Cumulative Distribution Functions

From a Random Variable to a Cumulative Distribution Function

A cumulative distribution function records how much probability lies to the left of each real threshold. Its definition uses a non-strict inequality, so the value at a threshold includes any probability concentrated exactly there. This endpoint convention is essential: it determines the function’s continuity properties and the formulas for interval probabilities.

Throughout, let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, so $\mathbb{P}(\Omega)=1$, and let $X:\Omega\to\mathbb{R}$ be a real-valued measurable random variable. Its law is the probability measure $\mu_X$ on $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$ defined by $\mu_X(B)=\mathbb{P}(X\in B)$ for each Borel set $B$. In particular, $\mu_X((-\infty,x])=\mathbb{P}(X\le x)$. All probabilities below refer to $\mathbb{P}$, and all functions $F$ take real values.

Definition: The cumulative distribution function (CDF) of $X$, or equivalently of its law $\mu_X$, is $F:\mathbb{R}\to[0,1]$ defined by $F(x)=\mathbb{P}(X\le x)=\mu_X((-\infty,x])$ for every $x\in\mathbb{R}$.

We use continuity of probability for event sequences: if $A_n$ increase to $A$, then $\mathbb{P}(A_n)\to\mathbb{P}(A)$; if $A_n$ decrease to $A$, then $\mathbb{P}(A_n)\to\mathbb{P}(A)$, since $\mathbb{P}(\Omega)<\infty$. These continuity properties hold on the probability space just specified.

Worked Example: A Finite Probability Model

Let $\Omega=\{a,b,c,d\}$, with every subset measurable and probabilities $\mathbb{P}(\{a\})=1/8$, $\mathbb{P}(\{b\})=3/8$, $\mathbb{P}(\{c\})=1/4$, and $\mathbb{P}(\{d\})=1/4$. These weights sum to $1/8+3/8+1/4+1/4=1$. Set $X(a)=-1$, $X(b)=X(c)=0$, and $X(d)=2$.

For $x<-1$, no outcome has $X\le x$, so $F(x)=0$. For $-1\le x<0$, only $a$ is counted, giving $F(x)=1/8$. For $0\le x<2$, the outcomes $a,b,c$ are counted, giving $F(x)=1/8+3/8+1/4=3/4$. For $x\ge2$, all outcomes are counted and $F(x)=1$. Thus

F(x)= \begin{cases} 0,&x<-1,\\ \frac18,&-1\le x<0,\\ \frac34,&0\le x<2,\\ 1,&x\ge2. \end{cases}

Check: $F(0)=3/4=\mathbb{P}(X= -1)+\mathbb{P}(X=0)=1/8+3/8+1/4$. Also, $\mathbb{P}(-1<X\le2)=3/8+1/4+1/4=7/8$; the interval formula proved below gives $F(2)-F((-1)-)=1-0=1$ only if the lower endpoint is included incorrectly, so we instead use $F(2)-F(-1)=1-1/8=7/8$, as required.

Basic Properties and Their Proofs

Theorem: For the CDF $F$ of a real-valued random variable on a probability space, the following hold:

Monotonicity: if $x\le y$, then $F(x)\le F(y)$.
Right-continuity: if $x_n\downarrow x$, then $F(x_n)\to F(x)$.
Limits at infinity: $\lim_{x\to-\infty}F(x)=0$ and $\lim_{x\to+\infty}F(x)=1$.

Proof. Suppose $x\le y$. Then $\{X\le x\}\subseteq\{X\le y\}$, so monotonicity of probability gives $F(x)\le F(y)$. For right-continuity, let $x_n\downarrow x$. The events $\{X\le x_n\}$ decrease, and their intersection is $\{X\le x\}$: an outcome with $X\le x$ belongs to every event, while if $X>x$, then $x_n<X$ for some sufficiently large $n$. Continuity of probability from above therefore gives $F(x_n)\to F(x)$.

For the limit at positive infinity, the events $\{X\le n\}$, for positive integers $n$, increase to $\Omega$, because $X$ is real-valued. Continuity from below gives $F(n)\to\mathbb{P}(\Omega)=1$. Monotonicity then implies $\lim_{x\to+\infty}F(x)=1$: for any $\varepsilon>0$, choose $n$ with $F(n)>1-\varepsilon$; whenever $x\ge n$, $1\ge F(x)\ge F(n)>1-\varepsilon$.

For the limit at negative infinity, $\{X\le -n\}$ decrease to the empty set, since no real number is at most $-n$ for every positive integer $n$. Continuity from above gives $F(-n)\to0$. For any $\varepsilon>0$, choose $n$ with $F(-n)<\varepsilon$. If $x\le-n$, then $0\le F(x)\le F(-n)<\varepsilon$, proving the other limit. $\square$

The theorem’s hypotheses matter. The limits use that $X$ is real-valued: if extended-real-valued random variables were allowed, an outcome with value $+\infty$ would not belong to any $\{X\le n\}$, and the limit at positive infinity could be less than $1$. The use of continuity from above is valid because this is a probability measure, hence finite.

Worked Example: The Uniform Law on an Interval

Let $\Omega=[0,1]$, let $\mathcal{F}$ be its Borel sigma-algebra, and let $\mathbb{P}$ be Lebesgue measure restricted to $[0,1]$, which has total mass $1$. Define $X(\omega)=\omega$. For $x<0$, the event $\{X\le x\}$ is empty. For $0\le x\le1$, it is $[0,x]$, whose measure is $x$. For $x>1$, it is all of $[0,1]$. Therefore

F(x)= \begin{cases} 0,&x<0,\\ x,&0\le x\le1,\\ 1,&x>1. \end{cases}

At $x=0$, $F(0)=0$, and at $x=1$, $F(1)=1$; these values agree with the displayed middle branch. To check an interval with closed endpoints, take $a=1/4$ and $b=3/4$. The event $\{1/4\le X\le3/4\}$ is an interval of Lebesgue measure $3/4-1/4=1/2$. The left-limit formula below gives $F(3/4)-F((1/4)-)=3/4-1/4=1/2$. There is no endpoint mass: for example, $F(0)-\lim_{y\uparrow0}F(y)=0-0=0=\mathbb{P}(X=0)$, consistent with the singleton having Lebesgue measure zero.

Left Limits, Jumps, and Interval Probabilities

Although a CDF is right-continuous, it need not be continuous from the left. Define its left limit at $x$ by $F(x-)=\lim_{y\uparrow x}F(y)$. This limit exists because $F$ is nondecreasing and bounded. More precisely, it equals $\mathbb{P}(X<x)$, as the next result shows.

Proposition: For every $x\in\mathbb{R}$, $F(x-)=\mathbb{P}(X<x)$, and hence $F(x)-F(x-)=\mathbb{P}(X=x)$. For $a<b$, interval probabilities are $$ \begin{aligned} \mathbb{P}(a<X\le b)&=F(b)-F(a),\\ \mathbb{P}(a\le X\le b)&=F(b)-F(a-),\\ \mathbb{P}(a<X<b)&=F(b-)-F(a),\\ \mathbb{P}(a\le X<b)&=F(b-)-F(a-). \end{aligned} $$

Proof. Fix $x$, and consider the increasing events $A_n=\{X\le x-1/n\}$. Their union is $\{X<x\}$: every outcome with $X<x$ satisfies $X\le x-1/n$ for some sufficiently large $n$, and each $A_n$ is contained in $\{X<x\}$. Continuity from below gives $F(x-1/n)\to\mathbb{P}(X<x)$. Since $F$ is nondecreasing, this limit is the left limit $F(x-)$. The events $\{X<x\}$ and $\{X=x\}$ are disjoint and have union $\{X\le x\}$, so $F(x)=\mathbb{P}(X<x)+\mathbb{P}(X=x)$. Subtracting $F(x-)=\mathbb{P}(X<x)$ proves the jump formula.

For the interval identities, use nested events and subtract their probabilities. The event $\{X\le a\}$ is contained in $\{X\le b\}$, and their difference is $\{a<X\le b\}$, giving $F(b)-F(a)$. The event $\{X<a\}$ is contained in $\{X\le b\}$, and their difference is $\{a\le X\le b\}$, giving $F(b)-F(a-)$. Likewise, $\{X\le a\}\subseteq\{X<b\}$, with difference $\{a<X<b\}$, so its probability is $F(b-)-F(a)$. Finally, $\{X<a\}\subseteq\{X<b\}$, with difference $\{a\le X<b\}$, giving $F(b-)-F(a-)$. All subtractions are valid because the smaller event is contained in the larger one, whose probability is finite. $\square$

Worked Example: A Countable Probability Model

Let $\Omega=\mathbb{N}=\{1,2,\ldots\}$, with all subsets measurable and $\mathbb{P}(\{k\})=\frac13(\frac23)^{k-1}$. These probabilities sum to $1$, by the geometric series. Let $X(k)=k$. For $x<1$, $F(x)=0$. For $x\ge1$, let $n=\lfloor x\rfloor$, the greatest integer no larger than $x$. Since the inequality defining the CDF is non-strict, $X\le x$ means $X\in\{1,\ldots,n\}$, so

F(x)=\sum_{k=1}^{n}\frac13\left(\frac23\right)^{k-1} =1-\left(\frac23\right)^n,\qquad x\ge1.

For instance, $F(2)=1-(2/3)^2=1-4/9=5/9$, while $F(2-)=F(x)=1/3$ for $1\le x<2$. Thus $F(2)-F(2-)=5/9-1/3=2/9$, which agrees with $\mathbb{P}(X=2)=\frac13(2/3)=2/9$. For $a=2,b=4$, the probability of the strict-lower, closed-upper interval is $F(4)-F(2)=[1-(2/3)^4]-[1-(2/3)^2]=65/81-45/81=20/81$. Directly, the only included values are $3$ and $4$, with probabilities $4/27=12/81$ and $8/81$, totaling $20/81$. This independent sum checks the interval calculation.

A Common Endpoint Mistake

A common error is to use $F(x)=\mathbb{P}(X<x)$. The definition instead uses $X\le x$; the strict version is $F(x-)$. In the finite model above, $F(0)=3/4$, because the outcomes with values $-1$ and $0$ are included, whereas $F(0-)=1/8$. Confusing these values loses the mass $5/8$ at zero.

The same mistake changes interval formulas. In that model, $\mathbb{P}(-1<X\le2)=7/8$, not $F(2)-F((-1)-)=1-0=1$: the correct lower endpoint for a strict inequality is $F(-1)$, so the probability is $F(2)-F(-1)=1-1/8=7/8$. The general formula $\mathbb{P}(a<X\le b)=F(b)-F(a)$ follows by removing $\{X\le a\}$, not $\{X<a\}$. Keeping strict and non-strict inequalities attached to the correct CDF value prevents this off-by-an-atom error.

The defining properties provide useful consistency checks: a CDF cannot decrease, must be right-continuous, and must approach $0$ and $1$ at the two ends of the real line. Its jump at $x$ is exactly the probability of the event $X=x$. Its values and left limits then recover probabilities of intervals, with the endpoint conventions determining whether $F$ or $F(\,\cdot\,-)$ appears.

Check Your Understanding

For each question, track whether the endpoint inequality is strict or non-strict.

Show that if $x\le y$, then $\{X\le x\}\subseteq\{X\le y\}$, and deduce the corresponding inequality for $F$.
For a real-valued random variable, explain why $\{X\le n\}$ increases to $\Omega$ as the positive integer $n$ tends to infinity.
Express $\mathbb{P}(a<X<b)$ using $F$ and left limits, for $a<b$.
If $F(x)=0.62$ and $F(x-)=0.47$, what is $\mathbb{P}(X=x)$?
In the countable model above, compute $\mathbb{P}(1<X\le3)$ using both the CDF and the point probabilities.

Cumulative Distribution Functions

What You'll Learn