Strict and Weak Threshold Events

Why the Endpoint Matters

A cumulative distribution function gives the probability that a random variable is at most a specified value. But a question about a threshold can use either “less than” or “less than or equal to”; similarly, an upper-tail question can include or exclude equality. These distinctions make no difference at some thresholds and a substantial difference at others.

The key is to use both the CDF value at the threshold and its left limit. At a point with positive probability, the CDF includes the probability at that point, while the left limit does not. Keeping track of that distinction gives the probabilities of \(X\le x\), \(X<x\), \(X\ge x\), and \(X>x\) without ambiguity.

Definition: For a real-valued random variable \(X\), its cumulative distribution function is \(F(x)=P(X\le x)\). The left limit of \(F\) at \(x\) is denoted by \(F(x-)\) and is defined by \(F(x-)=\lim_{t\uparrow x}F(t)\). It equals the probability \(P(X<x)\).

What the CDF and Its Left Limit Count

The distinction is exactly which outcomes at the boundary are included. The event \(\{X\le x\}\) contains outcomes where \(X=x\), but the event \(\{X<x\}\) does not. Consequently, \(F(x)\) counts probability at \(x\), whereas \(F(x-)\) counts only probability strictly below \(x\).

The equality \(F(x-)=P(X<x)\) follows from continuity of probability for increasing sequences of events. To see this directly, let \(A_n=\{X\le x-1/n\}\). The events \(A_n\) increase as \(n\) increases, and their union is \(\{X<x\}\): every value strictly below \(x\) is at most \(x-1/n\) for some sufficiently large \(n\), while no value at or above \(x\) belongs to any \(A_n\). Therefore,

F(x-)=\lim_{n\to\infty}F(x-1/n) =\lim_{n\to\infty}P(A_n) =P\left(\bigcup_{n=1}^{\infty}A_n\right) =P(X<x).

This argument uses only countable additivity of probability, through continuity from below. It also explains why taking a left limit, rather than simply substituting \(x\) into the CDF, is necessary for a strict lower-tail event.

The Four Threshold Probabilities

Once \(F(x)\) and \(F(x-)\) are identified with their respective events, the four formulas follow from the events themselves and from complements.

Theorem: For any real-valued random variable \(X\) and any real number \(x\),

\begin{aligned} P(X\le x)&=F(x),\\ P(X<x)&=F(x-),\\ P(X\ge x)&=1-F(x-),\\ P(X>x)&=1-F(x). \end{aligned}

Proof. The first identity is the definition of \(F\), and the second follows from the left-limit argument above. The events \(\{X<x\}\) and \(\{X\ge x\}\) are disjoint and together form the entire sample space, so their probabilities sum to one. Likewise, \(\{X\le x\}\) and \(\{X>x\}\) partition the sample space. Thus,

P(X\ge x)=1-P(X<x)=1-F(x-), \qquad P(X>x)=1-P(X\le x)=1-F(x).

Each complement must exclude exactly the event that is being complemented. In particular, the complement of \(X<x\) is \(X\ge x\), not \(X>x\). \(\square\)

The formulas also show how much the answer changes when the endpoint is included. Subtracting the strict lower-tail probability from the weak lower-tail probability gives

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

The difference between \(P(X\ge x)\) and \(P(X>x)\) is the same point probability. Thus, all four probabilities can differ in pairs only when the threshold has positive probability. A threshold with \(P(X=x)=0\) is a continuity point of the CDF, and at such a point \(F(x)=F(x-)\).

Worked Examples

Worked Example: A Threshold at a Discrete Atom

Suppose \(X\) takes the values \(-1\), \(0\), and \(2\), with probabilities \(0.20\), \(0.50\), and \(0.30\), respectively. Find the four threshold probabilities at \(x=0\).

The probabilities sum to \(0.20+0.50+0.30=1\), so this is a valid distribution. The weak lower-tail event includes both \(-1\) and \(0\):

F(0)=P(X\le 0)=P(X=-1)+P(X=0)=0.20+0.50=0.70.

The strict lower-tail event includes only \(-1\), so \(F(0-)=P(X<0)=0.20\). Applying the upper-tail formulas gives

\begin{aligned} P(X\ge 0)&=1-F(0-)=1-0.20=0.80,\\ P(X>0)&=1-F(0)=1-0.70=0.30. \end{aligned}

As a check, \(P(X\le 0)-P(X<0)=0.70-0.20=0.50=P(X=0)\). The inclusive lower tail contains the entire mass at zero; the strict upper tail excludes it.

Worked Example: A Threshold With No Point Mass

Let \(X\sim\operatorname{Uniform}(0,4)\), and take \(x=1.5\). The CDF on the support is \(F(t)=t/4\) for \(0\le t\le4\). Therefore,

F(1.5)=\frac{1.5}{4}=0.375.

A uniform distribution assigns probability zero to any single point, so \(P(X=1.5)=0\) and \(F(1.5-)=F(1.5)=0.375\). The four probabilities are

\begin{aligned} P(X\le1.5)&=0.375, & P(X<1.5)&=0.375,\\ P(X\ge1.5)&=1-0.375=0.625, & P(X>1.5)&=1-0.375=0.625. \end{aligned}

The agreement within each pair is not because the inequalities have become interchangeable. It is because the event \(X=1.5\) has probability zero. For example, the interval from \(0\) to \(1.5\) has length \(1.5\) out of a total length \(4\), giving \(1.5/4=0.375\), consistent with both lower-tail calculations.

Worked Example: A Mixed Distribution at a Jump

Construct \(X\) by choosing, with probability \(1/4\), the value \(1\), and with probability \(3/4\), an independent draw from a uniform distribution on \((0,2)\). Find the four threshold probabilities at \(x=1\).

The point-mass component contributes probability \(1/4\) at \(1\). In the uniform component, the probability of a value below \(1\) is \(1/2\), since half of the interval \((0,2)\) lies below \(1\). Thus,

P(X<1)=\frac{3}{4}\cdot\frac{1}{2}=\frac{3}{8}, \qquad P(X=1)=\frac{1}{4}.

The continuous component assigns zero probability to the single value \(1\). Including the point mass gives

F(1)=P(X\le1)=\frac{3}{8}+\frac{1}{4} =\frac{3}{8}+\frac{2}{8}=\frac{5}{8}, \qquad F(1-)=\frac{3}{8}.

Now use the complements:

P(X\ge1)=1-\frac{3}{8}=\frac{5}{8}, \qquad P(X>1)=1-\frac{5}{8}=\frac{3}{8}.

Here \(P(X\le1)=P(X\ge1)=5/8\), while \(P(X<1)=P(X>1)=3/8\). These equalities arise from the particular mixture probabilities; they are not general identities. The jump size is \(F(1)-F(1-)=5/8-3/8=1/4\), exactly the probability placed at \(1\).

A Reliable Calculation Routine

At a threshold, first decide whether the requested event includes equality. Then use the matching CDF quantity for a lower tail; for an upper tail, take the complement of the corresponding lower-tail event. This routine keeps endpoint inclusion explicit.

Identify the inequality.
Record whether the event is strict or weak and whether it is a lower or upper tail.

Choose the lower-tail quantity.
Use \(F(x)\) for \(X\le x\), and \(F(x-)\) for \(X<x\).

Complement for an upper tail.
Use \(1-F(x-)\) for \(X\ge x\), and \(1-F(x)\) for \(X>x\).

Check the boundary mass if useful.
The difference \(F(x)-F(x-)\) must equal \(P(X=x)\), which is nonnegative.

Common Pitfall and Why the Distinction Matters

A frequent error is to write \(P(X<x)=F(x)\), reasoning that both expressions describe values “up to \(x\).” That reasoning silently includes the boundary in the CDF. The correction is to use \(P(X<x)=F(x-)\); at a jump, \(F(x)\) is larger by exactly \(P(X=x)\). For the same reason, \(P(X\ge x)\) is \(1-F(x-)\), not \(1-F(x)\).

This distinction matters whenever an outcome can equal the threshold with positive probability, as in a discrete or mixed distribution. In applications, a threshold may represent a cutoff for eligibility, a detection limit, or an event time. Whether equality meets the criterion changes which outcomes are counted. For a distribution with no point mass at the threshold, the strict and weak versions happen to agree in probability, but it is still important to match the formula to the event as stated.

Key takeaway: Use \(F(x)\) for “at most” and \(F(x-)\) for “strictly less than.” Their difference is the probability at the threshold. The matching upper-tail probabilities are their respective complements.

Check Your Understanding

For each question, pay close attention to whether equality at the threshold is included.

If \(F(3)=0.72\) and \(F(3-)=0.60\), find \(P(X\le3)\), \(P(X<3)\), \(P(X\ge3)\), \(P(X>3)\), and \(P(X=3)\).
A random variable has no point mass at \(x\). Which pairs among the four threshold probabilities must agree, and why?
Suppose \(P(X<5)=0.41\) and \(P(X=5)=0.09\). Find \(F(5)\), \(P(X\ge5)\), and \(P(X>5)\).
Explain why \(1-F(x)\) gives \(P(X>x)\), not \(P(X\ge x)\).
At a threshold where \(F(x)=F(x-)\), does that imply that \(X\) cannot equal \(x\)? Explain what it does imply.