents a chance of an outcome, no weight can be negative — you cannot have "negative confidence" that the spinner lands on 4. This picture also explains why events are just sums: asking "what is P(X≤3)P(X \le 3)?" is asking "how much total weight sits on stopping positions at or below 3?" You simply add up the relevant engraved weights. This additive structure is what makes discrete probability computable by hand or by computer: it reduces to enumerating a countable set and summing a series (finite series in the simplest cases). The figure below shows the PMF of a specific, asymmetric example: the number of heads in three tosses of a biased coin with P(heads)=0.3P(\text{heads}) = 0.3 (this anticipates the Bernoulli and Binomial Distributions, where this shape is derived in general). Note the bars are of unequal height — deliberately asymmetric — and their heights sum to exactly 1.

Figure
PMF of X = number of heads in 3 tosses of a coin with p = 0.3
x (number of heads) p_X(x) 0 0.25 0.5 0 1 2 3 0.343 0.441 0.189 0.027
TheoremNormalization of the Probability Mass Function

Let (Ω,F,P)(\Omega, \mathcal{F}, P) be a probability space and let X:Ω→RX : \Omega \to \mathbb{R} be a discrete random variable with range RX={x1,x2,x3,… }R_X = \{x_1, x_2, x_3, \dots\} (finite or countably infinite), and let pXp_X be its PMF. Then: 1. pX(xi)≥0p_X(x_i) \geq 0 for every ii; 2. ∑ipX(xi)=1\displaystyle\sum_{i} p_X(x_i) = 1; 3. For every event A⊆RA \subseteq \mathbb{R} (more precisely, every AA such that X−1(A)∈FX^{-1}(A) \in \mathcal{F}),

P(X∈A)=∑xi∈ApX(xi).P(X \in A) = \sum_{x_i \in A} p_X(x_i).
Proof

We use the axioms of a probability space: P(Ω)=1P(\Omega)=1, P(E)≥0P(E) \geq 0 for all E∈FE \in \mathcal{F}, and countable additivity: for pairwise disjoint events E1,E2,⋯∈FE_1, E_2, \dots \in \mathcal{F},

P(⋃kEk)=∑kP(Ek).P\left(\bigcup_{k} E_k\right) = \sum_k P(E_k).

We assume as given (from Random Variables) that XX is measurable, so that for every x∈Rx \in \mathbb{R} the set {X=x}:={ω∈Ω:X(ω)=x}\{X = x\} := \{\omega \in \Omega : X(\omega) = x\} lies in F\mathcal{F}, and hence P(X=x)P(X=x) is well defined.

Step 1 (Nonnegativity). For each ii, pX(xi)=P({X=xi})p_X(x_i) = P(\{X = x_i\}), and {X=xi}∈F\{X=x_i\} \in \mathcal{F}. By the nonnegativity axiom of a probability measure, P({X=xi})≥0P(\{X=x_i\}) \geq 0. Hence pX(xi)≥0p_X(x_i) \geq 0.

Step 2 (Disjointness of the level sets). Define Ei={X=xi}={ω∈Ω:X(ω)=xi}E_i = \{X = x_i\} = \{\omega \in \Omega : X(\omega) = x_i\} for each xi∈RXx_i \in R_X. If i≠ji \neq j then xi≠xjx_i \neq x_j, so no ω\omega can satisfy X(ω)=xiX(\omega) = x_i and X(ω)=xjX(\omega) = x_j simultaneously, since X(ω)X(\omega) is a single real number. Hence Ei∩Ej=∅E_i \cap E_j = \emptyset for i≠ji \neq j: the events EiE_i are pairwise disjoint.

Step 3 (Covering Ω\Omega). By definition of the range, every ω∈Ω\omega \in \Omega satisfies X(ω)∈RXX(\omega) \in R_X, i.e. X(ω)=xiX(\omega) = x_i for some ii. Hence

Ω=⋃iEi,\Omega = \bigcup_i E_i,

a countable union (finite or countably infinite, matching RXR_X) of pairwise disjoint sets.

Step 4 (Countable additivity applied). Apply countable additivity of PP to the disjoint union from Step 3:

1=P(Ω)=P(⋃iEi)=∑iP(Ei)=∑ipX(xi).1 = P(\Omega) = P\left(\bigcup_i E_i\right) = \sum_i P(E_i) = \sum_i p_X(x_i).

This proves part 2. (If RXR_X is finite, countable additivity restricted to a finite disjoint union follows from finite additivity, itself a consequence of countable additivity applied to a sequence padded with ∅\emptyset's; either way the sum equals 1.)

Step 5 (General events). Let A⊆RA \subseteq \mathbb{R} be such that X−1(A)∈FX^{-1}(A) \in \mathcal{F}. Write A∩RX={xi:i∈I}A \cap R_X = \{x_{i} : i \in I\} for some index set I⊆NI \subseteq \mathbb{N}. Then

X−1(A)={ω∈Ω:X(ω)∈A}=⋃i∈IEi,X^{-1}(A) = \{\omega \in \Omega : X(\omega) \in A\} = \bigcup_{i \in I} E_i,

because X(ω)∈AX(\omega) \in A if and only if X(ω)∈A∩RXX(\omega) \in A \cap R_X (as X(ω)X(\omega) is always in RXR_X), which happens if and only if ω∈Ei\omega \in E_i for some i∈Ii \in I. The sets {Ei}i∈I\{E_i\}_{i \in I} are pairwise disjoint, being a subfamily of the disjoint family from Step 2. Applying countable additivity again,

P(X∈A)=P(⋃i∈IEi)=∑i∈IP(Ei)=∑i∈IpX(xi)=∑x∈ApX(x).P(X \in A) = P\left(\bigcup_{i\in I} E_i\right) = \sum_{i \in I} P(E_i) = \sum_{i \in I} p_X(x_i) = \sum_{x \in A} p_X(x).

This is exactly part 3, and it specializes to part 2 when A=RA = \mathbb{R} (or any set containing RXR_X), since then P(X∈A)=P(Ω)=1P(X \in A) = P(\Omega) = 1. ■\blacksquare

∎

Proof technique used: decomposition of the sample space into a countable disjoint partition indexed by the range of XX, followed by direct application of countable additivity — the defining axiom of a probability measure. No result depending on PMFs is used in the proof, so it is non-circular: this argument constructs normalization from Kolmogorov's axioms, rather than assuming it.

CorollaryConverse: PMFs generate probability spaces

Given any valid PMF pp (nonnegative, summing to 1 over a countable set R={x1,x2,… }R = \{x_1, x_2, \dots\}), there exists a probability space (Ω,F,P)(\Omega, \mathcal{F}, P) and a discrete random variable XX on it whose PMF is exactly pp.

Proof

Take Ω=R\Omega = R, F=2R\mathcal{F} = 2^{R} (the power set, which is a valid σ\sigma-algebra since RR is countable), and define P(E)=∑x∈Ep(x)P(E) = \sum_{x \in E} p(x) for E⊆RE \subseteq R. Countable additivity of PP holds because rearrangement of nonnegative series preserves sums (Tonelli's theorem for series, or simply: for disjoint E1,E2,⋯⊆RE_1, E_2, \dots \subseteq R, ∑x1x∈⋃kEkp(x)=∑k∑x∈Ekp(x)\sum_x \mathbf 1_{x \in \bigcup_k E_k} p(x) = \sum_k \sum_{x \in E_k} p(x) by absolute convergence of nonnegative terms). We have P(Ω)=∑x∈Rp(x)=1P(\Omega) = \sum_{x \in R} p(x) = 1 by hypothesis, and P(E)≥0P(E) \geq 0 since p≥0p \geq 0. Define X:Ω→RX : \Omega \to \mathbb{R} by X(ω)=ωX(\omega) = \omega (the identity map). Then RX=RR_X = R and for x∈Rx \in R,

pX(x)=P(X=x)=P({x})=p(x).p_X(x) = P(X = x) = P(\{x\}) = p(x).

So XX is a discrete random variable on (Ω,F,P)(\Omega,\mathcal F,P) with PMF equal to pp. ■\blacksquare

∎

This corollary explains why the two-axiom characterization in the Definition is not merely necessary but sufficient: any function satisfying nonnegativity and normalization is realizable as an actual PMF of some random variable, so these two properties are the complete and exact defining conditions.

Worked examples

Example 1: An asymmetric loaded die. A die is weighted so that the probability of face kk is proportional to kk, for k=1,…,6k = 1, \dots, 6. Let XX be the face shown. Find pXp_X, verify normalization, and compute P(X is even)P(X \text{ is even}) and P(2≤X≤4)P(2 \le X \le 4). Solution. We are told pX(k)=ckp_X(k) = ck for some constant c>0c > 0 and k∈{1,…,6}k \in \{1,\dots,6\}. By the normalization requirement of the Definition,

∑k=16pX(k)=c(1+2+3+4+5+6)=21c=1  ⟹  c=121.\sum_{k=1}^{6} p_X(k) = c(1+2+3+4+5+6) = 21c = 1 \implies c = \frac{1}{21}.

So

pX(k)=k21,k=1,2,…,6,p_X(k) = \frac{k}{21}, \qquad k = 1,2,\dots,6,

and pX(x)=0p_X(x) = 0 for all other xx. Check nonnegativity: each k/21>0k/21 > 0 for k=1,…,6k=1,\dots,6, satisfying part 1 of the theorem. Check normalization directly:

∑k=16k21=1+2+3+4+5+621=2121=1.✓\sum_{k=1}^6 \frac{k}{21} = \frac{1+2+3+4+5+6}{21} = \frac{21}{21} = 1. \checkmark

Now use part 3 of the theorem for events. Let A={2,4,6}A = \{2,4,6\} (even faces):

P(X∈A)=∑k∈{2,4,6}k21=2+4+621=1221=47.P(X \in A) = \sum_{k \in \{2,4,6\}} \frac{k}{21} = \frac{2+4+6}{21} = \frac{12}{21} = \frac{4}{7}.

Let B={2,3,4}B = \{2,3,4\}:

P(2≤X≤4)=∑k=24k21=2+3+421=921=37.P(2 \le X \le 4) = \sum_{k=2}^{4} \frac{k}{21} = \frac{2+3+4}{21} = \frac{9}{21} = \frac{3}{7}.

Both computed purely as sums of PMF values over subsets of the support, exactly as guaranteed by the theorem.

Example 2: A countably infinite support — a shifted geometric-type PMF. Let XX denote the number of failed attempts before the first success in a sequence of independent trials, each succeeding with probability θ=0.4\theta = 0.4 (so failure probability 1−θ=0.61-\theta = 0.6). This is a foreshadowing of the geometric distribution, but let us verify everything from scratch using only the theorem above, since the range here is countably infinite: RX={0,1,2,3,… }R_X = \{0, 1, 2, 3, \dots\}. The event {X=k}\{X = k\} means: the first kk trials fail, then trial k+1k+1 succeeds. By independence of trials,

pX(k)=(1−θ)kθ=(0.6)k(0.4),k=0,1,2,…p_X(k) = (1-\theta)^k \theta = (0.6)^k (0.4), \qquad k = 0, 1, 2, \dots

Nonnegativity: each term is a product of positive numbers, so pX(k)>0p_X(k) > 0 for all k≥0k \ge 0. Normalization: we must sum a geometric series:

∑k=0∞pX(k)=0.4∑k=0∞(0.6)k=0.4⋅11−0.6=0.4⋅10.4=1.\sum_{k=0}^{\infty} p_X(k) = 0.4 \sum_{k=0}^{\infty} (0.6)^k = 0.4 \cdot \frac{1}{1-0.6} = 0.4 \cdot \frac{1}{0.4} = 1.

This matches part 2 of the theorem exactly — note that here the sum is a genuine infinite series, converging by the ratio test since 0.6<10.6 < 1, and its convergence to 1 is not automatic; it is a real algebraic fact that must be, and has been, checked. Now compute P(X≥3)P(X \geq 3), the probability of at least 3 failures before the first success. Using part 3 of the theorem with A={3,4,5,… }A = \{3, 4, 5, \dots\}:

P(X≥3)=∑k=3∞(0.6)k(0.4)=0.4⋅(0.6)31−0.6=0.4⋅0.2160.4=0.216.P(X \ge 3) = \sum_{k=3}^{\infty} (0.6)^k (0.4) = 0.4 \cdot \frac{(0.6)^3}{1 - 0.6} = 0.4 \cdot \frac{0.216}{0.4} = 0.216.

Alternatively, via the complement using finite sums (a preview of the CDF machinery):

P(X≥3)=1−∑k=02pX(k)=1−[0.4+0.24+0.144]=1−0.784=0.216,P(X \ge 3) = 1 - \sum_{k=0}^{2} p_X(k) = 1 - \left[0.4 + 0.24 + 0.144\right] = 1 - 0.784 = 0.216,

matching, as it must, since both computations invoke the same theorem — total probability sums to 1, so the complement rule follows immediately from parts 2 and 3.

CounterexampleWhy normalization cannot be dropped

Consider dropping condition 2 (normalization) from the definition of a valid PMF, while keeping nonnegativity. Take the countable set R={1,2,3,… }R = \{1, 2, 3, \dots\} and define

q(k)=12kfor k=1,2,3,…q(k) = \frac{1}{2^{k}} \quad \text{for } k = 1, 2, 3, \dots

but suppose someone claims this is "the PMF" without checking the sum. In fact

∑k=1∞q(k)=∑k=1∞12k=1,\sum_{k=1}^{\infty} q(k) = \sum_{k=1}^\infty \frac{1}{2^k} = 1,

so this particular qq does happen to normalize — it is a legitimate PMF (in fact the PMF of a Geometric(1/2) random variable shifted to start at 1). The point of the counterexample is to see what happens if we perturb it and forget to renormalize. Suppose instead someone proposes

q~(k)=1k2,k=1,2,3,…\tilde q(k) = \frac{1}{k^2}, \qquad k = 1, 2, 3, \dots

This satisfies nonnegativity: q~(k)>0\tilde q(k) > 0 for every kk. But

∑k=1∞q~(k)=∑k=1∞1k2=π26≈1.6449≠1.\sum_{k=1}^\infty \tilde q(k) = \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6} \approx 1.6449 \neq 1.

If we nonetheless declared q~\tilde q to be a PMF and tried to use the Theorem's conclusion (3), we would get nonsense: for instance "P(X=1)=1P(X=1) = 1" combined with "P(X=2)=1/4P(X = 2) = 1/4" would force

P(X∈{1,2})=1.25>1,P(X \in \{1,2\}) = 1.25 > 1,

directly violating the axiom P(E)≤1P(E) \le 1 for any event EE. The entire machinery — disjoint decomposition summing via countable additivity to P(Ω)=1P(\Omega)=1 — collapses, because there is no probability space or random variable for which q~\tilde q is genuinely P(X=k)P(X=k): by the Corollary's converse construction, setting P(E)=∑x∈Eq~(x)P(E) = \sum_{x\in E}\tilde q(x) gives P(Ω)=π2/6≠1P(\Omega) = \pi^2/6 \ne 1, so PP is not even a probability measure. This shows normalization is not a cosmetic convenience; without it, "pp" fails to define any probability measure at all, and every derived probability statement becomes meaningless. To repair q~\tilde q, one must rescale: p(k)=q~(k)/(π2/6)=6π2k2p(k) = \tilde q(k) \big/ (\pi^2/6) = \dfrac{6}{\pi^2 k^2}, which does sum to 1 and is a genuine PMF (this is the discrete "zeta" or Zipf-type distribution).

Scope, edge cases & common traps

SituationWhat to checkCommon error
Finite supportSum is a finite algebraic sumForgetting a value in the support when summing
Countably infinite supportSum is a convergent infinite seriesAssuming convergence without checking the ratio/comparison test
Support with negative or non-integer valuesPMF still defined pointwise on R\mathbb{R}Assuming PMFs only live on {0,1,2,… }\{0,1,2,\dots\}
Events not subsets of supportP(X∈A)=P(X∈A∩RX)P(X \in A) = P(X \in A \cap R_X)Including points where pX=0p_X = 0 as if they contributed mass
Continuous random variablesPMF concept does not apply; use density insteadWriting "pX(x)p_X(x)" for a continuous XX (there, P(X=x)=0P(X=x)=0 for every xx)

The mathematician behind it

Origins · 1 minAndrei Nikolaevich Kolmogorov and how this was discovered
AN
Andrei Nikolaevich Kolmogorov
1903–1987 · Soviet Union, 20th century

Kolmogorov gave probability its modern axiomatic foundation in his 1933 monograph Grundbegriffe der Wahrscheinlichkeitsrechnung, building probability theory on measure theory. The normalization property of PMFs proved above — ∑xpX(x)=1\sum_x p_X(x)=1 — is a direct, elementary consequence of his axioms, particularly countable additivity, applied to the special case of a countable partition of the sample space induced by a discrete random variable. Kolmogorov's framework unified what had previously been scattered combinatorial and analytic techniques (from Bernoulli, Laplace, Poisson, and Chebyshev) into a single deductive system, of which discrete probability — the subject of this article — is the most concrete and computationally transparent instance.

History
- 1654 — Pascal and Fermat correspond on problems of chance, establishing early combinatorial probability calculations.
- 1713 — Jacob Bernoulli's Ars Conjectandi is published posthumously, formalizing what becomes the Bernoulli trial.
- 1837 — Poisson derives the limiting distribution for rare events, an early countably-infinite-support PMF.
- 1900 — Hilbert's sixth problem calls for the axiomatization of physics, including probability, as a mathematical theory.
- 1933 — Kolmogorov publishes the axiomatic foundations from which the normalization theorem above is rigorously derived.
What exactly counts as the "support" of a discrete random variable?

The support is the set of values xx for which pX(x)>0p_X(x) > 0; equivalently, it is (a countable superset of) the range RX=X(Ω)R_X = X(\Omega). Some authors define the support as literally the range, others as the closure of the set where the PMF is positive — for discrete variables these coincide since every point carrying positive mass is itself a value the variable can take. Values outside the support automatically have PMF equal to zero by convention, so summing "over the support" or "over all of R\mathbb{R}" gives the same total probability. Precisely specifying the support is essential: an incomplete support is the most common source of a PMF that seems not to normalize to 1.

Can a discrete random variable have uncountably many possible values?

No — this is excluded by definition. A random variable is called discrete precisely when its range is finite or countably infinite; if the range is uncountable (such as an interval of real numbers), the variable is not discrete, and the PMF framework does not apply directly. Continuous random variables, which typically take values in an interval, are instead described by a probability density function, for which P(X=x)=0P(X=x)=0 at every individual point and probabilities arise from integrating the density rather than summing a PMF. The distinction is fundamental and not merely technical: sums and integrals obey different convergence and manipulation rules.

Does every function satisfying nonnegativity and summing to 1 actually arise as the PMF of some random variable?

Yes, and this is exactly the content of the Corollary proved above. Given any countable set RR and any nonnegative function pp on RR summing to 1, one can explicitly construct a probability space — take Ω=R\Omega = R itself, all subsets as events, and P(E)=∑x∈Ep(x)P(E) = \sum_{x \in E} p(x) — on which the identity map X(ω)=ωX(\omega)=\omega is a discrete random variable with exactly that PMF. This converse result confirms that the two axioms (nonnegativity, normalization) are not just necessary consequences of the probability axioms but are jointly sufficient to characterize valid PMFs.

How does the PMF relate to the cumulative distribution function?

The CDF, FX(x)=P(X≤x)F_X(x) = P(X \le x), is obtained from the PMF by summing all mass at or below xx: FX(x)=∑xi≤xpX(xi)F_X(x) = \sum_{x_i \le x} p_X(x_i), a direct application of part 3 of the normalization theorem with A=(−∞,x]A = (-\infty, x]. Conversely, for a discrete random variable, the PMF can be recovered from the CDF by taking jump sizes: pX(xi)=FX(xi)−lim⁡x↑xiFX(x)p_X(x_i) = F_X(x_i) - \lim_{x \uparrow x_i} F_X(x). This equivalence is developed fully in Cumulative Distribution Functions, but the underlying summation principle is exactly the one established here.

Why must the PMF be defined on a countable set rather than an arbitrary set?

Countability is what makes the sum ∑xpX(x)\sum_x p_X(x) meaningful without invoking measure-theoretic integration. An uncountable sum of strictly positive terms is necessarily infinite (a classical fact: any uncountable set has, for some nn, uncountably many terms exceeding 1/n1/n, forcing divergence), so no PMF with uncountable support and all positive values could ever sum to a finite number like 1. This is precisely why discrete random variables are defined to have countable range — it is the exact threshold at which "summing probabilities" remains a well-posed operation.

How is expectation related to the PMF, and why is this article a prerequisite for it?

Expectation of a discrete random variable is defined as E[X]=∑xx pX(x)E[X] = \sum_x x\, p_X(x), a weighted average over the support using PMF values as weights. This definition presupposes exactly the machinery built here: a well-defined support, nonnegative weights, and (for the sum to even be a probability-weighted average) total weight equal to 1. Without the normalization theorem guaranteeing ∑xpX(x)=1\sum_x p_X(x) = 1, the formula for E[X]E[X] would not correspond to any meaningful "average," which is why Expectation is developed only after PMFs are rigorously established.

What goes wrong if I only check that 0≤p(x)≤10 \le p(x) \le 1 for each individual value?

This checks far too little: it is possible for every individual value to lie in [0,1][0,1] while the total sum is far from 1, as the Zipf-type counterexample q~(k)=1/k2\tilde q(k)=1/k^2 illustrates (it sums to π2/6≈1.645\pi^2/6 \approx 1.645, not 1). Individual bounds only reflect that no single outcome can have negative or more-than-certain probability; they say nothing about how the mass aggregates across the whole support. The full normalization condition — the sum over the entire countable support equals exactly 1 — is the property that actually certifies a function as a legitimate PMF, and it must be checked as a separate, global computation.

Before this

Where this leads