Left Limits and One-Sided Behavior

Why Look at a CDF from the Left?

A cumulative distribution function records probabilities up to and including a threshold: $F(x)=P(X\le x)$. Sometimes the relevant question is instead how much probability lies strictly below $x$. The left-hand limit of the CDF provides exactly that quantity. Keeping it distinct from $F(x)$ is especially important at thresholds where the CDF has a jump.

For a fixed value $x$, imagine moving a threshold $t$ toward $x$ while keeping $t<x$. The values $F(t)$ approach a limit, written $F(x-)$. The notation is a one-sided limit, not a value of the CDF at a new argument. In particular, $F(x-)$ and $F(x)$ need not be equal.

Definition: Let $F$ be the CDF of a real-valued random variable $X$. Its left limit at $x$ is $$ F(x-)=\lim_{t\uparrow x}F(t)=\lim_{\substack{t\to x\\t<x}}F(t). $$ The limit exists for every real $x$. The notation $t\uparrow x$ means that $t$ approaches $x$ from below.

Existence and Probability Interpretation

A CDF is nondecreasing and bounded between 0 and 1. These properties ensure that its values have a limit from the left at every $x$. More importantly, the limit has a direct probability interpretation: it is the probability that $X$ is strictly less than $x$.

Theorem: For every real-valued random variable $X$ with CDF $F$, and every real $x$, $$ F(x-)=P(X<x). $$

To justify the formula, choose the increasing sequence $x_n=x-1/n$, for $n=1,2,\ldots$. It approaches $x$ from below. The events $A_n=\{X\le x_n\}$ increase with $n$, and their union is $\{X<x\}$: every value strictly below $x$ is eventually no greater than $x_n$, whereas no value equal to or above $x$ belongs to any $A_n$. Probability is continuous for increasing sequences of events, so $P(A_n)$ tends to $P(\bigcup_n A_n)$. Since $P(A_n)=F(x_n)$, this gives the desired probability limit.

Proof of the Left-Limit Formula

Existence of the left limit.
For fixed $x$, let $L=\sup\{F(t):t<x\}$. This supremum exists because the set is nonempty and $0\le F(t)\le1$. Given $\epsilon>0$, choose $t_0<x$ with $F(t_0)>L-\epsilon$. For $t_0<t<x$, monotonicity gives $L-\epsilon<F(t_0)\le F(t)\le L$. Thus $F(t)\to L$ as $t\uparrow x$, so $F(x-)=L$.

Choose thresholds approaching $x$.
Set $x_n=x-1/n$. Then $x_n$ increases to $x$, and $F(x_n)\to F(x-)$. To see the latter, $F(x_n)\le L$; also, for every $t<x$, eventually $t<x_n$, so $F(t)\le F(x_n)$. Consequently the limit of $F(x_n)$ is the supremum $L$.

Use continuity of probability from below.
The events $A_n=\{X\le x_n\}$ form an increasing sequence and $\bigcup_{n=1}^{\infty}A_n=\{X<x\}$. Countable additivity implies $P(A_n)\to P(\bigcup_n A_n)$: write the union as the disjoint union of $A_1$ and the successive differences $A_n\setminus A_{n-1}$, for $n\ge2$, and take partial sums of their probabilities. Therefore $$ F(x-)=\lim_{n\to\infty}F(x_n)=\lim_{n\to\infty}P(A_n)=P(X<x). $$

The strict inequality in $P(X<x)$ is essential. By definition, $F(x)=P(X\le x)$ includes outcomes equal to $x$, while the event represented by $F(x-)$ excludes them. The two quantities can agree, but they refer to different events.

Worked Calculations

Worked Example: A Finite Discrete Distribution

Suppose $X$ takes the values $-1$, $2$, and $4$, with probabilities $0.2$, $0.5$, and $0.3$, respectively. Find $F(2-)$, $F(2)$, and $F(3-)$.

At $x=2$, the outcomes strictly below 2 consist only of $X=-1$. Therefore $$ F(2-)=P(X<2)=P(X=-1)=0.2. $$ The CDF at 2 includes both $-1$ and 2, so $$ F(2)=P(X\le2)=0.2+0.5=0.7. $$ Thus the left limit and the CDF value differ at this threshold.

For $x=3$, the outcomes strictly below 3 are again $-1$ and 2. Hence $$ F(3-)=P(X<3)=0.2+0.5=0.7. $$ There is no possible value of $X$ between 2 and 4, so the CDF is constant as its argument approaches 3 from below. Also $F(3)=0.7$. This calculation illustrates that a left limit can equal the CDF value even when the random variable is discrete.

Worked Example: A Continuous Distribution

Let $X\sim\operatorname{Uniform}(0,1)$. Its CDF is $F(x)=0$ for $x<0$, $F(x)=x$ for $0\le x\le1$, and $F(x)=1$ for $x>1$. Find $F(0.4-)$, $F(0-)$, and $F(1-)$.

For values $t<0.4$ sufficiently close to 0.4, $F(t)=t$. Thus $$ F(0.4-)=\lim_{t\uparrow0.4}t=0.4. $$ In probability terms, $P(X<0.4)=0.4$. Below zero the CDF is constantly zero, so $F(0-)=0$. For $t<1$ sufficiently close to 1, $F(t)=t$, giving $F(1-)=1$. At each of these thresholds the left limit agrees with the CDF value.

Worked Example: A Mixed Distribution at a Jump and a Continuity Point

Suppose $X=0$ with probability $0.25$, and with probability $0.75$, $X$ is uniformly distributed on $[1,3]$. Its CDF is $$ F(x)= \begin{cases} 0, & x<0,\\ 0.25, & 0\le x<1,\\ 0.25+0.75(x-1)/2, & 1\le x<3,\\ 1, & x\ge3. \end{cases} $$ Find the left limits at 0, 1, and 2, and compare each with the CDF value.

For $t<0$, $F(t)=0$, so $F(0-)=0$, while $F(0)=0.25$. At 1, values just below 1 lie in the region where $F(t)=0.25$, so $F(1-)=0.25$. Substitution into the branch applying at 1 gives $$ F(1)=0.25+0.75(1-1)/2=0.25. $$ At 2, the CDF follows the linear branch on both sides of the threshold. Therefore $$ F(2-)=0.25+0.75(2-1)/2=0.25+0.375=0.625, $$ and $F(2)=0.625$ as well. The left limit differs from the CDF value at 0, but agrees with it at 1 and 2.

Left Limits at Continuity Points

A function is continuous at $x$ when its values approach its value at $x$ as the argument approaches from either side. In particular, if the CDF is continuous at $x$, its values approaching from below must tend to $F(x)$. This gives a useful way to calculate left limits without separately evaluating a probability.

Theorem: If the CDF $F$ is continuous at $x$, then $$ F(x-)=F(x). $$

Proof. Continuity at $x$ means that for every $\epsilon>0$, there is a $\delta>0$ such that $|F(t)-F(x)|<\epsilon$ whenever $|t-x|<\delta$. This bound applies in particular to $x-\delta<t<x$, so the values $F(t)$ approach $F(x)$ as $t\uparrow x$. By the definition of the left limit, $F(x-)=F(x)$. $\square$

The condition matters: monotonicity alone does not imply that the left limit equals the value at the threshold. At a jump point the two can differ, as the discrete and mixed examples show. The left-limit formula still applies there, so $F(x-)$ remains a well-defined probability even when $F$ is not continuous at $x$.

Common Mistake and Practical Use

A frequent error is to read $F(x-)$ as $P(X\le x)$ because both expressions involve the same threshold. The correction is to keep the event inequalities attached to the notation: $F(x)=P(X\le x)$, whereas the theorem gives $F(x-)=P(X<x)$. If a distribution places probability at $x$, these two events treat outcomes equal to $x$ differently.

When calculating a left limit from a piecewise CDF, do not automatically substitute $x$ into the branch that defines $F(x)$. Instead, inspect the formula for arguments just less than $x$, or use $P(X<x)$ directly when the distribution is given as a probability model. At a continuity point, either approach must yield $F(x-)=F(x)$; at a jump point, the left-hand branch determines the limit.

Key takeaway: The left limit $F(x-)$ exists for every real threshold and equals $P(X<x)$. It equals $F(x)$ whenever $F$ is continuous at $x$, but the two values need not agree at a jump point.

Check Your Understanding

Answer each question using the distinction between a threshold approached from below and the CDF value at the threshold.

State the probability represented by $F(x-)$, and explain why its inequality is strict.
If $X$ takes values 1 and 5 with probabilities 0.4 and 0.6, calculate $F(5-)$ and $F(5)$.
Why does a CDF have a left limit at every real $x$?
If $F$ is continuous at $x$, what must $F(x-)$ equal?
For a piecewise CDF, which values should you examine to calculate $F(x-)$: the formula at $x$, or the behavior for arguments less than $x$?