Left Limits, Jumps, and Endpoint Conventions

Why Endpoint Conventions Matter

A cumulative distribution function can change abruptly at a value that has positive probability. At such a value, $F(a)$ includes the probability at $a$, while the value approached from the left does not. Confusing these quantities can include or exclude an endpoint incorrectly, even when the arithmetic that follows is flawless.

The distinction matters whenever an event uses strict or non-strict inequalities, such as $X<a$ versus $X\le a$. For a distribution with no probability concentrated at the endpoints, some of these distinctions make no numerical difference. For a discrete or mixed distribution, they can change the answer.

Definition: For a random variable $X$, its cumulative distribution function is $F(x)=P(X\le x)$. The left-hand limit at $a$ is $$ F(a-)=\lim_{x\uparrow a}F(x)=P(X<a). $$ Thus $F(a)$ includes the probability that $X=a$, while $F(a-)$ excludes it.

The notation $F(a-)$ means the value approached as the input comes up to $a$ from below; it does not mean $F(a)-1$, nor does it mean a value of $F$ at some fixed distance to the left. For example, if $X$ can equal 4, then $F(4-)=P(X<4)$, and $F(4)=P(X\le4)$.

Endpoint Rules for Intervals

The most reliable method is to translate each endpoint into a CDF value according to whether that endpoint is included. A closed upper endpoint uses $F(b)$; an open upper endpoint uses $F(b-)$. A closed lower endpoint is subtracted using $F(a-)$; an open lower endpoint is subtracted using $F(a)$.

Theorem: If $a<b$, then the four endpoint conventions are $$ \begin{aligned} P(a<X\le b)&=F(b)-F(a),\\ P(a\le X\le b)&=F(b)-F(a-),\\ P(a<X<b)&=F(b-)-F(a),\\ P(a\le X<b)&=F(b-)-F(a-). \end{aligned} $$ In addition, $P(X=a)=F(a)-F(a-)$.

These rules are valid for discrete, continuous, and mixed distributions. They do not require a density or a probability mass function. In particular, a density alone cannot describe an atom, so density-based calculations should not replace the CDF endpoint rules unless the model is known to have no atoms at the relevant endpoints.

Start with the event.
Write down whether each endpoint is included, using $<$ or $\le$.

Choose the CDF values.
Use $F(b)$ for an included upper endpoint and $F(b-)$ for an excluded upper endpoint. At the lower endpoint, subtract $F(a-)$ if it is included and $F(a)$ if it is excluded.

Subtract and check.
Confirm that endpoint masses are included or excluded as the original event requires.

Why the Rules Hold

The formulas follow from subtracting probabilities of nested events. For example, the event $\{X\le b\}$ contains every outcome in $\{X\le a\}$ when $a<b$. Removing the latter event leaves $\{a<X\le b\}$. Therefore,

P(a<X\le b)=P(X\le b)-P(X\le a)=F(b)-F(a).

To include $a$ instead, remove only outcomes with $X<a$. Since $P(X<a)=F(a-)$, this gives $P(a\le X\le b)=F(b)-F(a-)$. To make the upper endpoint strict, replace $P(X\le b)$ by $P(X<b)=F(b-)$. Applying both changes gives $P(a\le X<b)=F(b-)-F(a-)$. Finally, the outcomes with $X=a$ are those with $X\le a$ after removing those with $X<a$, so their probability is $F(a)-F(a-)$.

Worked Examples

Worked Example: A Claim Amount With an Atom at a Threshold

Consider a hypothetical insurance payment $X$, in dollars. Its distribution has probability 0.20 at zero, probability 0.30 at $100, and probability 0.50 spread uniformly over the interval from $0 to $200. The continuous component is uniform with total probability 0.50, so its density on $0<x<200$ is $0.50/200=0.0025$ per dollar. Find $P(50\le X\le100)$, $P(50<X\le100)$, and $P(X=100)$. No interest or mortality model is involved.

Method. Use the CDF endpoint rules. The continuous component contributes $0.0025$ times the interval length, and the atom at $100 is included only when the event includes 100.

First calculate the probability accumulated strictly below $50. It includes the mass at zero and the continuous probability from 0 to 50:

$$ F(50-)=0.20+(0.0025)(50)=0.20+0.125=0.325. $$

There is no atom at $50, so $F(50)=0.325$ as well. At $100, the probability strictly below 100 is

$$ F(100-)=0.20+(0.0025)(100)=0.20+0.250=0.450. $$

Including the atom at $100 gives $F(100)=0.450+0.300=0.750$. Therefore,

\begin{aligned} P(50\le X\le100)&=F(100)-F(50-)=0.750-0.325=0.425,\\ P(50<X\le100)&=F(100)-F(50)=0.750-0.325=0.425,\\ P(X=100)&=F(100)-F(100-)=0.750-0.450=0.300. \end{aligned}

The first two interval probabilities agree because there is no mass at $50. Both include the full atom at $100. As a check, the probability from the continuous component between 50 and 100 is $(0.0025)(50)=0.125$; adding the $100 atom gives $0.125+0.300=0.425$. The common mistake is to use $F(100-)$ as the upper-end value for an event that includes 100, which would omit the atom and give only 0.125.

Worked Example: Selecting Strict and Inclusive Endpoints From a PMF

A hypothetical discrete random variable $X$ records the number of hours a monitoring device remains in an alert state. Its distribution is $P(X=1)=0.10$, $P(X=3)=0.25$, $P(X=5)=0.40$, and $P(X=8)=0.25$. Find $P(1<X\le5)$, $P(1\le X<5)$, and $P(X=3)$.

Method. First accumulate the CDF at the relevant values and left limits. Since all probability is at the four listed values, the left limit immediately before a value excludes the probability at that value.

At 1, $F(1-)=0$ and $F(1)=0.10$. At 3, $F(3-)=0.10$ and $F(3)=0.10+0.25=0.35$. At 5, $F(5-)=0.10+0.25=0.35$ and $F(5)=0.35+0.40=0.75$.

For the first event, 1 is excluded and 5 is included:

P(1<X\le5)=F(5)-F(1)=0.75-0.10=0.65.

For the second event, 1 is included and 5 is excluded:

P(1\le X<5)=F(5-)-F(1-)=0.35-0=0.35.

For the point probability at 3:

$$ P(X=3)=F(3)-F(3-)=0.35-0.10=0.25. $$

Check against the PMF: $1<X\le5$ contains 3 and 5, with probability $0.25+0.40=0.65$; $1\le X<5$ contains 1 and 3, with probability $0.10+0.25=0.35$. The mistake to avoid is subtracting $F(1-)$ for the first event: that would include $X=1$, contrary to its strict lower endpoint.

Worked Example: Reading Left Limits From a Piecewise CDF

A hypothetical mixed random variable $X$ measures a processing delay in hours. Its CDF is

F(x)= \begin{cases} 0, & x<0,\\ 0.20+0.10x, & 0\le x<4,\\ 0.75+0.05(x-4), & 4\le x<9,\\ 1, & x\ge9. \end{cases}

Find $P(0<X\le4)$, $P(0\le X<4)$, $P(4\le X<7)$, and $P(X=4)$. No interest or mortality assumptions are used; all probabilities are specified by this CDF.

Method. Evaluate the CDF at each included endpoint and its left limit at each excluded endpoint. At zero, $F(0-)=0$, while $F(0)=0.20$. At four,

F(4-)=0.20+(0.10)(4)=0.60,\qquad F(4)=0.75.

At seven, the applicable second branch gives

$$ F(7-)=0.75+0.05(7-4)=0.75+0.15=0.90. $$

Now apply the endpoint rules:

\begin{aligned} P(0<X\le4)&=F(4)-F(0)=0.75-0.20=0.55,\\ P(0\le X<4)&=F(4-)-F(0-)=0.60-0=0.60,\\ P(4\le X<7)&=F(7-)-F(4-)=0.90-0.60=0.30,\\ P(X=4)&=F(4)-F(4-)=0.75-0.60=0.15. \end{aligned}

The first probability is the continuous increase from just above zero through four: the mass at zero is excluded, while the mass at four is included. The second includes the mass at zero but excludes the mass at four. The third includes the mass at four and the continuous increase between four and seven. As a check, that third probability is $0.15+0.05(7-4)=0.15+0.15=0.30$. A common error is to substitute $F(4)$ for $F(4-)$ in $P(0\le X<4)$, incorrectly including the atom at four.

When Endpoint Choices Make a Difference

An endpoint choice changes a probability only if there is positive probability at that endpoint. If $P(X=a)=0$, then $F(a)=F(a-)$, so including or excluding $a$ has no effect on the event probability. This is why interval conventions may appear interchangeable in a continuous model. They are not interchangeable as general rules: a distribution may be discrete or mixed, or the model may assign mass to a boundary such as zero.

For any calculation, it is useful to compare the interval formulas directly. For example,

P(a\le X\le b)-P(a<X\le b)=F(a)-F(a-)=P(X=a).

The difference is exactly the probability at the lower endpoint. Likewise,

P(a\le X\le b)-P(a\le X<b)=F(b)-F(b-)=P(X=b).

These comparisons provide an efficient check: changing an endpoint from open to closed should change the probability by the mass at that endpoint, not by an unrelated amount.

Common Exam Traps

Using $F(a)$ for a strict inequality. $F(a)=P(X\le a)$, not $P(X<a)$. For a strict upper bound, use $F(a-)$.
Subtracting the wrong lower-end value. An included lower endpoint requires subtracting $F(a-)$; an excluded lower endpoint requires subtracting $F(a)$.
Assuming every endpoint has zero probability. This may be true for a continuous distribution, but it cannot be assumed for a discrete or mixed distribution.
Treating $F(a-)$ as an ordinary value at a nearby point. It is the limit from below, and it excludes any probability concentrated exactly at $a$.
Using a density to account for an atom. A density describes continuous probability, not a point probability. Use the CDF difference $F(a)-F(a-)$ to identify the mass at $a$.

Key takeaway: Translate endpoint symbols before calculating. For $P(a\le X\le b)$, use $F(b)-F(a-)$; for $P(a<X<b)$, use $F(b-)-F(a)$. The left limit is always the probability of a strict inequality.

Check Your Understanding

For each problem, identify the endpoint convention before using the CDF or the stated distribution.

A discrete random variable has probabilities $P(X=1)=0.20$, $P(X=3)=0.50$, and $P(X=6)=0.30$. Find $P(1\le X<6)$. (Answer: 0.7000)
At $a=2$, a CDF has $F(2-)=0.42$ and $F(2)=0.62$. Find $P(X<2)$, $P(X=2)$, and $P(X\le2)$. (Answer: 0.4200, 0.2000, and 0.6200)
A CDF satisfies $F(2-)=0.28$ and $F(5)=0.81$. Find $P(2<X\le5)$. (Answer: 0.5300)
A random variable has CDF $F(x)=0$ for $x<0$, $F(x)=0.10+0.20x$ for $0\le x<3$, $F(x)=0.80+0.10(x-3)$ for $3\le x<5$, and $F(x)=1$ for $x\ge5$. Find $P(0\le X<3)$ and $P(3\le X<4)$. (Answer: 0.7000 and 0.2000)
For a mixed distribution, $F(2-)=0.30$, $F(2)=0.50$, $F(6-)=0.85$, and $F(6)=0.90$. Find $P(2\le X<6)$ and $P(2<X\le6)$. (Answer: 0.5500 and 0.4000)

What You'll Learn