CDF Jumps and Point Probabilities

Why a CDF Jump Measures a Point Probability

A cumulative distribution function records how much probability has accumulated up to each threshold. At a threshold $x$, its value $F(x)$ includes outcomes equal to $x$, as well as outcomes below $x$. The value immediately to the left, $F(x-)$, accounts only for outcomes strictly below $x$. The difference isolates the probability at exactly $x$.

This link is especially useful for discrete and mixed distributions, where some outcomes can carry positive probability at individual points. A CDF may rise gradually over a range, jump at a point, or do both. To calculate the probability of one exact value, focus on the jump there, not on the overall height of the CDF.

Definition: For a real-valued random variable $X$, its cumulative distribution function is $F(x)=P(X\le x)$. The left-hand limit at $x$ is denoted by $F(x-)=\lim_{t\uparrow x}F(t)$. The quantity $F(x)-F(x-)$ is the jump size of $F$ at $x$.

The limit $F(x-)$ exists because $F$ is nondecreasing and bounded between 0 and 1. The key point is that this left-hand limit corresponds to a strict inequality:

Theorem: For every real number $x$, $$ P(X=x)=F(x)-F(x-). $$ Thus, a point has positive probability exactly when the CDF has a positive jump there.

Why the Formula Holds

To justify the interpretation of the left limit, approach $x$ from below with the thresholds $x-1/n$, for positive integers $n$. The events $\{X\le x-1/n\}$ increase as $n$ increases, and their union is $\{X<x\}$. Every value less than $x$ is eventually at most $x-1/n$, while $x$ itself is never included.

We use continuity from below for probability: if events $A_n$ satisfy $A_1\subseteq A_2\subseteq\cdots$, then $P(\bigcup_{n=1}^{\infty}A_n)=\lim_{n\to\infty}P(A_n)$. This follows by writing the union as the disjoint union of $A_1$ and the increments $A_n\setminus A_{n-1}$ for $n\ge2$. Countable additivity gives the probability of the union as the sum of the probabilities of these disjoint pieces; the finite partial sums are $P(A_n)$, so their limit is the probability of the union.

Applying this result to $A_n=\{X\le x-1/n\}$ shows that $\lim_{n\to\infty}F(x-1/n)=P(X<x)$. Since these thresholds increase to $x$, their CDF values converge to the left-hand limit $F(x-)$. Therefore $F(x-)=P(X<x)$.

Proof of the jump formula: The event $\{X\le x\}$ is the disjoint union of $\{X<x\}$ and $\{X=x\}$. Hence $$ F(x)=P(X\le x)=P(X<x)+P(X=x). $$ Using $P(X<x)=F(x-)$ and rearranging gives $$ P(X=x)=F(x)-F(x-). $$ $\square$

Reading the Jump in Discrete CDFs

In a discrete distribution, the CDF is constant between possible values of $X$, then rises at those values. The jump at a support point is the probability assigned to that point. If there is no rise at a particular threshold, the difference $F(x)-F(x-)$ is zero.

Worked Example: A Three-Value Distribution

Suppose $X$ takes the values $0$, $2$, and $5$, with probabilities $0.2$, $0.5$, and $0.3$, respectively. Find $P(X=2)$, $P(X=3)$, and $P(X=5)$ using CDF jumps.

The CDF is $0$ for $x<0$, $0.2$ for $0\le x<2$, $0.7$ for $2\le x<5$, and $1$ for $x\ge5$. At $x=2$, the CDF value is $F(2)=0.7$, while the left-hand limit is $F(2-)=0.2$. Thus $$ P(X=2)=F(2)-F(2-)=0.7-0.2=0.5. $$

At $x=3$, the CDF stays at $0.7$ on both sides: $F(3)=0.7$ and $F(3-)=0.7$. Therefore $P(X=3)=0.7-0.7=0$. At $x=5$, $F(5)=1$ and $F(5-)=0.7$, so $P(X=5)=1-0.7=0.3$. These differences agree with the probabilities specified for the distribution.

Jumps in a Piecewise CDF

A CDF need not be a staircase throughout its domain. A CDF that increases continuously over part of its range and jumps at particular points can represent both continuously spread probability and point masses. To find a point probability from a piecewise CDF, evaluate the function at the point and take the limit from below. In particular, use the value assigned at the point for $F(x)$; do not substitute a formula that applies only to values strictly below it.

Worked Example: Point Masses and a Rising Segment

Consider the CDF $$ F(x)= \begin{cases} 0, & x<0,\\ 0.25+0.15x, & 0\le x<2,\\ 0.80, & 2\le x<4,\\ 1, & x\ge4. \end{cases} $$ Find the point probabilities at $x=0$, $x=1$, $x=2$, and $x=4$.

At $x=0$, the CDF value is $F(0)=0.25+0.15(0)=0.25$. Since $F(x)=0$ for $x<0$, its left-hand limit is $F(0-)=0$. Therefore $P(X=0)=0.25-0=0.25$.

At $x=1$, $F(1)=0.25+0.15(1)=0.40$. Approaching 1 from below uses the same linear segment, so $F(1-)=0.25+0.15(1)=0.40$. Thus $P(X=1)=0.40-0.40=0$.

At $x=2$, the value is given by the third case, $F(2)=0.80$. From below, the linear part approaches $0.25+0.15(2)=0.55$. Consequently $P(X=2)=0.80-0.55=0.25$.

Finally, $F(4)=1$, and the values just below 4 are $0.80$, so $F(4-)=0.80$. This gives $P(X=4)=1-0.80=0.20$. The CDF rises from $0.25$ toward $0.55$ on the segment before $2$, but that gradual rise does not create a jump at $1$; the point probability there is zero.

An Infinite Discrete Distribution

The same calculation applies when a distribution has infinitely many possible values. One can use a CDF formula to find the value immediately before a given support point and subtract it from the CDF value at that point.

Worked Example: A Geometric Distribution

Let $X$ have a geometric distribution on $0,1,2,\ldots$, with $P(X=k)=0.4(0.6)^k$ for each integer $k\ge0$. Find $P(X=2)$ from the CDF jump.

For an integer $k\ge0$, summing the probabilities from 0 through $k$ gives $$ F(k)=\sum_{j=0}^{k}0.4(0.6)^j=1-(0.6)^{k+1}. $$ For $x<0$, $F(x)=0$; between consecutive nonnegative integers, the CDF is constant. Hence $F(2)=1-(0.6)^3=1-0.216=0.784$, and the left-hand limit at 2 is $F(2-)=F(1)=1-(0.6)^2=1-0.36=0.64$. The jump is $$ P(X=2)=0.784-0.64=0.144. $$

Directly from the specified mass function, $P(X=2)=0.4(0.6)^2=0.4(0.36)=0.144$, which checks the CDF calculation.

Common Mistake: Using the CDF Height as the Mass

A frequent error is to say that $P(X=x)=F(x)$. That confuses the probability accumulated up to $x$ with the probability at $x$. For example, in the three-value distribution above, $F(2)=0.7$, but only $0.5$ of the probability is located at 2; the remaining $0.2$ lies below 2. Subtracting $F(2-)=0.2$ isolates the mass at 2.

Another error is to use the formula for the CDF immediately to the right of a point when computing the left-hand limit. For a piecewise CDF, first identify which case applies for arguments just less than $x$. At a boundary where the formula changes, that left-hand formula may give a different value from the CDF assigned at the boundary itself.

The jump formula also gives a useful consistency check. Since probabilities are nonnegative, $F(x)-F(x-)$ cannot be negative. A negative difference signals either an arithmetic error or an invalid proposed CDF. A zero difference means there is no probability concentrated at that exact point, even if the CDF itself is well above zero there.

Key takeaway: To find the probability at an exact value $x$, calculate $F(x)$, calculate the limit $F(x-)$ from strictly smaller arguments, and subtract: $P(X=x)=F(x)-F(x-)$. The jump size, not the CDF's height, is the point probability.

Check Your Understanding

Use the jump formula in each question.

A CDF has $F(3-)=0.42$ and $F(3)=0.67$. What is $P(X=3)$?
A CDF is constant at $0.6$ on an interval containing $x=4$, including a neighborhood immediately to its left. What is $P(X=4)$?
For the piecewise CDF in the worked example, why is $F(2-)$ not equal to $F(2)$?
If $F(x)=0.91$ and $F(x-)=0.76$, explain what each number represents and find the point probability at $x$.
Can a CDF value be positive while the point probability at that value's threshold is zero? Explain using the jump formula.