Why Tail Endpoints Matter
A cumulative distribution function records probabilities up to and including a threshold. That “including” is consequential: if a random variable can equal the threshold with positive probability, then \(P(X\le x)\) and \(P(X<x)\) differ. Likewise, \(P(X>x)\) and \(P(X\ge x)\) differ at the same points. Reliable tail calculations begin by identifying whether the endpoint belongs to the event.
For a continuous distribution, a single point has probability zero, so strict and inclusive versions often agree. That shortcut is not valid for a general distribution: discrete and mixed distributions can have jumps in their CDFs. The identities below work for all real-valued random variables and make the endpoint convention explicit.
CDF Values, Left Limits, and Jumps
The equality between the left limit and \(P(X<x)\) follows from continuity of probability for increasing events. For example, the events \(\{X\le x-1/n\}\) increase as \(n\) increases, and their union is \(\{X<x\}\). Consequently, their probabilities converge to \(P(X<x)\), giving the stated left limit. The jump formula then follows by subtracting \(P(X<x)\) from \(P(X\le x)\); the remaining event is \(\{X=x\}\).
The right-tail identities come from taking complements: the complement of \(\{X\le x\}\) is \(\{X>x\}\), while the complement of \(\{X<x\}\) is \(\{X\ge x\}\). A common convention defines the survival function as \(S(x)=P(X>x)=1-F(x)\). With this convention, \(S(x)\) gives the strict right tail; the inclusive right tail is \(P(X\ge x)=1-F(x-)\), not generally \(1-F(x)\).
Worked Example: Reading Both Tails at a CDF Jump
Suppose a CDF satisfies \(F(2-)=0.75\) and \(F(2)=0.82\). Find the probability at 2, each of the two left tails, and each of the two right tails.
The point probability is the jump: $$ P(X=2)=F(2)-F(2-)=0.82-0.75=0.07. $$
The strict and inclusive left tails are $$ P(X<2)=F(2-)=0.75,\qquad P(X\le 2)=F(2)=0.82. $$
The strict right tail is the survival-function value \(S(2)=1-F(2)\), and the inclusive right tail uses the left limit: $$ P(X>2)=1-0.82=0.18,\qquad P(X\ge 2)=1-0.75=0.25. $$
As a check, the inclusive tails \(P(X\le2)\) and \(P(X>2)\) sum to \(0.82+0.18=1\). The strict left and inclusive right tails also partition the outcomes: \(0.75+0.25=1\). The difference between the two right-tail probabilities is \(0.25-0.18=0.07\), exactly the mass at 2.
Intervals: Match Each Endpoint to the Event
The same endpoint rule applies when a probability is requested over an interval. For \(a<b\), subtracting CDF values gives the probability between the thresholds, but whether either endpoint is included determines whether a CDF value or its left limit is needed.
| Event | Probability from the CDF |
|---|---|
| \(a<X\le b\) | \(F(b)-F(a)\) |
| \(a\le X\le b\) | \(F(b)-F(a-)\) |
| \(a<X<b\) | \(F(b-)-F(a)\) |
| \(a\le X<b\) | \(F(b-)-F(a-)\) |
For the first row, \(\{a<X\le b\}\) is the part of \(\{X\le b\}\) left after removing \(\{X\le a\}\), so its probability is \(F(b)-F(a)\). To include \(a\), remove only \(\{X<a\}\), whose probability is \(F(a-)\). To exclude \(b\), use \(P(X<b)=F(b-)\) in place of \(F(b)\). These set differences establish all four formulas.
Worked Example: Choosing the Correct CDF Endpoints
Suppose \(F(3-)=0.35\), \(F(3)=0.40\), \(F(7-)=0.90\), and \(F(7)=0.94\). Calculate \(P(3<X<7)\) and \(P(3\le X\le7)\).
For the open interval at both ends, exclude probability at 7 by using \(F(7-)\), and exclude probability at or below 3 by subtracting \(F(3)\): $$ P(3<X<7)=F(7-)-F(3)=0.90-0.40=0.50. $$
For the closed interval, include probability at 7 using \(F(7)\), and include probability at 3 by subtracting only \(F(3-)\): $$ P(3\le X\le7)=F(7)-F(3-)=0.94-0.35=0.59. $$
The difference, \(0.59-0.50=0.09\), is the combined endpoint mass: \(P(X=3)=0.40-0.35=0.05\) and \(P(X=7)=0.94-0.90=0.04\). Their sum is \(0.05+0.04=0.09\), confirming the calculation.
Deriving the Tail Rules
The CDF convention fixes the endpoint algebra. In particular, subtracting the CDF value at \(x\) from 1 produces the strict right tail, whereas the left limit is required if equality at \(x\) should remain in the event.
To verify the first identity involving \(S(x-)\), take \(t\uparrow x\). Since \(S(t)=1-F(t)\), its left limit is \(1-F(x-)\), which is \(P(X\ge x)\) by the complement identity. For the second identity, \(\{X\ge x\}\) is the disjoint union of \(\{X=x\}\) and \(\{X>x\}\). Additivity therefore gives \(P(X\ge x)=P(X=x)+P(X>x)\). The jump formula identifies the difference with \(F(x)-F(x-)\).
Worked Example: A Discrete Distribution at a Threshold
Let \(X\) have probability \(0.2\) at \(-1\), \(0.3\) at \(0\), and \(0.5\) at \(2\). Calculate the tails at \(x=0\) and the probability of the interval \((-1,2]\).
At 0, the mass strictly below the threshold is \(0.2\), and the mass at or below it is \(0.2+0.3=0.5\). Thus \(F(0-)=0.2\) and \(F(0)=0.5\). The two right tails are $$ P(X>0)=1-F(0)=1-0.5=0.5,\qquad P(X\ge0)=1-F(0-)=1-0.2=0.8. $$
The difference, \(0.8-0.5=0.3\), is the probability at 0. For the interval, \(F(2)=1\) and \(F(-1)=0.2\), so $$ P(-1<X\le2)=F(2)-F(-1)=1-0.2=0.8. $$
This interval includes the mass at 0 and at 2, but not the mass at \(-1\). The included probabilities sum to \(0.3+0.5=0.8\), matching the CDF calculation.
Mixed Distributions and a Useful Endpoint Check
A distribution can have both a continuous component and an atom. Consider a random variable that is equal to 1 with probability \(0.4\), and otherwise has a Uniform distribution on \((0,2)\); the continuous component therefore has total probability \(0.6\). At the threshold 1, the continuous component contributes \(0.6(1/2)=0.3\) below 1 and \(0.3\) above 1. The atom adds \(0.4\) at 1. Consequently, $$ F(1-)=0.3,\qquad F(1)=0.3+0.4=0.7. $$
Worked Example: Strict and Inclusive Tails in a Mixed Law
Using the mixed distribution just described, find the probabilities \(P(X<1)\), \(P(X\le1)\), \(P(X>1)\), and \(P(X\ge1)\).
The left-tail probabilities are \(P(X<1)=F(1-)=0.3\) and \(P(X\le1)=F(1)=0.7\). The right-tail probabilities are $$ P(X>1)=1-F(1)=0.3,\qquad P(X\ge1)=1-F(1-)=0.7. $$
The strict tails are equal here because the continuous portions on either side each have probability \(0.3\). The inclusive tails are each \(0.7\) because they include the atom at 1. This example shows why knowing that a distribution has a continuous component does not justify treating its entire law as continuous.
Common Mistake: Using the Wrong Side of a Jump
A frequent error is to write \(P(X\ge x)=1-F(x)\). This subtracts all outcomes at or below \(x\), so it actually gives \(P(X>x)\). The correction is to use \(1-F(x-)\) for \(P(X\ge x)\). For example, if \(F(2-)=0.75\) and \(F(2)=0.82\), the incorrect expression gives \(0.18\), while the inclusive tail is \(1-0.75=0.25\).
A second version of the same mistake occurs for intervals: \(F(b)-F(a)\) is \(P(a<X\le b)\), not the probability of every interval with endpoints \(a\) and \(b\). Before subtracting, write the event in words: include or exclude the lower endpoint, and include or exclude the upper endpoint. Then choose \(F(a)\) or \(F(a-)\), and \(F(b)\) or \(F(b-)\), accordingly.
Check Your Understanding
For each question, identify the endpoint convention before substituting CDF values.
- If \(F(4-)=0.62\) and \(F(4)=0.70\), find \(P(X=4)\), \(P(X>4)\), and \(P(X\ge4)\).
- Write \(P(a<X\le b)\) in terms of \(F\), and explain why the lower endpoint uses \(F(a)\).
- If a random variable has no probability mass at \(x\), how do \(P(X>x)\) and \(P(X\ge x)\) compare?
- Given \(F(1-)=0.2\) and \(F(5)=0.9\), calculate \(P(1\le X\le5)\).
- A student claims that \(1-F(x)\) always gives the probability of being at least \(x\). State the correction and identify the event that \(1-F(x)\) does represent.