Endpoint Inclusion in Interval Probabilities

Why Endpoints Matter

Two intervals with the same numerical endpoints can describe different events. The difference is whether either endpoint is included. If a random variable has positive probability at an endpoint, including or excluding that endpoint changes the probability; if it has no point mass there, that particular inclusion choice makes no difference. The CDF records probabilities of events extending up to a point, while its left limit records probabilities strictly below that point.

For \(a<b\), the probability of \((a,b]\) is given by a difference of CDF values. The other three endpoint conventions require either replacing a CDF value with a left limit or accounting for the point mass at an endpoint. Keeping track of what each endpoint includes makes these formulas systematic rather than a matter of memorization.

Definition: The cumulative distribution function of a real-valued random variable \(X\) is \(F(x)=P(X\le x)\). Its left limit at \(x\) is denoted by \(F(x-)\), and satisfies \(F(x-)=P(X<x)\). The point mass at \(x\) is \(P(X=x)=F(x)-F(x-)\).

Left Limits and Endpoint Point Masses

The identity \(F(x-)=P(X<x)\) connects a one-sided limit of the CDF to a strict inequality. To see why it holds, take any increasing sequence \(x_n\) with \(x_n<x\) and \(x_n\to x\). The events \(\{X\le x_n\}\) increase, and their union is exactly \(\{X<x\}\). Indeed, if \(X<x\), then \(x_n\ge X\) for some sufficiently large \(n\); and every outcome in \(\{X\le x_n\}\) has \(X<x\). By continuity of probability for an increasing sequence of events,

F(x-)=\lim_{n\to\infty}F(x_n) =\lim_{n\to\infty}P(X\le x_n) =P(X<x).

Because \(\{X<x\}\) is contained in \(\{X\le x\}\), their probability difference is the probability of their set difference, \(\{X=x\}\). Thus

P(X=x)=P(X\le x)-P(X<x)=F(x)-F(x-).

This difference is the jump of the CDF at \(x\). When it is positive, \(x\) is an atom of the distribution. When it is zero, \(F(x)=F(x-)\), so including or excluding \(x\) does not change a probability.

Four Interval Formulas

For finite real numbers \(a<b\), each interval probability can be obtained by taking a larger event and removing the part that is not wanted. The resulting formulas are:

Theorem: If \(a<b\), then

\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}

Here is a direct derivation. The event \(\{a<X\le b\}\) is \(\{X\le b\}\) with \(\{X\le a\}\) removed. The event \(\{a\le X\le b\}\) is \(\{X\le b\}\) with \(\{X<a\}\) removed. Similarly, \(\{a<X<b\}\) is \(\{X<b\}\) with \(\{X\le a\}\) removed, and \(\{a\le X<b\}\) is \(\{X<b\}\) with \(\{X<a\}\) removed. In each case, the removed event is contained in the larger event, so subtracting its probability gives the probability of the difference. Substituting \(P(X\le t)=F(t)\) and \(P(X<t)=F(t-)\) gives all four identities. The events involved are measurable because \(X\) is a random variable and the intervals are Borel sets.

The formulas also show precisely how the endpoint choices affect the result. Starting with \(P(a<X\le b)=F(b)-F(a)\), including \(a\) adds \(P(X=a)\), while excluding \(b\) subtracts \(P(X=b)\). Including \(a\) and excluding \(b\) therefore gives

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

This adjustment is valid whether or not either endpoint has a point mass: a term is simply zero when the corresponding endpoint has no mass.

Interval	Probability	Endpoint effect relative to \((a,b]\)
\((a,b]\)	\(F(b)-F(a)\)	Baseline: excludes \(a\), includes \(b\)
\([a,b]\)	\(F(b)-F(a-)\)	Adds the mass at \(a\)
\((a,b)\)	\(F(b-)-F(a)\)	Removes the mass at \(b\)
\([a,b)\)	\(F(b-)-F(a-)\)	Adds the mass at \(a\) and removes the mass at \(b\)

Worked Examples

Worked Example: A Discrete Distribution With Mass at Both Endpoints

Suppose \(X\) takes the values \(0,1,3\), with probabilities \(0.2,0.5,0.3\), respectively. Find the probabilities of the four intervals with endpoints \(a=0\) and \(b=3\).

The CDF values and left limits needed are \(F(0-)=0\), \(F(0)=0.2\), \(F(3-)=0.7\), and \(F(3)=1\). For example, \(F(3-)=0.7\) because only the masses at 0 and 1 are strictly below 3. Substitution gives

\begin{aligned} P(0<X\le 3)&=F(3)-F(0)=1-0.2=0.8,\\ P(0\le X\le 3)&=F(3)-F(0-)=1-0=1,\\ P(0<X<3)&=F(3-)-F(0)=0.7-0.2=0.5,\\ P(0\le X<3)&=F(3-)-F(0-)=0.7-0=0.7. \end{aligned}

A direct check confirms each result. The first interval contains the values 1 and 3, with total mass \(0.5+0.3=0.8\). The closed interval contains all three possible values. The open interval contains only 1, and \([0,3)\) contains 0 and 1, with total mass \(0.2+0.5=0.7\). In particular, changing the inclusion of 0 changes the probability by \(0.2\), and changing the inclusion of 3 changes it by \(0.3\).

Worked Example: An Atom Next to a Continuous Component

Let \(X\) have a mixture distribution: with probability \(0.25\), it equals 1; with probability \(0.75\), it is uniformly distributed on \([0,2]\). Find the four interval probabilities for \(a=1\) and \(b=1.5\).

The continuous component contributes \(0.75\) times the uniform probability, and the atom contributes \(0.25\) at 1. Just below 1, the continuous component has accumulated \(1/2\) of its probability, so \(F(1-)=0.75(1/2)=0.375\). At 1, the atom is included, giving \(F(1)=0.375+0.25=0.625\). At 1.5 the continuous component has accumulated \(1.5/2=0.75\) of its probability, and there is no atom there. Consequently,

$$ F(1.5-)=F(1.5)=0.75(0.75)+0.25=0.8125. $$

Using the interval formulas,

\begin{aligned} P(1<X\le 1.5)&=0.8125-0.625=0.1875,\\ P(1\le X\le 1.5)&=0.8125-0.375=0.4375,\\ P(1<X<1.5)&=0.8125-0.625=0.1875,\\ P(1\le X<1.5)&=0.8125-0.375=0.4375. \end{aligned}

As a check, the atom at 1 contributes \(0.25\) to the two intervals that include 1 and contributes nothing to the two that exclude it. The continuous component contributes \(0.75(0.5/2)=0.1875\) to the portion between 1 and 1.5. There is no mass at 1.5, so including or excluding that endpoint makes no difference.

Worked Example: Checking Endpoint Adjustments Against the Masses

Suppose \(X\) has the following probability distribution: \(P(X=0)=0.35\), \(P(X=2)=0.15\), \(P(X=3)=0.30\), \(P(X=5)=0.10\), and \(P(X=7)=0.10\). Find the four probabilities for \(a=2\) and \(b=5\).

The mass strictly below 2 is \(0.35\), and adding the mass at 2 gives \(F(2-)=0.35\) and \(F(2)=0.50\). The mass strictly below 5 is \(0.35+0.15+0.30=0.80\), and adding the mass at 5 gives \(F(5-)=0.80\) and \(F(5)=0.90\). Thus

\begin{aligned} P(2<X\le 5)&=0.90-0.50=0.40,\\ P(2\le X\le 5)&=0.90-0.35=0.55,\\ P(2<X<5)&=0.80-0.50=0.30,\\ P(2\le X<5)&=0.80-0.35=0.45. \end{aligned}

The mass strictly between the endpoints is \(P(X=3)=0.30\). The endpoint masses are \(0.15\) at 2 and \(0.10\) at 5. Therefore the four answers are, respectively, \(0.30+0.10=0.40\), \(0.15+0.30+0.10=0.55\), \(0.30\), and \(0.15+0.30=0.45\), agreeing with the CDF calculations.

When the Endpoints Are Equal

The four formulas above assume \(a<b\). If \(a=b\), the intervals must be considered from their definitions: \((a,a]\), \((a,a)\), and \([a,a)\) are empty, whereas \([a,a]\) contains exactly the value \(a\). Hence

P(a<X\le a)=P(a<X<a)=P(a\le X<a)=0, \qquad P(a\le X\le a)=F(a)-F(a-).

This is also a useful endpoint check. Applying the open-interval formula without first checking \(a<b\) would produce \(F(a-)-F(a)\), which is nonpositive and can be negative when there is an atom. The probability of an empty interval is zero, not a negative number.

Common Mistake and Practical Check

A frequent error is to use \(F(a)\) whenever a formula has a lower endpoint and \(F(b)\) whenever it has an upper endpoint, regardless of whether the interval includes those endpoints. For example, someone might reason that \(P(a\le X\le b)=F(b)-F(a)\). This subtracts the probability of \(X=a\), even though the interval is supposed to include that value. The correction is to remove only values strictly below \(a\), so the appropriate subtraction is \(F(a-)\). The correct expression is \(F(b)-F(a-)\).

A reliable check is to ask what happens if one endpoint is changed from excluded to included. Including \(a\) should add \(P(X=a)\); excluding \(b\) should subtract \(P(X=b)\). If the calculation does not change by the relevant endpoint mass, the wrong CDF value or left limit has likely been used. If there is no point mass at an endpoint, the two choices coincide there, which is why endpoint errors can go unnoticed for distributions with no atoms.

Key takeaway: Use \(F(x)\) for probabilities up to and including \(x\), and \(F(x-)\) for probabilities strictly below \(x\). For \(a<b\), the lower endpoint is represented by \(F(a)\) when excluded and \(F(a-)\) when included; the upper endpoint is represented by \(F(b)\) when included and \(F(b-)\) when excluded.

Check Your Understanding

For each question, identify the endpoint choices before substituting CDF values.

Write a CDF formula for \(P(a\le X<b)\), assuming \(a<b\).
If \(F(c-)=0.42\) and \(F(c)=0.57\), what is \(P(X=c)\)?
For an interval \((a,b]\), which endpoint point masses are included?
If \(P(X=b)=0\), explain why \(P(a<X\le b)=P(a<X<b)\).
What is \(P(a\le X\le a)\) in terms of \(F(a)\) and \(F(a-)\)?