Discrete Laws and Probability Mass Functions

Discrete Laws as Probability Mass Functions

A discrete random variable assigns its probability to individual values. A probability mass function records these point probabilities, making it possible to recover the probability of an event by adding the masses of the values it contains. The essential checks are that no mass is negative, the total mass is one, and the values receiving mass are the stated possible values.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \(X:\Omega\to\mathbb{R}\) be a random variable. Call \(X\) discrete if there is a finite or countable set \(D\subseteq\mathbb{R}\) such that \(\mathbb{P}(X\in D)=1\). This definition allows a null set of outcomes on which \(X\) takes values outside \(D\). For each real number \(x\), define the probability mass function (pmf) by \(p_X(x)=\mathbb{P}(X=x)\).

The sum over a countable set means the sum of its nonnegative terms, independent of the order of enumeration. Probabilities of events are obtained by summing point masses, with a small measurability qualification: for a Borel set \(A\subseteq\mathbb{R}\), the events \(\{X\in A\}\) and \(\bigcup_{x\in A\cap D}\{X=x\}\) agree except possibly on \(\{X\notin D\}\). The latter event has probability zero, so the two events have the same probability. In particular,

If instead \(X\in D\) surely, the two events are equal as sets for every \(A\); probabilities are asserted only when the events in question are measurable. Using Borel sets avoids asking for the measurability of \(\{X\in A\}\) when \(A\) is an arbitrary non-Borel set.

Why the pmf conditions hold. The events \(\{X=x\}\), for distinct \(x\in D\), are disjoint. Their union is \(\{X\in D\}\), so countable additivity gives \(\sum_{x\in D}p_X(x)=\mathbb{P}(X\in D)=1\). Also, each mass is a probability and hence is nonnegative. If \(x\notin D\), then \(\{X=x\}\subseteq\{X\notin D\}\), so \(p_X(x)=0\). Therefore the sum over \(D_+\) is also one.

Proof. For the first statement, the events \(\{X=x\}\) for \(x\in A\cap D\) are pairwise disjoint and measurable. Their union differs from \(\{X\in A\}\) only on outcomes with \(X\notin D\), a null event. Countable additivity therefore gives the stated sum. For the converse, let \(D\) be finite or countable and let \(p(x)\ge 0\) for \(x\in D\), with \(\sum_{x\in D}p(x)=1\). Take \(\Omega=D\), let \(\mathcal{F}\) be its power set, assign each singleton \(\{x\}\) probability \(p(x)\), and define \(X(x)=x\). For every \(A\subseteq D\), countable additivity gives \(\mathbb{P}(X\in A)=\sum_{x\in A}p(x)\), so \(X\) has the required pmf. \(\square\)

This proposition gives a practical test. A proposed pmf must have nonnegative entries and total mass one. Once those conditions hold, its support identifies the values that actually have positive probability. For a requested event, include the masses of the points in the event and no others.

Three Basic Discrete Models

A binomial random variable counts successes in a fixed number of trials. In the usual trial model, the trials have the same success probability and are independent; here the formula below is the definition of the binomial pmf. The number of trials \(n\) is a positive integer and \(0\le p\le1\). For \(0<p<1\), the count \(X\) has possible values \(0,1,\ldots,n\), with mass \(\binom{n}{k}p^k(1-p)^{n-k}\) at \(k\). At the endpoints, the law is degenerate: \(p=0\) gives mass one at 0, and \(p=1\) gives mass one at \(n\).

Justification. For \(0<p<1\), apply the binomial theorem to the two numbers \(p\) and \(1-p\); their sum is one, so the displayed sum is \(1^n=1\). Each term is nonnegative because \(p\) and \(1-p\) are positive. At \(p=0\), the specified endpoint law has its single positive mass, of size one, at 0; at \(p=1\), it has its single positive mass, of size one, at \(n\). These cases verify normalization without relying on ambiguous expressions such as \(0^0\).

A geometric random variable, under the convention used here, counts the trial on which the first success occurs. Thus its possible values are the positive integers. If the success probability satisfies \(0<p\le1\), then

In this formula \(k\) means that there are exactly \(k-1\) failures followed by a success. The event \(X\ge k\), by contrast, means that the first \(k-1\) trials are all failures; it does not require a success on trial \(k\).

Proof. For \(0<p<1\), the finite geometric sum satisfies \(\sum_{r=0}^{m}q^r=(1-q^{m+1})/(1-q)\). Since \(0<q<1\), \(q^{m+1}\) tends to zero as \(m\) tends to infinity. Consequently, \(\sum_{r=0}^{\infty}q^r=1/(1-q)\). Multiplying by \(p\), and using \(1-q=p\), gives total mass \(p/p=1\). Each term \(q^{k-1}p\) is nonnegative. If \(p=1\), then the first trial succeeds with probability one, so the specified masses are one at \(k=1\) and zero thereafter; they also sum to one. \(\square\)

Reading Supports and Avoiding Tail Errors

Support is part of a pmf, not an optional afterthought. A formula may be valid only for particular values: a Bernoulli variable has possible values 0 and 1; a binomial count with \(n\) trials has possible values from 0 through \(n\); and the geometric waiting time in this convention takes positive integer values. When computing an event probability, sum over the values in the event that can carry mass. Values outside the support contribute zero. Also distinguish the positive-mass support from a larger set on which a formula is stated: for endpoint Bernoulli or binomial parameters, some nominal values have zero mass.

Common mistake: confusing an exact value with a tail. With a fair success probability \(p=\tfrac12\), the probability of exactly one failure followed by success is \(\mathbb{P}(X=2)=(\tfrac12)^1(\tfrac12)=\tfrac14\). It is incorrect to use this as the probability of at least one failure before eventual success. That latter event includes two failures then success, three failures then success, and every longer finite string of failures before success. Its probability is \(\sum_{k=2}^{\infty}(\tfrac12)^{k-1}(\tfrac12)=\sum_{r=1}^{\infty}(\tfrac12)^{r+1}=\tfrac12\), not \(\tfrac14\). Equivalently, the event is that the first trial fails, which has probability \(\tfrac12\). The expression \(q^{k-1}p\) is the probability of exactly \(k-1\) failures followed by success, namely \(X=k\); a tail probability requires summing all relevant masses.

The same discipline applies to any discrete law: first state the values the variable can take, then translate the event into those values, and finally add the corresponding masses. Checking that the full list sums to one—and, where useful, checking an event against its complement—catches many errors before they propagate into later calculations.