# 14 Discrete Random Variables: Probability Mass Functions

- The
**(probability) distribution**of a random variable specifies the possible values of the random variable and a way of determining corresponding probabilities. - A
**discrete**random variable can take on only countably many isolated points on a number line. These are often counting type variables. Note that “countably many” includes the case of countably infinite, such as \(\{0, 1, 2, \ldots\}\). - We often specify the distribution of a discrete with a
*probability mass function*. - Certain common distributions have special names and properties.
**Do not confuse a random variable with its distribution.**- A random variable measures a numerical quantity which depends on the outcome of a random phenomenon
- The distribution of a random variable specifies the long run pattern of variation of values of the random variable over many repetitions of the underlying random phenomenon.

## 14.1 Probability mass functions

- The
**probability mass function (pmf)**(a.k.a., density (pdf)) of a*discrete*RV \(X\), defined on a probability space with probability measure \(\text{P}\), is a function \(p_X:\mathbb{R}\mapsto[0,1]\) which specifies each possible value of the RV and the probability that the RV takes that particular value: \(p_X(x)=\text{P}(X=x)\) for each possible value of \(x\).

**Example 14.1 **Let \(Y\) be the larger of two rolls of a fair four-sided die. Donny Dont says the following is the probability mass function of \(Y\); do you agree?

\[ p_Y(y) = \frac{2y-1}{16} \]

**Example 14.2 **Randomly select a county in the U.S. Let \(X\) be the leading digit in the county’s population. For example, if the county’s population is 10,040,000 (Los Angeles County) then \(X=1\); if 3,170,000 (Orange County) then \(X=3\); if 283,000 (SLO County) then \(X=2\); if 30,600 (Lassen County) then \(X=3\). The possible values of \(X\) are \(1, 2, \ldots, 9\). You might think that \(X\) is equally likely to be any of its possible values. However, a more appropriate model is to assume that \(X\) has pmf

\[ p_X(x) = \begin{cases} \log_{10}(1+\frac{1}{x}), & x = 1, 2, \ldots, 9,\\ 0, & \text{otherwise} \end{cases} \]

This distribution^{1} is known as Benford’s law.

- Construct a table specifying the distribution of \(X\), and the corresponding spinner.

- Find \(\text{P}(X \ge 3)\)

## 14.2 Poisson distributions

**Example 14.3 **Let \(X\) be the number of home runs hit (in total by both teams) in a randomly selected Major League Baseball game. Technically, there is no fixed upper bound on what \(X\) can be, so mathematically it is convenient to consider \(0, 1, 2, \ldots\) as the possible values of \(X\). Assume that the pmf of \(X\) is

\[ p_X(x) = \begin{cases} e^{-2.3} \frac{2.3^x}{x!}, & x = 0, 1, 2, \ldots\\ 0, & \text{otherwise.} \end{cases} \]

This is known as the Poisson(2.3) distribution.

Verify that \(p_X\) is a valid pmf.

Compute \(\text{P}(X = 3)\), and interpret the value as a long run relative frequency.

Construct a table and spinner corresponding to the distribution of \(X\).

Find \(\text{P}(X \le 13)\), and interpret the value as a long run relative frequency. (The most home runs ever hit in a baseball game is 13.)

Use simulation to approximate the long run average value of \(X\), and interpret this value.

Use simulation to approximate the variance and standard deviation of \(X\).

- Poisson distributions are often used to model random variables that count “relatively rare events”.
- A discrete random variable \(X\) has a
**Poisson distribution**with parameter \(\mu>0\) if its probability mass function \(p_X\) satisfies \[ p_X(x) = \frac{e^{-\mu}\mu^x}{x!}, \quad x=0,1,2,\ldots \] - The function \(\mu^x / x!\) defines the shape of the pmf. The constant \(e^{-\mu}\) ensures that the probabilities sum to 1.
- If \(X\) has a Poisson(\(\mu\)) distribution then \[\begin{align*} \text{Long run average value of $X$} & = \mu\\ \text{Variance of $X$} & = \mu\\ \text{SD of $X$} & = \sqrt{\mu} \end{align*}\]

## 14.3 Binomial distributions

**Example 14.4 **Five messages of roughly equal length are to be transmitted across a noisy communication system. Assume the probability of any single message being transmitted successfully is 0.75, and that messages are transmitted independently of each other.

Let \(X\) be the number of messages (out of 5) that are transmitted successfully.

Compute \(\text{P}(X=0)\).

Compute the probability that the first message is transmitted successfully but the rest are not.

Compute \(\text{P}(X=1)\).

Compute \(\text{P}(X=2)\).

Find the pmf of \(X\).

Construct a table, plot, and spinner representing the distribution of \(X\).

Make an educated guess for the long run average value of \(X\).

What does the random variable \(Y = 5 - X\) measure? What is the distribution of \(Y\)?

- A discrete random variable \(X\) has a
**Binomial distribution**with parameters \(n\), a nonnegative integer, and \(p\in[0, 1]\) if its probability mass function is \[\begin{align*} p_{X}(x) & = \binom{n}{x} p^x (1-p)^{n-x}, & x=0, 1, 2, \ldots, n \end{align*}\] - If \(X\) has a Binomial(\(n\), \(p\)) distribution then \[\begin{align*} \text{Long run average value of $X$} & = np\\ \text{Variance of $X$} & = np(1-p)\\ \text{SD of $X$} & = \sqrt{np(1-p)} \end{align*}\]
- Imagine a box containing tickets
- With \(p\) representing the proportion of tickets in the box labeled 1 (“success”); the rest are labeled 0 (“failure”).
- Randomly select \(n\) tickets from the box
*with replacement* - Let \(X\) be the number of tickets in the sample that are labeled 1.
- Then \(X\) has a Binomial(\(n\), \(p\)) distribution.
- Since the tickets are labeled 1 and 0, the random variable \(X\) which counts the number of successes is equal to the sum of the 1/0 values on the tickets.
- If the selections are made with replacement, the draws are independent, so it is enough to just specify the population proportion \(p\) without knowing the population size \(N\).

- The situation in the previous bullets and the message example involves a sequence of
**Bernoulli trials**.- There are only two possible outcomes, “success” (1) and “failure” (0), on each trial.
- The unconditional/marginal probability of success is the same on every trial, and equal to \(p\)
- The trials are independent.

- If \(X\) counts the number of successes in a
*fixed number*, \(n\), of Bernoulli(\(p\)) trials then \(X\) has a Binomial(\(n, p\)) distribution.