Deriving Atom Sizes from CDF Jumps

Why a CDF Jump Measures a Point Probability

A cumulative distribution function records probabilities up to and including each threshold: $F(x)=P(X\le x)$. If the CDF rises abruptly at a particular value $x$, that rise records probability assigned to $X=x$. The size of the rise—not the height of the CDF at $x$—is the probability at that point.

This relationship is useful because a CDF can describe discrete, continuous, or mixed behavior. In every case, comparing its value at a point with the limit from the left gives the probability concentrated exactly there. We will establish the comparison from the probability axioms, then use it to read atoms from several CDFs.

Definition: For a random variable $X$ with CDF $F(x)=P(X\le x)$, the left limit at $x$ is $$ F(x-)=\lim_{t\uparrow x}F(t). $$ An atom at $x$ is a point with positive probability $P(X=x)>0$. The jump size of $F$ at $x$ is $F(x)-F(x-)$.

The notation $t\uparrow x$ means that $t$ approaches $x$ through values strictly less than $x$. In particular, $F(x-)$ is not generally the same as $F(x)$: the value $F(x)$ includes the event $X=x$, while the left limit does not.

The Jump Formula

The event $\{X<x\}$ contains all outcomes for which $X$ is at or below a threshold approaching $x$ from below. The formal connection follows by applying continuity from below of probability to an increasing sequence of such events.

Theorem: For every real-valued random variable $X$ and every $x\in\mathbb{R}$, $$ P(X=x)=F(x)-F(x-). $$ Consequently, $x$ is an atom if and only if the CDF has a positive jump at $x$.

Proof. Fix $x$, and for each positive integer $n$ define $x_n=x-1/n$. The events $\{X\le x_n\}$ increase as $n$ increases, and $$ \bigcup_{n=1}^{\infty}\{X\le x_n\}=\{X<x\}. $$ Indeed, if $X<x$, then for sufficiently large $n$, $1/n\le x-X$, so $X\le x-1/n$. Conversely, $X\le x-1/n$ implies $X<x$. Continuity from below therefore gives $$ \lim_{n\to\infty}F(x_n) =\lim_{n\to\infty}P(X\le x_n) =P(X<x). $$ The sequence $x_n$ approaches $x$ from below, so this limit is $F(x-)$. Since $\{X\le x\}$ is the disjoint union of $\{X<x\}$ and $\{X=x\}$, additivity gives $$ F(x)=P(X\le x)=P(X<x)+P(X=x). $$ Subtracting $F(x-)=P(X<x)$ from both sides proves the formula. A point has positive probability exactly when this difference is positive. $\square$

The limit defining $F(x-)$ exists: CDFs are nondecreasing, so their values approaching $x$ from below have a well-defined supremum. The sequence in the proof calculates that left limit. Its particular choice is convenient, but the resulting limit is the same for any sequence increasing to $x$.

Useful consequence: A CDF has no discontinuity at $x$ other than a jump from its left. In particular, $F$ is continuous at $x$ if and only if $P(X=x)=0$.

For completeness, right-continuity follows from continuity from above of probability. The events $\{X\le x+1/n\}$ decrease to $\{X\le x\}$. Since probabilities are finite, continuity from above applies, yielding $\lim_{n\to\infty}F(x+1/n)=F(x)$. Thus the CDF is already continuous from the right. Its only possible failure of continuity is a difference between its left limit and its value. By the jump formula, that difference is precisely $P(X=x)$.

Reading Jumps in Step CDFs

For a step CDF, the constant stretches mark intervals across which the accumulated probability does not change. At each step up, compare the value immediately before the jump with the value at the jump. The difference is the atom’s probability.

Worked Example: Finding Every Atom in a Step CDF

Suppose the CDF is $$ F(x)= \begin{cases} 0, & x<-1,\\ 0.2, & -1\le x<2,\\ 0.7, & 2\le x<5,\\ 1, & x\ge 5. \end{cases} $$ Find the atoms and their probabilities.

At $x=-1$, the CDF is $F(-1)=0.2$, while $F(-1-)=0$. Therefore $$ P(X=-1)=0.2-0=0.2. $$ At $x=2$, $F(2)=0.7$ and $F(2-)=0.2$, so $$ P(X=2)=0.7-0.2=0.5. $$ At $x=5$, $F(5)=1$ and $F(5-)=0.7$, giving $$ P(X=5)=1-0.7=0.3. $$ There are no other atoms: away from these three step locations, the CDF is locally constant and thus has no jump. The atom probabilities sum to $0.2+0.5+0.3=1$, which checks that the full probability is accounted for.

The value $F(x)$ includes all probability accumulated before $x$, as well as any probability at $x$ itself. That is why subtracting the left limit isolates the point probability. A CDF value alone generally cannot do so.

Atoms in a Hybrid CDF

A CDF need not be a staircase. It may increase gradually over some ranges and jump at particular points. The same jump formula applies without change: compare the CDF’s value at the point with the limit of its values to the left. Gradual change across an interval does not itself create a point atom.

Worked Example: Atoms at the Ends of a Hybrid CDF

Consider the CDF $$ F(x)= \begin{cases} 0, & x<0,\\ 0.3+0.25x, & 0\le x<2,\\ 1, & x\ge 2. \end{cases} $$ Locate and quantify its atoms.

At $x=0$, the left-hand limit is $F(0-)=0$, and $F(0)=0.3$. Thus $P(X=0)=0.3$. At $x=2$, the left-hand limit is $$ F(2-)=0.3+0.25(2)=0.8, $$ while $F(2)=1$. Hence $P(X=2)=1-0.8=0.2$.

For an interior point $0<x<2$, the formula $0.3+0.25x$ is continuous, so its value agrees with its left limit and there is no atom there. The CDF is also constant and continuous at points outside the interval other than $0$ and $2$. Therefore the only atoms are $0$ and $2$, with probabilities $0.3$ and $0.2$. The remaining probability, $1-0.3-0.2=0.5$, is accumulated through the gradual increase between them.

A useful check is to use CDF differences to compute interval probabilities. In this example, $F(2-)=P(X<2)=0.8$, and $F(0)=P(X\le0)=0.3$. Therefore $$ P(0<X<2)=P(X<2)-P(X\le0)=0.8-0.3=0.5. $$ The interval calculation agrees with the gradual increase across the middle range, while excluding the atoms at both endpoints.

Worked Example: Comparing Both Sides of Each Knot

Suppose $$ F(x)= \begin{cases} 0, & x<-1,\\ 0.2+0.3(x+1), & -1\le x<0,\\ 0.65+0.1x, & 0\le x<2,\\ 1, & x\ge2. \end{cases} $$ Identify all points with positive probability and calculate their masses.

At the first knot, $x=-1$, $F(-1-)=0$ and $F(-1)=0.2$, so $P(X=-1)=0.2$.

At $x=0$, the first formula approaches $0.2+0.3(0+1)=0.5$ from the left, while the second formula gives $F(0)=0.65$. Thus $$ P(X=0)=0.65-0.5=0.15. $$

At $x=2$, the second formula approaches $0.65+0.1(2)=0.85$, while $F(2)=1$. Therefore $P(X=2)=1-0.85=0.15$. Between the knots, each formula is continuous, so there are no other jumps and hence no other atoms. The total mass at the identified points is $0.2+0.15+0.15=0.5$; the rest of the CDF’s increase occurs between those points.

Common Mistake: Treating the CDF Value as the Atom

A frequent error is to claim that $P(X=x)=F(x)$. That would be true only when $F(x-)=0$, because in general the probability at $x$ is the increase above everything accumulated strictly before $x$. For example, in the hybrid CDF above, $F(0)=0.3$ and $P(X=0)=0.3$, but at $x=2$, $F(2)=1$ while $P(X=2)=0.2$.

Another tempting but incorrect approach is to use the value of the formula immediately to the right of a knot as the atom size. The atom is not the right-hand value by itself, nor is it a slope or an interval’s total increase. The relevant subtraction is always $$ \text{value at the point}-\text{left-hand limit at the point}. $$ Check both quantities at the same point, and retain the endpoint convention in the CDF definition: $F(x)$ includes $X=x$; $F(x-)$ does not.

Key takeaway: For any CDF, the mass at $x$ is $F(x)-F(x-)$. Locate atoms by finding where the CDF jumps, and quantify each one by subtracting its left-hand limit from its value at the jump.

Check Your Understanding

For each question, use the difference between a CDF value and its left-hand limit.

A CDF has $F(3-)=0.42$ and $F(3)=0.60$. What is $P(X=3)$?
Why does $F(x-)$ equal $P(X<x)$ rather than $P(X\le x)$?
A CDF is continuous at $x=4$. What can you conclude about $P(X=4)$?
For a step CDF, what feature of the graph identifies a point with positive probability?
A CDF increases continuously on an interval and has a jump at its right endpoint. How would you calculate the probability at that endpoint?