Probability Mass Function | PMF (2024)

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3.1.3 Probability Mass Function (PMF)

If $X$ is a discrete random variable then its range $R_X$ is a countable set, so, we can listthe elements in $R_X$. In other words, we can write$$R_X=\{x_1,x_2,x_3,...\}.$$Note that here $x_1, x_2,x_3,...$ are possible values of the random variable $X$. While randomvariables are usually denoted by capital letters, to represent the numbers in the range we usuallyuse lowercase letters such as $x$, $x_1$, $y$, $z$, etc. For a discrete random variable $X$, weare interested in knowing the probabilities of $X=x_k$. Note that here, the event $A=\{X=x_k\}$is defined as the set of outcomes $s$ in the sample space $S$ for which the corresponding valueof $X$ is equal to $x_k$. In particular,$$A=\{s \in S | X(s)=x_k\}.$$The probabilities of events $\{X=x_k\}$ are formally shown by the probability mass function (pmf) of $X$.

Definition
Let $X$ be a discrete random variable with range $R_X=\{x_1,x_2,x_3, ...\}$ (finite or countably infinite).The function$$P_X(x_k)=P(X=x_k), \textrm{ for } k=1,2,3,...,$$is called the probability mass function (PMF) of $X$.


Thus, the PMF is a probability measure that gives us probabilities of the possible values for arandom variable. While the above notation is the standard notation for the PMF of $X$, it mightlook confusing at first. The subscript $X$ here indicates that this is the PMF of the random variable $X$.Thus, for example, $P_X(1)$ shows the probability that $X=1$. To better understand all of the above concepts,let's look at some examples.


Example

I toss a fair coin twice, and let $X$ be defined as the number of heads I observe. Find the range of $X$,$R_X$, as well as its probability mass function $P_X$.

  • Solution
    • Here, our sample space is given by$$S=\{HH,HT,TH,TT\}.$$The number of heads will be $0$, $1$ or $2$. Thus$$R_X=\{0,1,2\}.$$Since this is a finite (and thus a countable) set, the random variable $X$ is a discrete randomvariable. Next, we need to find PMF of $X$. The PMF is defined as$$P_X(k)=P(X=k) \textrm{ for } k=0,1,2.$$We have$$P_X(0)=P(X=0)=P(TT)=\frac{1}{4},$$$$P_X(1) =P(X=1)=P(\{HT,TH\})=\frac{1}{4}+\frac{1}{4}=\frac{1}{2},$$$$P_X(2)=P(X=2)=P(HH)=\frac{1}{4}.$$

Although the PMF is usually defined for values in the range, it is sometimes convenient to extendthe PMF of $X$ to all real numbers. If $x \notin R_X$, we can simply write $P_X(x)=P(X=x)=0$. Thus,in general we can write\begin{equation}\nonumber P_X(x) = \left\{\begin{array}{l l}P(X=x) & \quad \text{if $x$ is in } R_X\\0 & \quad \text{otherwise}\end{array} \right.\end{equation}

To better visualize the PMF, we can plot it. Figure 3.1 shows the PMF of the aboverandom variable $X$. As we see, the random variable can take three possible values $0,1$ and $2$.The figure also clearly indicates that the event $X=1$ is twice as likely as the other two possiblevalues. The Figure can be interpreted in the following way: If we repeat the random experiment(tossing a coin twice) a large number of times, then about half of the times we observe $X=1$,about a quarter of times we observe $X=0$, and about a quarter of times we observe $X=2$.

Probability Mass Function | PMF (2)

For discrete random variables, the PMF is also called the probability distribution. Thus, whenasked to find the probability distribution of a discrete random variable $X$, we can do this byfinding its PMF. The phrase distribution function is usually reserved exclusively for the cumulativedistribution function CDF (as defined later in the book). The word distribution, on the otherhand, in this book is used in a broader sense and could refer to PMF, probability density function (PDF),or CDF.


Example

I have an unfair coin for which $P(H)=p$, where $0 < p < 1$. I toss the coin repeatedly until I observe aheads for the first time. Let $Y$ be the total number of coin tosses. Find the distribution of $Y$.

  • Solution
    • First, we note that the random variable $Y$ can potentially take any positive integer, so wehave $R_Y=\mathbb{N}=\{1,2,3,...\}$. To find the distribution of $Y$, we need tofind $P_Y(k)=P(Y=k)$ for $k=1,2,3,...$. We have$$P_Y(1) =P(Y=1)=P(H)=p,$$$$P_Y(2) =P(Y=2)=P(TH)=(1-p)p,$$$$P_Y(3) =P(Y=3)=P(TTH)=(1-p)^2 p,$$$$. \hspace{50pt} . \hspace{50pt} . \hspace{50pt} .$$$$. \hspace{50pt} . \hspace{50pt} . \hspace{50pt} .$$$$. \hspace{50pt} . \hspace{50pt} . \hspace{50pt} .$$$$P_Y(k) =P(Y=k)=P(TT...TH)=(1-p)^{k-1} p.$$Thus, we can write the PMF of $Y$ in the following way\begin{equation}\nonumber P_Y(y) = \left\{\begin{array}{l l}(1-p)^{y-1} p& \quad \text{for } y=1,2,3,...\\0 & \quad \text{otherwise}\end{array} \right.\end{equation}

Consider a discrete random variable $X$ with Range$(X)=R_X$. Note that by definition the PMFis a probability measure, so it satisfies all properties of a probability measure. In particular, we have

  • $0\leq P_X(x) \leq 1$ for all $x$, and
  • $\sum_{x \in R_X} P_X(x)=1$.

Also note that for any set $A \subset R_X$, we can find the probability that $X \in A$ using the PMF$$P(X \in A)=\sum_{x \in A} P_X(x).$$

Properties of PMF:

  • $0\leq P_X(x) \leq 1$ for all $x$;
  • $\sum_{x \in R_X} P_X(x)=1$;
  • for any set $A \subset R_X, P(X \in A)=\sum_{x \in A} P_X(x)$.


Example

For the random variable $Y$ in Example 3.4,

  1. Check that $\sum_{y \in R_Y} P_Y(y)=1$.
  2. If $p=\frac{1}{2}$, find $P(2\leq Y <5)$.
  • Solution
    • In Example 3.4, we obtained$$P_Y(k) =P(Y=k)=(1-p)^{k-1} p, \textrm{ for } k=1,2,3,...$$Thus,
      1. to check that $\sum_{y \in R_Y} P_Y(y)=1$, we have
        $\sum_{y \in R_Y} P_Y(y)$$= \sum_{k=1}^{\infty} (1-p)^{k-1} p$
        $= p \sum_{j=0}^{\infty} (1-p)^{j}$
        $= p \frac{1}{1-(1-p)}$$\hspace{30pt}\textrm{ Geometric sum}$
        $= 1$;

      2. if $p=\frac{1}{2}$, to find P$(2\leq Y < 5)$, we can write
        $\textrm{P}(2\leq Y < 5)$$= \sum_{k=2}^{4} P_Y(k)$
        $= \sum_{k=2}^{4} (1-p)^{k-1} p$
        $= \frac{1}{2}\bigg(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\bigg)$
        $=\frac{7}{16}$.

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