Testing Whether a Piecewise Function Is a CDF

Why Piecewise CDF Checks Need More Than Endpoint Values

A formula can look like a cumulative distribution function because its values stay between 0 and 1 and it eventually reaches 1. Those observations are not enough. A valid CDF must also be nondecreasing, right-continuous, and approach 0 in the far left tail. A single failure of any one condition rules out the entire candidate.

For a random variable $X$, its CDF is $F_X(x)=P(X\leq x)$. When a proposed formula changes at specified breakpoints, check the behavior inside each interval and at every breakpoint. In particular, the value assigned exactly at a breakpoint matters: it determines right-continuity and can affect the size of a jump.

Definition: A function $F$ is a valid CDF if it is nondecreasing, is right-continuous at every real number, and satisfies $$ \lim_{x\to-\infty}F(x)=0 \qquad\text{and}\qquad \lim_{x\to+\infty}F(x)=1. $$ For a valid CDF, $0\leq F(x)\leq1$ for every $x$. These conditions are both necessary and sufficient: any function satisfying them is the CDF of some probability distribution.

In practice, the range check is a useful quick screen, but it does not replace the four defining checks. For a piecewise function with finitely many breakpoints, monotonicity can be examined by checking each interval and comparing the values or one-sided limits at its breakpoints. Right-continuity must be checked separately at each breakpoint, including any endpoint where the formula changes.

A Reliable Checklist for Piecewise Candidates

Write the candidate as $F(x)$, and proceed in an order that makes failures easy to locate. The distinction between the left-hand limit and the value at a breakpoint is especially important.

Check the tails.
Evaluate the limiting value as $x$ tends to negative infinity and as $x$ tends to positive infinity. They must be 0 and 1, respectively.

Check the range.
Look for values below 0 or above 1. A candidate outside this range cannot be a CDF.

Check nondecreasing behavior.
Within each interval, verify that the formula does not decrease. At each breakpoint, compare the values on the left and right; the function cannot drop as $x$ increases.

Check right-continuity.
At every breakpoint $a$, verify that $F(a)=\lim_{x\to a^+}F(x)$. The value at $a$ must agree with the formula immediately to its right.

Inspect jumps.
Calculate $F(a)-F(a^-)$, where $F(a^-)=\lim_{x\to a^-}F(x)$. A jump cannot be negative; for a valid CDF, it is the probability $P(X=a)$.

A jump is allowed. A downward jump is not. The left-hand limit need not equal the value at the breakpoint: an upward jump is precisely how a CDF represents positive probability at one point. Right-continuity says the function’s value equals its limit from the right, not its limit from the left.

Why Jumps Have a Probability Interpretation

The relationship between a jump and a point probability follows directly from the definition of a CDF. The events $\{X<a\}$ and $\{X=a\}$ are disjoint, and together they make up $\{X\leq a\}$. Also, $F_X(a^-)=P(X<a)$, so subtracting the probability to the left of $a$ from the probability at or below $a$ isolates the probability at $a$.

F_X(a)-F_X(a^-) =P(X\leq a)-P(X<a) =P(X=a)\geq 0.

This also explains why a downward jump cannot occur in a CDF: it would imply a negative probability at that point. More generally, for $a<b$, the increase in a CDF is $F_X(b)-F_X(a)=P(a<X\leq b)$, which cannot be negative. This is the probability reason the CDF must be nondecreasing.

Worked Example: Checking a Candidate With Two Jumps

Method: Check the tails and range, then verify monotonicity and right-continuity interval by interval. Finally, compute the jumps. The common mistake to avoid is rejecting every discontinuous candidate: upward jumps are permitted.

Consider the candidate

F(x)= \begin{cases} 0, & x<0,\\ x/4, & 0\leq x<2,\\ 3/4, & 2\leq x<3,\\ 1, & x\geq3. \end{cases}

Tail limits. For all $x<0$, $F(x)=0$, so $\lim_{x\to-\infty}F(x)=0$. For all $x\geq3$, $F(x)=1$, so $\lim_{x\to+\infty}F(x)=1$.

Range and behavior within intervals. On $0\leq x<2$, $x/4$ increases from 0 toward $1/2$. On the other intervals the function is constant. All values lie between 0 and 1.

Breakpoints. At $x=0$, $F(0)=0$, matching the limit from the right. At $x=2$, the left-hand limit is $2/4=1/2$, while $F(2)=3/4$, which matches the value just to the right. At $x=3$, the left-hand limit is $3/4$, while $F(3)=1$, again matching the value just to the right. The function never decreases.

Jumps and conclusion. The jump at 2 is $3/4-1/2=1/4$, and the jump at 3 is $1-3/4=1/4$. Both are nonnegative. All CDF conditions hold, so this is a valid CDF. The jumps correspond to point probabilities of $1/4$ at each of 2 and 3.

Worked Example: Detecting a Decrease at a Breakpoint

Method: Compare the function immediately before and at each breakpoint. The common mistake to avoid is checking only that the limiting values at the two ends of the whole real line are correct.

Suppose

G(x)= \begin{cases} 0, & x<0,\\ x/2, & 0\leq x<1,\\ 2/5, & 1\leq x<2,\\ 1, & x\geq2. \end{cases}

The left tail is 0 and the right tail is 1. The function takes values between 0 and 1. Within each listed interval it is nondecreasing, and at $x=0$ it is right-continuous.

At $x=1$, however, the left-hand limit is $1/2$, but $G(1)=2/5$. Since $2/5<1/2$, the function falls as $x$ passes 1. For example, $G(9/10)=9/20$, while $G(1)=2/5=8/20$, so $G(9/10)>G(1)$. This directly contradicts the requirement that $G$ be nondecreasing.

The value at 1 does match the formula to its right, so right-continuity at 1 is not the problem. The jump calculation is $G(1)-G(1^-)=2/5-1/2=-1/10$, a negative amount. Therefore $G$ is not a valid CDF.

Worked Example: Separating Right-Continuity From Monotonicity

Method: Evaluate the function at a breakpoint and compare it with the limit from the right. The common mistake to avoid is checking the left-hand limit instead; a CDF is allowed to jump upward from the left.

Consider

H(x)= \begin{cases} 0, & x<0,\\ 1/5, & 0\leq x<1,\\ 1/2, & x=1,\\ 7/10, & 1<x<2,\\ 1, & x\geq2. \end{cases}

The tail limits are 0 and 1, and all values lie in $[0,1]$. The function is nondecreasing: its successive levels are 0, $1/5$, $1/2$, $7/10$, and 1. The value at 1 is also no smaller than the left-hand limit $1/5$, so there is no downward jump there.

But $\lim_{x\to1^+}H(x)=7/10$, while $H(1)=1/2$. Since $1/2\ne7/10$, the function is not right-continuous at 1 and therefore is not a valid CDF. The left-hand jump at 1 is not the defect; the mismatch with the right-hand limit is.

Worked Example: Catching an Incorrect Left-Tail Limit

Method: Test the limits at both infinities even when the candidate is nondecreasing and has plausible jumps. The common mistake to avoid is assuming that a function that eventually reaches 1 must be a CDF.

Let

K(x)= \begin{cases} 1/10, & x<0,\\ 1/2, & 0\leq x<1,\\ 1, & x\geq1. \end{cases}

The function is nondecreasing and right-continuous at 0 and 1. Its values lie in $[0,1]$, and its right-tail limit is 1. Its jumps are $1/2-1/10=2/5$ at 0 and $1-1/2=1/2$ at 1; neither jump is negative.

However, $K(x)=1/10$ for every $x<0$, so $\lim_{x\to-\infty}K(x)=1/10$, not 0. This alone rules it out as a CDF. A CDF must assign probability approaching zero to the event $X\leq x$ as the cutoff $x$ moves arbitrarily far to the left.

Common Exam Traps

Checking only the tails. Correct limits at infinity do not prevent a decrease or a right-continuity failure in the middle.
Treating every discontinuity as invalid. A CDF can have upward jumps. The jump at $a$ is $F(a)-F(a^-)$, and it must be nonnegative.
Using the wrong side for continuity. Right-continuity requires $F(a)=\lim_{x\to a^+}F(x)$. It does not require the value to match the left-hand limit.
Ignoring the value assigned at a breakpoint. A formula specified for $x<a$ does not determine $F(a)$. Read the piecewise conditions carefully and check the equality case.
Checking only the displayed levels. If a formula varies within an interval, it may decrease there even if the neighboring pieces appear to be in the right order.

Key takeaway: For each piecewise candidate, check both tail limits, the range, nondecreasing behavior within and across intervals, and right-continuity at every breakpoint. A positive jump is allowed and represents a point probability; a negative jump is impossible.

Check Your Understanding

For each candidate, decide whether it is a valid CDF. Check all relevant conditions, not just the values at the breakpoints.

A candidate is $F(x)=0$ for $x<-1$, $F(x)=2/5$ for $-1\leq x<2$, and $F(x)=1$ for $x\geq2$. Is it a valid CDF, and what are its jump sizes? (Answer: Yes; the jumps are $2/5$ at $-1$ and $3/5$ at 2.)
A candidate is $G(x)=0$ for $x<0$, $G(x)=x/3$ for $0\leq x<1$, $G(x)=1/4$ for $1\leq x<3$, and $G(x)=1$ for $x\geq3$. Identify the condition it fails. (Answer: It is not nondecreasing; $G(9/10)=3/10>G(1)=1/4$.)
A candidate is $H(x)=0$ for $x<0$, $H(x)=1/5$ for $0\leq x<2$, and $H(x)=3/5$ for $x\geq2$. Is it valid, and what is its jump at 2? (Answer: Yes; the jump at 2 is $3/5-1/5=2/5$, and the jump at 0 is $1/5$.)
A candidate is $J(x)=0$ for $x<0$, $J(x)=3/10$ for $0\leq x<1$, $J(1)=2/5$, $J(x)=7/10$ for $1<x<2$, and $J(x)=1$ for $x\geq2$. Which condition fails? (Answer: Right-continuity fails at 1 because $J(1)=2/5$, while $\lim_{x\to1^+}J(x)=7/10$.)
A candidate is $L(x)=1/8$ for $x<0$, $L(x)=1/2$ for $0\leq x<2$, and $L(x)=1$ for $x\geq2$. It is nondecreasing and right-continuous. Is it a valid CDF? (Answer: No; $\lim_{x\to-\infty}L(x)=1/8$, not 0.)

What You'll Learn