Uniqueness of a Distribution from Its CDF

Why a CDF Identifies a Distribution

A cumulative distribution function records the probability that a random variable lies at or below each real threshold. Knowing those probabilities at every threshold does more than answer questions about half-lines: it determines probabilities for intervals and, ultimately, for every Borel set. This is why the CDF is a complete description of a distribution on the real line.

The statement concerns probability laws, not whether two random variables are the same object or take the same value outcome by outcome. A law describes probabilities assigned to sets of real numbers. We will show that if two laws have the same CDF at every real number, they are the same probability measure.

Definition: For a real-valued random variable \(X\), its law (or distribution) is the probability measure \(\mu_X\) on the Borel subsets \(B\) of \(\mathbb{R}\) defined by \(\mu_X(B)=P(X\in B)\). Its cumulative distribution function is \(F_X(x)=\mu_X((-\infty,x])=P(X\le x)\), for \(x\in\mathbb{R}\).

The Borel sets are the sets in the smallest sigma-algebra containing all open subsets of \(\mathbb{R}\). In particular, intervals and half-lines are Borel sets. Two random variables, even if defined on different probability spaces, can therefore have laws that are compared as probability measures on the same collection of subsets of \(\mathbb{R}\).

From CDF Values to Interval Probabilities

Let \(\mu\) be a law with CDF \(F\). For \(a<b\), the half-open interval \((a,b]\) is the difference between the events \((-\infty,b]\) and \((-\infty,a]\). Subtracting their probabilities gives

\mu((a,b])=F(b)-F(a).

The other endpoint conventions are determined by the same CDF values together with the left limit \(F(t-)=\lim_{x\uparrow t}F(x)\). This left limit equals \(\mu((-\infty,t))\). Thus:

\begin{aligned} \mu((a,b])&=F(b)-F(a),\\ \mu([a,b])&=F(b)-F(a-),\\ \mu((a,b))&=F(b-)-F(a),\\ \mu([a,b))&=F(b-)-F(a-). \end{aligned}

For example, the closed-interval formula follows by taking \((-\infty,b]\) and removing \((-\infty,a)\). The formula also shows why endpoint inclusion matters: the probability at \(a\) is \(F(a)-F(a-)\), so including \(a\) can change the answer. These formulas immediately imply that two laws with the same CDF assign the same probability to every interval.

Worked Example: Recovering a Discrete Law from Its CDF

Suppose a law \(\mu\) assigns probability \(1/4\) to \(-2\), probability \(1/2\) to \(1\), and probability \(1/4\) to \(4\). Its CDF is

F(x)= \begin{cases} 0, & x<-2,\\ 1/4, & -2\le x<1,\\ 3/4, & 1\le x<4,\\ 1, & x\ge 4. \end{cases}

The jump at \(-2\) is \(F(-2)-F(-2-)=1/4-0=1/4\). At \(1\), the jump is \(F(1)-F(1-)=3/4-1/4=1/2\), and at \(4\), it is \(F(4)-F(4-)=1-3/4=1/4\). These are exactly the three point probabilities.

The CDF also gives interval probabilities. For example,

\mu((-2,4])=F(4)-F(-2)=1-\frac14=\frac34.

This interval excludes \(-2\) and includes \(4\), so its probability is \(1/2+1/4=3/4\), as calculated directly from the masses. Any other law with this same CDF must have the same three jumps and the same interval probability.

Worked Example: A CDF with an Atom and Continuous Mass

Consider a law that puts mass \(1/3\) at \(0\), and has density \(1/3\) on \([1,3]\), with no other mass. The density contributes \((3-1)(1/3)=2/3\), so the total probability is \(1/3+2/3=1\). Its CDF is

F(x)= \begin{cases} 0, & x<0,\\ 1/3, & 0\le x<1,\\ x/3, & 1\le x<3,\\ 1, & x\ge 3. \end{cases}

The atom at \(0\) is visible as the jump from \(F(0-)=0\) to \(F(0)=1/3\). On the other hand, \(F\) does not jump at \(1\): the density begins there, but a single point has probability zero. Using the CDF, the interval \((1,3]\) has probability

\mu((1,3])=F(3)-F(1)=1-\frac13=\frac23.

Including \(0\) changes a different interval probability. Since \(F(1)=1/3\) and \(F(0-)=0\), we have \(\mu([0,1])=F(1)-F(0-)=1/3\). This probability consists entirely of the atom at \(0\); the continuous part gives no mass to \([0,1]\).

The Uniqueness Theorem

Theorem: Let \(\mu\) and \(\nu\) be probability measures on the Borel subsets of \(\mathbb{R}\), with CDFs \(F_\mu(x)=\mu((-\infty,x])\) and \(F_\nu(x)=\nu((-\infty,x])\). If \(F_\mu(x)=F_\nu(x)\) for every \(x\in\mathbb{R}\), then \(\mu(B)=\nu(B)\) for every Borel set \(B\). In particular, the two measures assign the same probability to every interval.

The interval formulas establish the last claim for intervals. To establish equality on all Borel sets, we use the following measure-uniqueness argument. A collection \(\Pi\) is a pi-system if it is closed under finite intersections. A collection \(\mathcal{D}\) of subsets of a space \(\Omega\) is a lambda-system if it contains \(\Omega\), is closed under complements relative to \(\Omega\), and is closed under countable disjoint unions. The pi-lambda theorem says that a lambda-system containing a pi-system \(\Pi\) contains the sigma-algebra generated by \(\Pi\).

For completeness, the key step of that theorem is as follows. Let \(\mathcal{L}\) be the smallest lambda-system containing \(\Pi\). For fixed \(A\in\Pi\), the collection of \(B\in\mathcal{L}\) such that \(A\cap B\in\mathcal{L}\) is a lambda-system: it contains the whole space because \(A\in\mathcal{L}\), and its complement and disjoint-union properties follow from the corresponding properties of \(\mathcal{L}\). It contains \(\Pi\), since \(\Pi\) is closed under intersections. It therefore contains \(\mathcal{L}\). Now fix \(B\in\mathcal{L}\) and apply the same argument to the collection of \(A\in\mathcal{L}\) such that \(A\cap B\in\mathcal{L}\). This collection contains \(\Pi\) by the first step, so it contains \(\mathcal{L}\). Hence \(\mathcal{L}\) is closed under finite intersections. A lambda-system closed under finite intersections is a sigma-algebra: it is closed under complements and finite unions, and any countable union can be written as a countable disjoint union by removing the preceding sets. Thus \(\mathcal{L}\) is the sigma-algebra generated by \(\Pi\), which gives the stated theorem.

Choose a generating pi-system.
Take \(\Pi=\{(-\infty,x]:x\in\mathbb{R}\}\). The intersection of two such rays is the ray with the smaller endpoint, so \(\Pi\) is closed under finite intersections.

Check that it generates the Borel sets.
Each ray is Borel. Conversely, \((-\infty,b)=\bigcup_{n\ge1}(-\infty,b-1/n]\), so open intervals can be formed from rays and their complements. Every open set is a countable union of open intervals with rational endpoints. Therefore \(\Pi\) generates the Borel sigma-algebra.

Form the class on which the measures agree.
Let \(\mathcal{D}=\{B:\mu(B)=\nu(B)\}\), where \(B\) ranges over Borel sets. This is a lambda-system: both measures assign probability \(1\) to \(\mathbb{R}\); equality is preserved by complements because \(\mu(B^c)=1-\mu(B)\); and equality is preserved by countable disjoint unions by countable additivity.

Apply the pi-lambda theorem.
The equal-CDF assumption says that \(\mu\) and \(\nu\) agree on every ray in \(\Pi\), so \(\mathcal{D}\) contains \(\Pi\). The theorem then gives equality on the sigma-algebra generated by \(\Pi\), which is the Borel sigma-algebra. Therefore \(\mu(B)=\nu(B)\) for every Borel set \(B\). \(\square\)

Using the Result and Avoiding a Common Mistake

Worked Example: Comparing Intervals with Endpoint Masses

Suppose two laws have the same CDF \(F\) at every real number, and their common values satisfy \(F(-1-)=0.10\), \(F(-1)=0.20\), \(F(2-)=0.75\), and \(F(2)=0.90\). The theorem says the laws agree on every Borel set. The interval formulas make the endpoint effects explicit:

\begin{aligned} \mu([-1,2))&=F(2-)-F(-1-)=0.75-0.10=0.65,\\ \mu((-1,2])&=F(2)-F(-1)=0.90-0.20=0.70,\\ \mu([-1,2])&=F(2)-F(-1-)=0.90-0.10=0.80. \end{aligned}

The mass at \(-1\) is \(0.20-0.10=0.10\), and the mass at \(2\) is \(0.90-0.75=0.15\). Thus including \(-1\) but excluding \(2\) gives probability \(0.65\), while including both endpoints gives \(0.65+0.15=0.80\). The other law must give exactly the same probabilities, since its CDF agrees everywhere.

Common mistake: Two CDFs that match at a few selected values have not thereby been shown to describe the same law. For instance, agreement at \(x=-1\) and \(x=2\) establishes equality of the probabilities of the two corresponding half-lines, but says nothing by itself about the CDF between those thresholds. The uniqueness theorem requires agreement at every real number. Once that condition is met, the theorem is stronger than interval-by-interval subtraction: it guarantees agreement on every Borel set, including sets that are not intervals.

The practical importance is that a CDF can serve as a complete identifier for a probability law. It records point masses through jumps and also determines probabilities accumulated between jumps. When comparing two proposed distributions, equality of their CDFs at all real thresholds is enough; no separate check of every interval or Borel set is needed.

Check Your Understanding

Use the definitions and uniqueness argument above to answer these questions.

Two laws have the same CDF \(F\), with \(F(3)=0.8\) and \(F(1)=0.25\). What probability does each law assign to \((1,3]\)?
Why can the probabilities of \([a,b]\) and \((a,b]\) differ, even when the upper endpoint is the same?
In the uniqueness proof, why is the class of Borel sets on which two probability measures agree a lambda-system?
Why does equality of two CDFs at only finitely many points not establish equality of the laws?
What role does the pi-lambda theorem play in extending equality from half-lines to all Borel sets?

Uniqueness of a Distribution from Its CDF

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