What a Cumulative Distribution Function Records
For a real-valued random variable, the cumulative distribution function (CDF) records how much probability lies at or to the left of each threshold. Its values can answer more than questions of the form “what is the probability of being at most \(x\)?” By taking differences and left limits, we can recover probabilities of intervals with different endpoint conventions, detect point masses, and describe where the law is concentrated. These distinctions matter: a point can belong to the support without having positive probability itself.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \(X:\Omega\to\mathbb{R}\) be a random variable. Its induced law assigns probability \(\mathbb{P}(X\in B)\) to each Borel set \(B\subseteq\mathbb{R}\). The CDF of \(X\) is the function \(F:\mathbb{R}\to[0,1]\) defined by \(F(x)=\mathbb{P}(X\le x)\). All probabilities below refer to this law. We use continuity of probability: probabilities of an increasing sequence of events converge to the probability of their union, and probabilities of a decreasing sequence converge to the probability of their intersection.
Proof. If \(x\le y\), then \(\{X\le x\}\subseteq\{X\le y\}\), so \(F(x)\le F(y)\). If \(x_n\downarrow x\), then the events \(\{X\le x_n\}\) decrease and their intersection is \(\{X\le x\}\). Continuity of probability for decreasing events gives \(F(x_n)\to F(x)\), which is right-continuity. As \(n\to\infty\), the events \(\{X\le -n\}\) decrease to the empty event, while \(\{X\le n\}\) increase to \(\Omega\). Continuity of probability therefore gives the two limits at infinity.
For the left limit, take any sequence \(x_n\uparrow x\). The events \(\{X\le x_n\}\) increase to \(\{X<x\}\), so \(F(x-)=\mathbb{P}(X<x)\). The event \(\{a<X\le b\}\) is the difference between \(\{X\le b\}\) and \(\{X\le a\}\); these are nested events, so subtracting their probabilities gives the first formula. Replacing \(\{X\le a\}\) by \(\{X<a\}\), or \(\{X\le b\}\) by \(\{X<b\}\), gives the other three formulas. Finally, \(\{X=x\}=\{X\le x\}\setminus\{X<x\}\), so its probability is \(F(x)-F(x-)\). \(\square\)
The jump formula identifies atoms: a point \(x\) is an atom of the law if \(\mathbb{P}(X=x)>0\), equivalently if \(F(x)>F(x-)\). At a point with no atom, the CDF has no jump, even if it changes rapidly nearby. The left limit is essential whenever an interval includes its left endpoint or excludes its right endpoint.
Worked Example: Reading Atoms from a Step CDF
Suppose \(X\) takes the values \(-1,0,2\) with probabilities \(\tfrac12,\tfrac13,\tfrac16\), respectively. These probabilities are nonnegative and sum to \(\tfrac12+\tfrac13+\tfrac16=1\), so they define a probability law. For \(x<-1\), \(F(x)=0\). For \(-1\le x<0\), \(F(x)=\tfrac12\); for \(0\le x<2\), \(F(x)=\tfrac12+\tfrac13=\tfrac56\); and for \(x\ge2\), \(F(x)=1\).
The jumps are \(\tfrac12\) at \(-1\), \(\tfrac56-\tfrac12=\tfrac13\) at \(0\), and \(1-\tfrac56=\tfrac16\) at \(2\). Thus each jump recovers the assigned probability at that point. To compute \(\mathbb{P}(-1<X\le2)\), use \(F(2)-F(-1)=1-\tfrac12=\tfrac12\). Directly, the interval contains \(0\) and \(2\), but not \(-1\), and their probabilities sum to \(\tfrac13+\tfrac16=\tfrac12\). The two calculations agree.
Support Is Not the Same as the Set of Atoms
The support of a real-valued law is the set of points whose every open neighborhood has positive probability. Explicitly, \(x\) belongs to the support if \(\mathbb{P}(x-\varepsilon<X<x+\varepsilon)>0\) for every \(\varepsilon>0\). A support point need not be an atom: every small interval around a point may have positive probability even though the point itself has probability zero.
Proof. If \(x\) is outside the support, then some open interval containing \(x\) has probability zero. Every point sufficiently close to \(x\) has a smaller open neighborhood inside that interval, also of probability zero, so it too lies outside the support. Thus the complement of the support is open, and the support is closed.
To show that the complement has probability zero, consider all open intervals with rational endpoints that have probability zero. There are only countably many such intervals. For every point outside the support, there is a zero-probability open neighborhood around it. Inside that neighborhood one can choose rational endpoints \(r<s\) so that the point lies in \((r,s)\) and \((r,s)\) remains inside the neighborhood. This rational interval also has probability zero. Consequently, the complement of the support is covered by a countable union of zero-probability rational intervals, which has probability zero. Hence the support has probability one. \(\square\)
Worked Example: A Continuous Law with No Atoms
Let \(X\) have the uniform law on \([0,2]\), so its CDF is \(0\) for \(x<0\), \(x/2\) for \(0\le x<2\), and \(1\) for \(x\ge2\). This function is nondecreasing and right-continuous, with limits \(0\) and \(1\) at the two ends of the real line. It has no jumps: at \(0\), the left limit and value are both \(0\); at \(2\), both are \(1\); and inside \((0,2)\) the function is continuous. Therefore \(\mathbb{P}(X=x)=0\) at every point.
For example, the probability of \(\tfrac12\le X<\tfrac32\) is \(F((\tfrac32)-)-F((\tfrac12)-)=\tfrac34-\tfrac14=\tfrac12\). The endpoint values do not affect the result because neither endpoint is an atom. Every neighborhood of any point in \([0,2]\), including either endpoint, intersects the interval \([0,2]\) in a segment of positive length, and thus has positive probability. A point outside \([0,2]\) has a sufficiently small neighborhood disjoint from \([0,2]\), with probability zero. The support is therefore \([0,2]\), although none of its points is an atom.
Discrete, Atomless, and Mixed Laws
A law is discrete if some countable set \(D\) satisfies \(\mathbb{P}(X\in D)=1\). A law is atomless if \(\mathbb{P}(X=x)=0\) for every real \(x\). “Continuous law” is sometimes used as a synonym for atomless law; here we use “atomless” to avoid suggesting that every such law has a probability density. A law is mixed if it has positive total atomic probability but is not concentrated on its atoms.
There are at most countably many atoms. Indeed, for each positive integer \(m\), there can be at most \(m\) points with probability at least \(1/m\), because the probabilities of distinct points add and their total cannot exceed \(1\). Every atom has probability at least \(1/m\) for some \(m\). Thus the atoms form a countable set. Their probabilities have a well-defined sum, called the total atomic mass. If this mass is \(1\), the law is discrete; if it is \(0\), the law is atomless; and if it lies strictly between \(0\) and \(1\), the law is mixed. This classification concerns the law’s probability, not simply the shape of its support.
Worked Example: Locating the Atoms and Support of a Mixed Law
Let \(X\) have probability \(\tfrac13\) at \(0\), and with probability \(\tfrac23\) have the uniform law on \([1,2]\). The weights sum to \(\tfrac13+\tfrac23=1\). Its CDF is \(0\) for \(x<0\), \(\tfrac13\) for \(0\le x<1\), \(\tfrac13+\tfrac23(x-1)\) for \(1\le x<2\), and \(1\) for \(x\ge2\). At \(x=1\), the formula gives \(\tfrac13+\tfrac23(1-1)=\tfrac13\); at \(x=2\), its left limit is \(\tfrac13+\tfrac23(2-1)=1\). There is a jump of size \(\tfrac13\) at \(0\), and no jump at \(1\) or \(2\). The only atom is \(0\), so the total atomic mass is \(\tfrac13\), strictly between \(0\) and \(1\).
For the interval \((\tfrac32,2]\), the probability is \(F(2)-F(\tfrac32)=1-[\tfrac13+\tfrac23(\tfrac12)]=1-\tfrac23=\tfrac13\). Directly, this is the mixture weight \(\tfrac23\) times the uniform interval probability \(\tfrac12\), also \(\tfrac13\). The support is \(\{0\}\cup[1,2]\): every neighborhood of \(0\) contains its atom, every neighborhood of a point in \([1,2]\) contains a positive-probability part of the uniform component, and points outside this set have a small neighborhood with probability zero. In particular, \(1\) and \(2\) are support points but not atoms.
Endpoint Pitfalls and a Reliable Procedure
A frequent mistake is to use \(F(b)-F(a)\) for \(\mathbb{P}(a\le X\le b)\). The difference actually gives \(\mathbb{P}(a<X\le b)\), so it excludes any mass at \(a\). In the step-CDF example, the incorrect calculation for \(\mathbb{P}(-1\le X\le0)\) would be \(F(0)-F(-1)=\tfrac56-\tfrac12=\tfrac13\). But the interval includes both \(-1\) and \(0\), so its probability is \(\tfrac12+\tfrac13=\tfrac56\). The correction is \(F(0)-F((-1)-)=\tfrac56-0=\tfrac56\).
For any interval calculation, first identify whether each endpoint is included. Then use \(F(x)\) for an included upper endpoint and \(F(x-)\) for an excluded upper endpoint; at the lower endpoint, subtract \(F(x)\) when it is excluded and \(F(x-)\) when it is included. Finally, check whether an endpoint is an atom: if it is not, switching its inclusion does not change the probability. Keep support and atoms separate as well: support asks whether neighborhoods carry probability, while an atom asks whether the single point carries probability.
Check Your Understanding
Use CDF values and left limits, keeping the strictness of each endpoint explicit.
- If \(F(a)=\tfrac25\), \(F(a-)=\tfrac1{10}\), and \(F(b-)=\tfrac9{10}\), find \(\mathbb{P}(a\le X<b)\).
- What is the probability of an atom at \(x\) in terms of \(F(x)\) and \(F(x-)\)?
- Can a point belong to the support while having probability zero itself? Explain using the uniform law on \([0,2]\).
- A law has atoms with total probability \(\tfrac34\). Is it discrete, atomless, or mixed under the definitions in this tutorial?
- For the mixed law with an atom at \(0\) and a uniform component on \([1,2]\), identify one support point that is not an atom and justify both properties.