The Defining Properties of a CDF

Why CDFs Have Defining Properties

A cumulative distribution function (CDF) records how probability accumulates as a threshold moves from left to right. That interpretation imposes strict mathematical conditions on its graph. A candidate function that looks plausible, or stays between zero and one, is not necessarily a CDF: its values must change in a way that is consistent with probabilities of nested events.

The conditions are useful both for understanding genuine distributions and for checking proposed functions. The central idea is that the events $\{X\le x\}$ are nested as $x$ increases. Probability also behaves predictably when such events increase or decrease toward a limiting event.

Definition: For a real-valued random variable $X$, its cumulative distribution function is $$ F_X(x)=P(X\le x), \qquad -\infty<x<\infty. $$ Every CDF is nondecreasing, right-continuous, and has limits zero and one at negative and positive infinity, respectively.

Here, “real-valued” means that $X$ takes a finite real value with probability one. The limits at infinity express that a sufficiently low threshold eventually includes almost none of the probability, while a sufficiently high threshold eventually includes almost all of it.

The Four Conditions to Check

A function $F$ is a CDF precisely when it satisfies the following three defining properties. Its range condition, $0\le F(x)\le 1$, follows from these properties, but it is also useful to check explicitly when examining a candidate.

Nondecreasing: if $a<b$, then $F(a)\le F(b)$.
Right-continuous: for every real $a$, $\lim_{h\downarrow 0}F(a+h)=F(a)$.
Limits at infinity: $\lim_{x\to-\infty}F(x)=0$ and $\lim_{x\to+\infty}F(x)=1$.
Range: $F(x)$ must lie between zero and one for every real $x$.

Right-continuity concerns values approached from the right of a point. It does not require continuity from the left. A CDF can have a discontinuity, but the value assigned at that location must agree with its limiting value from the right.

Why the Conditions Hold

Each property follows from the defining probability $P(X\le x)$. In particular, event inclusion establishes monotonicity, while continuity of probability for increasing and decreasing sequences of events establishes the two limit statements and right-continuity.

Monotonicity from inclusion.
If $a<b$, then every outcome with $X\le a$ also has $X\le b$. Thus $\{X\le a\}\subseteq\{X\le b\}$, so taking probabilities gives $F_X(a)\le F_X(b)$.

Right-continuity from decreasing events.
Take any sequence $x_n$ that decreases to $a$. The events $\{X\le x_n\}$ decrease, and their intersection is $\{X\le a\}$. Continuity of probability for decreasing events gives $\lim_{n\to\infty}F_X(x_n)=F_X(a)$. This is right-continuity.

Limits from increasing and decreasing events.
The events $\{X\le n\}$ increase to the whole sample space because $X$ is finite, so $F_X(n)\to1$. The events $\{X\le -n\}$ decrease to the empty event, so $F_X(-n)\to0$. These are the limits at positive and negative infinity.

For completeness, continuity of probability for an increasing sequence $A_n$ states that $P(A_n)$ tends to $P(\bigcup_n A_n)$. For a decreasing sequence $B_n$, it states that $P(B_n)$ tends to $P(\bigcap_n B_n)$. The latter uses that probabilities are finite, which is true for probabilities. These facts explain why the corresponding set limits yield limits of CDF values.

The range condition follows as well: $F_X(x)$ is a probability, so $0\le F_X(x)\le1$. Conversely, a nondecreasing, right-continuous function with limits zero and one is a CDF of some real-valued random variable. Therefore, the defining conditions are a complete test, not merely necessary checks.

Testing a Candidate Function

A reliable check examines each condition separately. Do not infer that a function is a CDF just because it is increasing or has values between zero and one. A single failed condition is enough to rule it out.

Worked Example: Checking a Smooth Candidate

A proposed function is $$ G(x)= \begin{cases} 0, & x<0,\\ 1-\dfrac{1}{(1+x)^2}, & x\ge 0. \end{cases} $$ Determine whether it is a CDF, and calculate $G(1)$ and $G(3)$. Use the defining-properties test.

1. Range. For $x<0$, $G(x)=0$. For $x\ge0$, $1+x\ge1$, so $0<(1+x)^{-2}\le1$. Therefore $0\le 1-(1+x)^{-2}<1$, and all values are in $[0,1]$.

2. Monotonicity. The function is constant on $x<0$. For $0\le a<b$, $1+a<1+b$, so $(1+a)^{-2}>(1+b)^{-2}$. Subtracting from one gives $G(a)<G(b)$. Across the boundary, $G(a)=0$ for $a<0$, whereas $G(b)\ge0$ for $b\ge0$. Thus $G$ is nondecreasing everywhere.

3. Right-continuity. Away from zero the formula is continuous on each applicable interval. At zero, $G(0)=1-1=0$, and for $h>0$, $$ \lim_{h\downarrow0}G(0+h) =\lim_{h\downarrow0}\left(1-\frac{1}{(1+h)^2}\right)=0=G(0). $$ So the boundary also satisfies right-continuity.

4. Limits. For all negative $x$, $G(x)=0$, so the limit at negative infinity is zero. As $x\to+\infty$, $(1+x)^{-2}\to0$, so $G(x)\to1$.

All defining conditions hold, so $G$ is a CDF. The requested values are $$ G(1)=1-\frac{1}{(1+1)^2}=1-\frac14=\frac34=0.75, \qquad G(3)=1-\frac{1}{(1+3)^2}=1-\frac1{16}=\frac{15}{16}=0.9375. $$ As a check, both values lie in $[0,1]$, and $G(1)<G(3)$, consistent with monotonicity. Method and trap avoided: testing each defining property separately avoids accepting a candidate based only on its formula or its range.

Worked Example: A Decrease Rules Out a CDF

Consider the proposed function $$ H(x)= \begin{cases} 0, & x<0,\\ 0.8, & 0\le x<2,\\ 0.6, & 2\le x<4,\\ 1, & x\ge4. \end{cases} $$ Check its range, limits, right-continuity at the boundaries, and monotonicity. Is it a CDF?

Range. Its only values are $0$, $0.8$, $0.6$, and $1$, all in $[0,1]$.

Limits. It is zero for every $x<0$, so its limit at negative infinity is zero. It is one for every $x\ge4$, so its limit at positive infinity is one.

Right-continuity at the boundaries. At zero, $H(0)=0.8$, and values immediately to the right are $0.8$. At two, $H(2)=0.6$, and values immediately to the right are $0.6$. At four, $H(4)=1$, and values immediately to the right are $1$. It is constant between boundaries, so it is right-continuous everywhere.

Monotonicity. Compare the two ordered inputs $1$ and $2$. Since $1<2$, $$ H(1)=0.8,\qquad H(2)=0.6,\qquad 0.8>0.6. $$ This contradicts the requirement $H(1)\le H(2)$. Therefore $H$ is not a CDF, despite meeting the range, limit, and right-continuity checks. Method and trap avoided: checking only endpoint limits or monotonicity within each constant segment misses the downward change between segments.

Worked Example: Monotonicity Is Not Enough

A different candidate is $$ K(x)= \begin{cases} 0, & x<0,\\ 0.4, & 0\le x\le1,\\ 1, & x>1. \end{cases} $$ Determine which conditions it satisfies and whether it is a CDF.

Range and monotonicity. The values are $0$, $0.4$, and $1$, so the range is valid. The function never decreases as $x$ increases.

Limits. It is zero sufficiently far to the left and one sufficiently far to the right. Thus its limits are zero and one, respectively.

Right-continuity at $x=1$. The assigned value is $K(1)=0.4$. But for every positive integer $n$, $1+1/n>1$, so $K(1+1/n)=1$. Consequently, $$ \lim_{n\to\infty}K(1+1/n)=1\ne0.4=K(1). $$ Thus it is not right-continuous at one and is not a CDF. Method and trap avoided: monotonicity and correct limits do not repair a value at a boundary that disagrees with values approached from the right.

Common Mistakes and a Practical Check

A CDF audit is most dependable when each condition is checked explicitly. In particular, point values at boundaries matter: the definition uses $P(X\le x)$, so the value at a boundary cannot be chosen independently of the behavior immediately to its right.

Wrong calculation: For $K$, one might write, “$K(1)=0.4$, so the values just to the right of one are also $0.4$.” This incorrectly substitutes the value at one for values at inputs greater than one.
Correction: Take actual inputs to the right, such as $1+1/n$. Each gives $K(1+1/n)=1$, so the right-hand limiting value is $1$, not $K(1)=0.4$.
Checking only whether values lie between zero and one: The function $H$ passes that check but decreases from $0.8$ to $0.6$, which is impossible for a CDF.
Confusing right-continuity with ordinary continuity: A CDF need not be continuous from both sides. The condition to test is the limit from the right at every point.

For a piecewise candidate, identify each boundary and test the value there against the formula immediately to its right. Then check for any decrease across boundaries, verify the two limits, and confirm the range. If every defining condition holds, the function is a valid CDF; if one fails, it is not.

Key takeaway: A CDF cannot decrease, must match its limit from the right at each point, and must approach zero and one at negative and positive infinity. These properties follow from event inclusion and continuity of probability.

Check Your Understanding

For each candidate, apply the defining conditions rather than relying on its appearance.

Let $A(x)=0$ for $x<-2$, $A(x)=(x+2)/4$ for $-2\le x<2$, and $A(x)=1$ for $x\ge2$. Is $A$ a CDF, and what is $A(0)$? (Answer: Yes; $A(0)=0.5$.)
Let $B(x)=0$ for $x<0$, $0.25$ for $0\le x<3$, $0.20$ for $3\le x<5$, and $1$ for $x\ge5$. Identify the failed defining condition and give a pair of inputs demonstrating the failure. (Answer: Monotonicity fails; $B(2)=0.25>B(3)=0.20$.)
Let $C(x)=0$ for $x<0$, $0.3$ for $0\le x<2$, $0.7$ for $2\le x\le4$, and $0.9$ for $x>4$. Identify every defining condition it fails. (Answer: It fails right-continuity at $4$, since $C(4)=0.7$ but the right-hand limit is $0.9$; it also fails the limit-at-positive-infinity condition because that limit is $0.9$, not $1$.)
Let $D(x)=0$ for $x<1$, and $D(x)=1-1/x$ for $x\ge1$. Determine whether it is a CDF and calculate $D(3)$. (Answer: Yes; it is nondecreasing and right-continuous, its limits are $0$ and $1$, and $D(3)=2/3$.)
Suppose a candidate $Q$ is nondecreasing, right-continuous, and takes values in $[0,1]$, but $\lim_{x\to-\infty}Q(x)=0.1$ and $\lim_{x\to+\infty}Q(x)=1$. Is it a CDF? State the failed condition. (Answer: No; the limit at negative infinity must be $0$, not $0.1$.)

What You'll Learn