• 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 distribution1 is known as Benford’s law.

1. Construct a table specifying the distribution of $$X$$, and the corresponding spinner.

2. 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.

1. Verify that $$p_X$$ is a valid pmf.

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

3. Construct a table and spinner corresponding to the distribution of $$X$$.

4. 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.)

5. Use simulation to approximate the long run average value of $$X$$, and interpret this value.

6. 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.

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

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

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

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

5. Find the pmf of $$X$$.

6. Construct a table, plot, and spinner representing the distribution of $$X$$.

7. Make an educated guess for the long run average value of $$X$$.

8. 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.

1. You can find a nice explanation of Benford’s law here.↩︎