Constructing a CDF from a PMF

From Probability Masses to a Cumulative Distribution

A probability mass function describes how probability is assigned to each possible value of a discrete random variable. A cumulative distribution function answers a different question: how much probability has accumulated at or below a given cutoff? Constructing the CDF is therefore an ordered accumulation task. The support values must be put in increasing order, and the masses at all values no greater than the cutoff must be added.

This procedure works for finite supports and for countably infinite supports, provided the probabilities form a valid PMF. It is useful whenever a problem gives probabilities value by value but asks for a function of a real cutoff, or asks for a probability involving a threshold. The key is to keep the cutoff $x$ distinct from the possible values of $X$.

Definition: For a discrete random variable $X$, its cumulative distribution function is $F_X(x)=P(X\leq x)$, for every real number $x$. If the possible values of $X$ are $a_1<a_2<\cdots$, with masses $p_j=P(X=a_j)$, then $$ F_X(x)=\sum_{j:\,a_j\leq x}p_j. $$ For a finite support, this is a finite sum. For a countably infinite support, it is the sum of all masses at support values no greater than $x$.

The inequality in the definition is $\leq$, so a mass at a support value is included when the cutoff equals that value. Between consecutive support values, the set of included support points does not change, so the CDF remains constant there. Below the smallest support value, the CDF is zero; at or above the largest value of a finite support, it is one.

A Reliable Construction Method

Start by checking that the listed masses are nonnegative and add to one. Then sort the support values from smallest to largest, even if the PMF is presented in a different order. Calculate running totals in that order. Each running total applies from its support value up to, but not including, the next support value. This produces a piecewise function defined for every real $x$, not just for the possible values of $X$.

Order the support.
Write the distinct possible values as $a_1<a_2<\cdots$. Preserve any gaps between values.

Check the PMF.
Confirm that each mass is nonnegative and that the total probability is one.

Accumulate the masses.
At $a_k$, add the masses at $a_1,\ldots,a_k$. This running total is $F_X(x)$ for $a_k\leq x<a_{k+1}$.

Include the outer intervals.
Set $F_X(x)=0$ below the smallest support value. For a finite support, set $F_X(x)=1$ at and above the largest support value.

Worked Examples

Worked Example: Constructing a CDF on a Finite Support

A hypothetical device records $X$, the number of warning signals in a test cycle. Its PMF is $P(X=0)=0.15$, $P(X=1)=0.25$, $P(X=3)=0.40$, and $P(X=5)=0.20$. Construct $F_X(x)$, then find $F_X(2.5)$ and $P(1<X\leq 5)$.

Method and PMF check. The support is already ordered: $0,1,3,5$. The masses are nonnegative, and $0.15+0.25+0.40+0.20=1.00$, so they define a valid PMF.

Running totals. At $x=0$, the accumulated mass is $0.15$. At $x=1$, it is $0.15+0.25=0.40$. At $x=3$, it is $0.40+0.40=0.80$. At $x=5$, it is $0.80+0.20=1.00$. Thus

F_X(x)= \begin{cases} 0, & x<0,\\ 0.15, & 0\leq x<1,\\ 0.40, & 1\leq x<3,\\ 0.80, & 3\leq x<5,\\ 1, & x\geq 5. \end{cases}

Since $2.5$ lies between $1$ and $3$, $F_X(2.5)=0.40$. For the event $1<X\leq 5$, the included support values are $3$ and $5$, so its probability is $0.40+0.20=0.60$. As a check using the CDF, $P(1<X\leq 5)=F_X(5)-F_X(1)=1.00-0.40=0.60$.

Common mistake avoided: Using only $P(X=3)=0.40$ for $F_X(3.5)$ would omit the masses at 0 and 1. The cutoff includes every support value at or below it, so $F_X(3.5)=0.15+0.25+0.40=0.80$.

Worked Example: Constructing a CDF on an Infinite Support

A hypothetical counter records a nonnegative integer $X$. Its PMF is $P(X=k)=0.4(0.6)^k$ for $k=0,1,2,\ldots$. Find a formula for $F_X(x)$, and calculate $F_X(2.4)$.

Method and PMF check. The support is the nonnegative integers. The masses are positive, and their total is the geometric series

\sum_{k=0}^{\infty}0.4(0.6)^k =0.4\frac{1}{1-0.6} =0.4\frac{1}{0.4} =1.

So this is a valid PMF. If $x<0$, no support value is at or below $x$, and $F_X(x)=0$. If $x\geq0$, the included values are $0,1,\ldots,\lfloor x\rfloor$, where $\lfloor x\rfloor$ is the greatest integer no greater than $x$. Accumulating those masses gives

F_X(x)=\sum_{k=0}^{\lfloor x\rfloor}0.4(0.6)^k =1-(0.6)^{\lfloor x\rfloor+1},\qquad x\geq 0.

This is a finite geometric sum: the first term is $0.4$, the common ratio is $0.6$, and the last included index is $\lfloor x\rfloor$. For $x=2.4$, $\lfloor 2.4\rfloor=2$. Therefore

$$ F_X(2.4)=0.4+0.4(0.6)+0.4(0.6)^2 =0.4+0.24+0.144 =0.784. $$

The closed-form expression gives the same result: $1-(0.6)^3=1-0.216=0.784$. The check confirms that the cutoff includes $0,1,2$, but not $3$.

Common mistake avoided: Using $1-(0.6)^{\lfloor x\rfloor}$ would give $1-(0.6)^2$ at $x=2.4$, omitting the mass at zero. Since the sum begins at $k=0$, it contains $\lfloor x\rfloor+1$ terms.

Worked Example: Using the CDF to Read an Interval Probability

A hypothetical inventory sample has a discrete measurement $X$ with PMF $P(X=-2)=\frac18$, $P(X=1)=\frac14$, $P(X=4)=\frac38$, and $P(X=10)=\frac14$. Construct its CDF and find $P(1<X\leq 10)$.

Method and PMF check. The support is ordered as $-2,1,4,10$. The mass total is $\frac18+\frac14+\frac38+\frac14=\frac18+\frac28+\frac38+\frac28=\frac88=1$. The successive cumulative totals are $\frac18$, $\frac18+\frac28=\frac38$, $\frac38+\frac38=\frac68=\frac34$, and $\frac34+\frac14=1$.

F_X(x)= \begin{cases} 0, & x<-2,\\ \frac18, & -2\leq x<1,\\ \frac38, & 1\leq x<4,\\ \frac34, & 4\leq x<10,\\ 1, & x\geq 10. \end{cases}

The event $1<X\leq10$ includes $X=4$ and $X=10$, with total probability $\frac38+\frac14=\frac38+\frac28=\frac58=0.625$. The CDF calculation agrees: $F_X(10)-F_X(1)=1-\frac38=\frac58$.

Common mistake avoided: The value $F_X(1)=\frac38$ includes the mass at $X=1$. Subtracting it removes that mass, as required by the strict inequality $1<X$. The final probability includes only the masses at 4 and 10.

Why the Accumulation Formula Holds

The formula for constructing a CDF follows directly from the event it measures. For a discrete random variable with support values $a_j$, the event $\{X\leq x\}$ consists exactly of the outcomes for which $X=a_j$ for some $a_j\leq x$. These events $\{X=a_j\}$ are mutually exclusive: $X$ cannot equal two distinct support values at once. The probability of their union is therefore the sum of their probabilities.

\{X\leq x\}=\bigcup_{j:\,a_j\leq x}\{X=a_j\}, \qquad P(X\leq x)=\sum_{j:\,a_j\leq x}P(X=a_j).

For an infinite support, the same reasoning uses a countable union of mutually exclusive events, whose probability is the sum of their probabilities. Because all masses are nonnegative and the total mass is one, the partial sums are well-defined and cannot exceed one. This also explains why the CDF cannot decrease as $x$ increases: a larger cutoff includes all the values counted by a smaller cutoff, and possibly additional ones.

Reading and Checking the Result

A constructed CDF should cover every real value of $x$, including values that are not in the support. It should also agree with the PMF when evaluated through cumulative totals. Useful checks are:

The first interval has CDF value zero below the smallest support point.
Each new interval adds exactly the mass at the support value just reached.
The running totals never decrease and never exceed one.
For a finite support, the CDF equals one from the largest support value onward.
For an infinite support, the CDF at a finite cutoff sums only the support values at or below that cutoff; it does not sum the entire PMF unless all support values are included.

A frequent exam error is to report the PMF value $P(X=x)$ when asked for $F_X(x)$. For instance, in the first example $P(X=3)=0.40$, but $F_X(3)=0.80$, because the CDF includes the probabilities at 0, 1, and 3. Another error is to place an interval boundary incorrectly: the mass at $a_k$ belongs in the CDF for $x=a_k$, so the interval with the new cumulative total begins at $a_k$, not immediately after it.

The distinction matters because a PMF answers a single-value question, while a CDF summarizes all probability up to a threshold. Once the support is ordered, the CDF is obtained by accumulating the PMF, one mass at a time. This construction is aligned with the SOA VEE Mathematical Statistics syllabus and provides a practical way to move between point probabilities and cumulative probabilities.

Check Your Understanding

For each problem, order the support before accumulating probabilities. Answers are provided for checking.

A discrete variable $X$ has support $-3,2,7$, with masses $0.20,0.50,0.30$, respectively. Find $F_X(4)$. (Answer: $0.7000$)
A variable $X$ has support $1,4,9$, with masses $0.10,0.35,0.55$. Write $F_X(x)$ as a piecewise function. (Answer: $0$ for $x<1$; $0.10$ for $1\leq x<4$; $0.45$ for $4\leq x<9$; $1$ for $x\geq9$)
A nonnegative integer-valued variable has $P(X=k)=0.25(0.75)^k$ for $k=0,1,\ldots$. Find $F_X(3.2)$. (Answer: $0.6836$)
A variable has masses $P(X=-1)=\frac16$, $P(X=2)=\frac12$, and $P(X=5)=\frac13$. Find $P(-1<X\leq5)$ using its CDF. (Answer: $\frac56$)
A variable has support $k=1,2,\ldots$ and PMF $P(X=k)=\frac13(\frac23)^{k-1}$. First verify the total mass is one, then find $F_X(4.5)$. (Answer: total mass $1$; $F_X(4.5)=\frac{65}{81}\approx0.8025$)

What You'll Learn