Classifying a Law from Its Formula
A formula for a distribution may present its structure in different ways: as a probability mass function, a density, a CDF, or a combination of these. Classification means identifying which parts of the law assign probability to individual points, which spread probability according to a density, and whether any remaining probability has a different structure. The distinction matters because a CDF can look smooth without having a density, and a formula involving a density may also include point masses.
We will use four labels. A law is purely discrete if all its probability is concentrated on a finite or countable set of points. It is absolutely continuous if its probabilities are given by integrating a density. It is singular continuous if it has no point masses but places all its probability on a set of Lebesgue measure zero. A law with more than one of these types of component is called mixed. These labels describe the probability law, not just the appearance of a formula.
The CDF is \(F(x)=P(X\le x)\). It records all probabilities up to each threshold, so its jumps reveal atoms. In an absolutely continuous region, CDF increments are integrals of the density. A singular continuous component, by contrast, can increase the CDF without jumps even though no density accounts for that increase.
What Jumps and Increments Tell Us
Proof. Choose any sequence \(x_n<a\) increasing to \(a\). The events \(\{X\le x_n\}\) increase to \(\{X<a\}\). Continuity of probability for increasing sequences gives \[ F(a-)=\lim_{n\to\infty}P(X\le x_n)=P(X<a). \] The event \(\{X\le a\}\) is the disjoint union of \(\{X<a\}\) and \(\{X=a\}\). Therefore \[ F(a)-F(a-)=P(X\le a)-P(X<a)=P(X=a). \] This also shows that \(F\) is continuous at \(a\) exactly when \(a\) is not an atom. \(\square\)
A density gives a different CDF signature. If \(X\) has density \(f\), then for \(a<b\), \[ F(b)-F(a)=P(a<X\le b)=\int_{(a,b]} f(t)\,dt. \] In particular, the CDF is continuous. The converse is not true: continuity rules out jumps, but does not establish that the increments come from integrating a density.
Worked Example: A Purely Discrete Law
Suppose \(X\) takes the values \(0,1,3\), with probabilities \(P(X=0)=1/2\), \(P(X=1)=1/3\), and \(P(X=3)=1/6\). First check the total probability: \[ \frac12+\frac13+\frac16=\frac{3+2+1}{6}=1. \] The law is purely discrete because all probability is assigned to these three points.
Its CDF is \[ F(x)= \begin{cases} 0, & x<0,\\ 1/2, & 0\le x<1,\\ 5/6, & 1\le x<3,\\ 1, & x\ge 3. \end{cases} \] The jumps are \(1/2\) at 0, \(5/6-1/2=1/3\) at 1, and \(1-5/6=1/6\) at 3. Each agrees with the corresponding point probability. For example, \(P(0<X\le 3)=F(3)-F(0)=1-1/2=1/2\), which is the mass at 1 and 3 combined.
Worked Example: An Absolutely Continuous Law
Consider the proposed density \(f(x)=2x\) for \(0\le x\le 1\), and \(f(x)=0\) otherwise. It is nonnegative, and its total integral is \[ \int_{\mathbb R}f(x)\,dx=\int_0^1 2x\,dx=[x^2]_0^1=1. \] It therefore defines an absolutely continuous probability law.
For \(0\le x\le1\), its CDF is \[ F(x)=\int_{-\infty}^x f(t)\,dt=\int_0^x2t\,dt=x^2. \] Including the two outer regions gives \(F(x)=0\) for \(x<0\), \(F(x)=x^2\) for \(0\le x\le1\), and \(F(x)=1\) for \(x>1\). There are no jumps, so there are no atoms. For instance, \[ P(1/4<X\le3/4)=F(3/4)-F(1/4)=\frac{9}{16}-\frac{1}{16}=\frac12. \] The same probability is \(\int_{1/4}^{3/4}2x\,dx=[x^2]_{1/4}^{3/4}=1/2\).
Reading a Formula with More Than One Component
When a formula includes both point masses and a density, calculate their contributions separately. For example, a CDF of the form \[ F(x)=\sum_{j:x_j\le x}p_j+\int_{-\infty}^x f(t)\,dt \] has a discrete contribution from the points \(x_j\) and an absolutely continuous contribution from \(f\), provided the \(x_j\) are distinct, \(p_j\ge0\), \(f\ge0\), and \(\sum_jp_j+\int_{\mathbb R}f(t)\,dt=1\). The sum is over a finite or countable collection. If both contributions have positive total probability, the law is mixed.
Proof. The formula is nondecreasing because increasing \(x\) can only add nonnegative terms to the sum and enlarge the integration region. Its limits at \(-\infty\) and \(+\infty\) are 0 and \(\sum_jp_j+\int_{\mathbb R}f(t)\,dt=1\), respectively. Both terms are right-continuous: for the sum, this follows by applying dominated convergence to the indicators \(1\{x_j\le x\}\), bounded by the summable sequence \(p_j\); for the integral, it follows from integrability of \(f\). Thus the formula is a CDF.
For \(a<b\), subtracting its values gives \[ F(b)-F(a)=\sum_{j:a<x_j\le b}p_j+\int_{(a,b]}f(t)\,dt. \] The jump at any \(x_j\) is therefore \(p_j\); the integral term has no jump because integrals of an integrable function over intervals shrinking to a point tend to zero. The interval probabilities consist of exactly the stated point-mass and density contributions, as claimed. \(\square\)
Worked Example: An Atom Together with a Density
Let \(P(X=0)=1/4\), and let the remaining probability \(3/4\) have constant density \(3/8\) on \((0,2)\). The density contributes \((3/8)(2)=3/4\), so the atom and density together have total probability \(1/4+3/4=1\). This is a mixed law: it has a positive atom and a positive density contribution.
Its CDF is \[ F(x)= \begin{cases} 0, & x<0,\\ \frac14+\frac{3x}{8}, & 0\le x<2,\\ 1, & x\ge2. \end{cases} \] At zero, \(F(0)-F(0-)=1/4\), identifying the atom. Between zero and two, the CDF rises at rate \(3/8\), the density. At \(x=2\), there is no jump: the formula approaches \(1/4+(3/8)(2)=1\). For \(0<a<b<2\), \(P(a<X\le b)=3(b-a)/8\). For example, \(P(1/2<X\le3/2)=3(1)/8=3/8\).
The atom at zero does not prevent the law from also having a density contribution. Conversely, writing down a density alone would miss the probability \(1/4\) at zero.
A Continuous CDF Without a Density
A CDF with no jumps need not be absolutely continuous. One important example is the Cantor distribution. Let \(B_1,B_2,\ldots\) be independent Bernoulli random variables with \(P(B_k=0)=P(B_k=1)=1/2\), and define \[ X=\sum_{k=1}^{\infty}\frac{2B_k}{3^k}. \] The series converges because \(0\le X\le\sum_{k=1}^{\infty}2/3^k=1\). Its values lie in the Cantor set, which has Lebesgue measure zero.
Worked Example: Classifying the Cantor Distribution
After fixing the first \(n\) digits \(B_1,\ldots,B_n\), the possible values of \(X\) lie in one of the \(2^n\) remaining level-\(n\) intervals. Each specified digit sequence has probability \(2^{-n}\). Any single point belongs to at most two of these intervals, so \[ P(X=x)\le 2\cdot 2^{-n} \] for every \(n\). Letting \(n\) increase shows \(P(X=x)=0\): the law has no atoms.
The level-\(n\) Cantor construction covers the Cantor set with \(2^n\) intervals of length \(3^{-n}\). Their total length is \((2/3)^n\), which tends to zero. Thus the Cantor set has Lebesgue measure zero, while \(P(X\text{ lies in the Cantor set})=1\). The law is singular continuous: it has no atoms but is concentrated on a null set.
Its CDF is continuous because it has no atoms, by the jump identity. Yet it cannot have a density: if it had one, the probability assigned to the Cantor set would be the integral of that density over a set of Lebesgue measure zero, hence zero. That contradicts the probability one just established. This example shows why “continuous CDF” and “density” are not interchangeable descriptions.
A Practical Classification Check
When given an unfamiliar formula, first determine what it asserts about probabilities, then check whether the contributions add to one. Use CDF jumps to identify atoms; use interval increments or an integral to identify density contributions. If the CDF is continuous but the formula concentrates probability on a null set, the law is singular continuous rather than absolutely continuous.
Verify that the total point mass, integrated density, and any other stated component sum to one.
For a CDF, compute its jump sizes. For a mass-function or hybrid formula, identify the point probabilities directly.
Check whether probabilities over intervals are given by integrating a nonnegative function.
If there are no atoms but probability is concentrated on a null set, the component is singular continuous. If distinct types both carry positive probability, the law is mixed.
Common mistake: “The CDF is continuous, so the law has a density.” Continuity only says there are no jumps, and therefore no atoms. The Cantor example has a continuous CDF but no density. To claim absolute continuity, verify that interval probabilities are integrals of a density—or otherwise establish the density representation. A related error is to differentiate a CDF where possible and assume the derivative alone accounts for all probability; the singular continuous example shows that CDF increase need not be captured by a density.
Check Your Understanding
For each question, justify the classification using the formula or CDF behavior.
- A CDF jumps by \(0.2\) at \(x=4\) and has no other jumps. What probability is assigned to \(X=4\)?
- A nonnegative function is proposed as a density but integrates to \(0.85\). What must be checked before it can define a probability law by itself?
- A law has mass \(1/3\) at zero and a density contribution of total mass \(2/3\). How should it be classified?
- Why does a continuous CDF not, by itself, establish that a law has a density?
- A law has no atoms and assigns probability one to a set of Lebesgue measure zero. Which classification applies?