Atoms and Flat Stretches in CDF Graphs

What Jumps and Flat Stretches Tell Us

A cumulative distribution function records how probability accumulates as the threshold moves from left to right. Its graph therefore has a useful visual interpretation: a jump marks probability concentrated at one exact value, while a flat stretch marks an interval over which no additional probability accumulates. Reading these features correctly depends on keeping track of which level applies at each endpoint.

The CDF of a random variable $X$ is $F(x)=P(X\le x)$. Because the event $X\le x$ includes equality, the value of the CDF at a jump location is the level after the jump. The level just before that location is its left limit, written $F(x-)$. These two levels, not the apparent height of a graph in isolation, determine the probability at the point.

Definition: A point $x_0$ is an atom of the distribution of $X$ if $P(X=x_0)>0$. The size of the jump of $F$ at $x_0$ is $F(x_0)-F(x_0-)$. A stretch on which $F$ is constant indicates that the probability of moving through that stretch is zero; endpoint probabilities must be checked separately.

Reading a Jump as a Point Probability

Suppose a graph rises from a level $r$ immediately before $x_0$ to a level $s$ at $x_0$. The probability at that point is $s-r$. A jump of height zero means there is no point mass there; a positive jump means there is an atom. A jump to the value 1 is still interpreted by the same rule: its size is the amount by which the CDF rises, not automatically 1.

The jump formula follows from the definition of the CDF. To make the left limit precise, take any increasing sequence $x_n$ with $x_n<x_0$ and $x_n\to x_0$. The events $\{X\le x_n\}$ increase to $\{X<x_0\}$. Continuity of probability for increasing sequences of events gives $F(x_0-)=P(X<x_0)$. Since $\{X\le x_0\}$ is the disjoint union of $\{X<x_0\}$ and $\{X=x_0\}$, subtracting yields the result.

Result: For every real $x_0$, $$ P(X=x_0)=F(x_0)-F(x_0-). $$ Thus the vertical distance from the level just before the jump to the CDF value at the jump is exactly the point probability.

Worked Example: Finding Three Atoms in a Step Graph

A labeled CDF is 0 for $x<-1$, 0.30 for $-1\le x<2$, 0.75 for $2\le x<4$, and 1 for $x\ge4$. Find the point probabilities at the jump locations.

At $x=-1$, the level just before the jump is 0 and the value at the jump is 0.30. Therefore $P(X=-1)=0.30-0=0.30$.

At $x=2$, the left-hand level is 0.30 and $F(2)=0.75$. Thus $P(X=2)=0.75-0.30=0.45$.

At $x=4$, the left-hand level is 0.75 and $F(4)=1$. Thus $P(X=4)=1-0.75=0.25$. As a check, the three jump sizes sum to $0.30+0.45+0.25=1$, which matches the total rise from the far-left level 0 to the far-right level 1.

Why a Flat Stretch Means No Probability in Its Interior

For $u<v$, subtracting the nested events $\{X\le u\}\subseteq\{X\le v\}$ gives $P(u<X\le v)=F(v)-F(u)$. Consequently, if the CDF has the same level at $u$ and $v$, the interval $(u,v]$ has probability zero. In particular, a horizontal stretch rules out probability accumulating between its thresholds.

It is important to read the endpoints precisely. If $F$ is constant on $[a,b)$, then for any $a<u<v<b$, $P(u<X\le v)=0$; taking an increasing union of such intervals shows $P(a<X<b)=0$. The atom at $a$, if any, is already included in $F(a)$ and is not ruled out. A jump at $b$, if present, is also not ruled out: $b$ is outside the flat stretch $[a,b)$.

Result: If $F$ is constant on $[a,b)$, then $P(a<X<b)=0$. If it is constant on the closed interval $[a,b]$, then $P(a<X\le b)=0$. Neither statement rules out an atom at $a$. A jump at $b$ is possible in the first case, but not in the second.

For a direct justification of the open-interval statement, choose sequences $u_n\downarrow a$ and $v_n\uparrow b$, with $a<u_n<v_n<b$. Constancy gives $P(u_n<X\le v_n)=F(v_n)-F(u_n)=0$. The intervals $(u_n,v_n]$ can be chosen to increase to $(a,b)$, so continuity of probability for increasing events implies $P(a<X<b)=0$. This also shows why a flat interior and its endpoints should not be conflated.

Worked Example: Locating Probability-Free Gaps

A CDF is 0 for $x<0$, 0.20 for $0\le x<1$, 0.50 for $1\le x<3$, 0.80 for $3\le x<5$, and 1 for $x\ge5$. Identify the atoms and the open intervals between jumps that receive no probability.

The jumps occur at $0,1,3,$ and $5$. Their sizes are, respectively, $0.20-0=0.20$, $0.50-0.20=0.30$, $0.80-0.50=0.30$, and $1-0.80=0.20$. These four point probabilities sum to 1.

The graph is flat from 0 up to but not including 1, from 1 up to but not including 3, and from 3 up to but not including 5. The corresponding open intervals $(0,1)$, $(1,3)$, and $(3,5)$ each have probability zero. The open intervals exclude both endpoint atoms. In contrast, $(1,3]$ includes the atom at 3, so its probability is 0.30, not zero. Similarly, $(3,5]$ includes the atom at 5 and has probability 0.20.

Reading Both Features Together

A graph can combine flat stretches and jumps, and the two features answer different questions. A jump tells how much probability is placed at its location. A flat stretch tells that there is no probability accumulating through its interior. When a flat stretch ends at a jump, the jump endpoint may itself have positive probability. The left limit at that endpoint gives the level across the preceding flat stretch; the CDF value at the endpoint includes the jump.

Worked Example: A Flat Stretch Ending in a Jump

Suppose a labeled CDF has $F(x)=0$ for $x<0$, $F(x)=0.20+0.30x$ for $0\le x<1$, $F(x)=0.60$ for $1\le x<3$, $F(x)=0.80$ for $3\le x<6$, and $F(x)=1$ for $x\ge6$. Read the atom probabilities and the flat intervals, then find $P(1<X\le6)$.

At 0, the left limit is 0 and $F(0)=0.20$, so $P(X=0)=0.20$. As $x$ approaches 1 from below, $0.20+0.30x$ approaches 0.50, while $F(1)=0.60$. Thus $P(X=1)=0.60-0.50=0.10$. At 3, the left limit is 0.60 and $F(3)=0.80$, so $P(X=3)=0.20$. For 6, the stated level is 0.80 for $3\le x<6$, while $F(6)=1$; hence $P(X=6)=1-0.80=0.20$.

The graph is flat on $[1,3)$ and $[3,6)$. Therefore $(1,3)$ and $(3,6)$ have probability zero. The point at 3 is an exception to the first open gap only if included: it has probability 0.20. The point at 6 also has probability 0.20.

For the requested interval, $P(1<X\le6)=F(6)-F(1)=1-0.60=0.40$. This comprises the atom at 3, of size 0.20, and the atom at 6, of size 0.20; the open stretches between them contribute no probability. In particular, using 0.80 as $F(6)$ would be incorrect: 0.80 is the level immediately before the jump, whereas right-continuity gives $F(6)=1$.

A Reliable Procedure for Labeled CDF Graphs

Mark each jump location.
Read the level immediately before the location and the CDF value at the location. The latter is the level after the jump.

Subtract to find the atom.
At a jump location $x_0$, compute $F(x_0)-F(x_0-)$.

Mark each horizontal stretch.
State its interval with endpoint brackets matching the graph. A constant stretch on $[a,b)$ means the open interior $(a,b)$ has probability zero, but says nothing against an atom at $b$.

Check any requested interval’s endpoints.
Use $P(u<X\le v)=F(v)-F(u)$, and inspect whether an endpoint jump is included in the event.

Common Mistakes and Why This Matters

Using the level before a jump as the CDF value at the jump. At $x_0$, the CDF includes $P(X=x_0)$, so its value is the level after the jump.
Calling a flat interval probability-free without specifying endpoints. A flat stretch on $[a,b)$ implies zero probability on $(a,b)$, but an atom at $b$ may occur. For example, $(a,b]$ includes it.
Assuming every point on a horizontal segment has an atom. A horizontal segment indicates no increase in cumulative probability across its interior. Atoms correspond to jumps, not flat portions.
Subtracting the wrong levels for a requested event. For $P(u<X\le v)$, subtract $F(u)$ from $F(v)$; then check that the event’s endpoint inclusion matches those CDF values.

These graph-reading skills are useful because they distinguish probability concentrated at exact values from probability accumulated elsewhere, and they make endpoint-sensitive event calculations transparent. Before reading a graph, identify its labeled levels and jump locations; after reading it, check that jump sizes and any other stated probability pieces are consistent with the total probability of 1.

Check Your Understanding

Use the jump and flat-stretch interpretations to answer each question.

A CDF has left limit 0.35 at $x=2$ and value $F(2)=0.62$. What is $P(X=2)$?
If $F$ is constant on $[4,7)$, which open interval is guaranteed to have probability zero? Is $P(X=7)$ necessarily zero?
A CDF is flat at level 0.40 from $x=1$ up to but not including $x=5$, and $F(5)=0.70$. Find $P(1<X<5)$ and $P(X=5)$.
Explain why a flat stretch on $[a,b)$ does not by itself imply $P(a\le X\le b)=0$.
A jump at $c$ rises from 0.72 to 1. What probability does the jump assign to $X=c$?