How to Find a PMF and CDF

A probability mass function (PMF) describes the probabilities of a discrete random variable. A cumulative distribution function (CDF) describes the probability that a random variable is less than or equal to a given number. The PMF is used only for discrete variables, while every random variable has a CDF.

The usual workflow is: identify the possible values of the random variable, find the probability of each value, and then accumulate those probabilities to obtain the CDF.

PMF definition

If $X$ is a discrete random variable, its PMF is

$$p_X(x)=P(X=x).$$

A valid PMF must satisfy two conditions:

$$p_X(x)\ge 0$$

for every $x$, and

$$\sum_x p_X(x)=1.$$

The sum is taken over all possible values of $X$. Values that $X$ cannot take have probability zero and may be omitted from the table.

CDF definition

The CDF of any random variable $X$ is

$$F_X(x)=P(X\le x).$$

For a discrete random variable, obtain it from the PMF by summing all probabilities at or below $x$:

$$F_X(x)=\sum_{t\le x}p_X(t).$$

The phrase “at or below” matters: the CDF uses $X\le x$, not $X<x$.

A CDF always has values between zero and one, is nondecreasing, approaches zero toward the left, and approaches one toward the right.

Worked example: two coin tosses

Suppose a fair coin is tossed twice, and let $X$ be the number of heads. The possible values are $0$, $1$, and $2$.

The four equally likely outcomes are

$$TT,\quad HT,\quad TH,\quad HH.$$

Here $T$ means tails and $H$ means heads. Count the outcomes corresponding to each value of $X$:

$x$	Outcomes	PMF $p_X(x)=P(X=x)$
0	$TT$	$1/4$
1	$HT,TH$	$2/4=1/2$
2	$HH$	$1/4$

Therefore,

$$p_X(x)= \begin{cases} \frac14, & x=0,\\ \frac12, & x=1,\\ \frac14, & x=2,\\ 0, & \text{otherwise}. \end{cases}$$

The probabilities add to one:

$$\frac14+\frac12+\frac14=1.$$

Now construct the CDF by accumulating the PMF values.

• If $x<0$, no possible value is at most $x$, so $F_X(x)=0$.
• If $0\le x<1$, only the value $0$ is included, so $F_X(x)=1/4$.
• If $1\le x<2$, the values $0$ and $1$ are included, so $F_X(x)=1/4+1/2=3/4$.
• If $x\ge2$, all possible values are included, so $F_X(x)=1$.

Thus,

$$F_X(x)= \begin{cases} 0, & x<0,\\ \frac14, & 0\le x<1,\\ \frac34, & 1\le x<2,\\ 1, & x\ge2. \end{cases}$$

The CDF is a step function. It jumps by $1/4$ at $x=0$, by $1/2$ at $x=1$, and by $1/4$ at $x=2$. In general, the jump at a point $x$ equals the PMF value $p_X(x)$.

Finding probabilities from the CDF

Once the CDF is known, you can calculate interval probabilities. For example,

$$P(a<X\le b)=F_X(b)-F_X(a).$$

For the coin-toss example,

$$P(0<X\le2)=F_X(2)-F_X(0)=1-\frac14=\frac34.$$

This correctly counts the cases $X=1$ and $X=2$ while excluding $X=0$.

For a discrete variable, a single-point probability can also be recovered from the CDF using the left-hand limit:

$$p_X(x)=F_X(x)-F_X(x^-),$$

where $F_X(x^-)$ denotes the value approached from the left. In the example, the jump at $x=1$ is

$$p_X(1)=F_X(1)-F_X(1^-)=\frac34-\frac14=\frac12.$$

If the variable is absolutely continuous

If $X$ is absolutely continuous—that is, its distribution has a probability density function (PDF)—then it does not have a PMF in the usual sense. Individual points have probability zero:

$$P(X=x)=0.$$

Its CDF is found by integrating its PDF $f_X(x)$:

$$F_X(x)=P(X\le x)=\int_{-\infty}^{x}f_X(t)\,dt.$$

Not every continuous random variable has a PDF, but the following formula applies whenever a PDF exists.

For example, if $X$ is uniform on $[0,2]$, its density is

$$f_X(x)= \begin{cases} \frac12, & 0\le x\le2,\\ 0, & \text{otherwise}. \end{cases}$$

Integrating gives

$$F_X(x)= \begin{cases} 0, & x<0,\\ \frac{x}{2}, & 0\le x\le2,\\ 1, & x>2. \end{cases}$$

For instance,

$$P(0.5\le X\le1.5)=F_X(1.5)-F_X(0.5) =\frac{1.5}{2}-\frac{0.5}{2}=\frac12.$$

A mixed random variable can have both point masses and a continuous component. In that case, the CDF still works, but there is no single PMF that describes the entire distribution.

Common mistakes

Confusing PMF and CDF

The PMF gives the probability of one exact value, $P(X=x)$. The CDF gives the probability of all values up to a point, $P(X\le x)$.

Forgetting cumulative addition

For a discrete variable, the CDF is not usually equal to the PMF. At $x=2$, for example, it includes the probabilities of $X=0$, $X=1$, and $X=2$.

Using a PMF for a continuous variable

For an absolutely continuous variable, use a PDF and integrate it. More generally, a continuous distribution need not have a PDF. The probability at a single point is zero, even though the density at that point may be positive.

Ignoring intervals between possible values

If a discrete variable takes integer values, $F_X(x)$ is constant between consecutive integers. For example, if $1\le x<2$, the CDF includes the probability of $X=1$ but not $X=2$.

Failing to check the total

A proposed discrete distribution is invalid if its PMF values do not sum to one. This check often catches missing outcomes or incorrect counting.