# 3.5: More Discrete Distributions (Special Topic)

- Page ID
- 276

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)## Negative Binomial Distribution

The geometric distribution describes the probability of observing the first success on the \(n^{th}\) trial. The negative binomial distribution is more general: it describes the probability of observing the \(k^{th}\) success on the \(n^{th}\) trial.

Example \(\PageIndex{1}\)

Each day a high school football coach tells his star kicker, Brian, that he can go home after he successfully kicks four 35 yard field goals. Suppose we say each kick has a probability \(p\) of being successful. If p is small - e.g. close to 0.1 - would we expect Brian to need many attempts before he successfully kicks his fourth field goal?

**Solution**

We are waiting for the fourth success (k = 4). If the probability of a success (p) is small, then the number of attempts (n) will probably be large. This means that Brian is more likely to need many attempts before he gets k = 4 successes. To put this another way, the probability of n being small is low.

To identify a negative binomial case, we check 4 conditions. The first three are common to the binomial distribution (see a similar guide for the binomial distribution on page 138).

Is it negative binomial? Four conditions to check.

- The trials are independent.
- Each trial outcome can be classified as a success or failure.
- The probability of a success (p) is the same for each trial.
- The last trial must be a success.

Exercise \(\PageIndex{1}\)

Suppose Brian is very diligent in his attempts and he makes each 35 yard field goal with probability \(p = 0.8\). Take a guess at how many attempts he would need before making his fourth kick.

**Answer**-
One possible answer: since he is likely to make each field goal attempt, it will take him at least 4 attempts but probably not more than 6 or 7.

Example \(\PageIndex{2}\)

In yesterday's practice, it took Brian only 6 tries to get his fourth field goal. Write out each of the possible sequence of kicks.

**Solution**

Because it took Brian six tries to get the fourth success, we know the last kick must have been a success. That leaves three successful kicks and two unsuccessful kicks (we label these as failures) that make up the first ve attempts. There are ten possible sequences of these first ve kicks, which are shown in Table 3.20. If Brian achieved his fourth success (k = 4) on his sixth attempt (n = 6), then his order of successes and failures must be one of these ten possible sequences.

Exercise \(\PageIndex{2}\)

Each sequence in Table \(\PageIndex{1}\) has exactly two failures and four successes with the last attempt always being a success. If the probability of a success is p = 0:8, find the probability of the first sequence.

**Answer**-
The first sequence: \(0.2 \times 0.2 \times 0.8 \times 0.8 \times 0.8 \times 0.8 = 0.0164\).

Kick | Attempt | |||||
---|---|---|---|---|---|---|

1 | 2 | 3 | 4 | 5 | 6 | |

1 |
F |
F |
\(\overset {1}{S}\) |
\(\overset {2}{S}\) |
\(\overset {3}{S}\) |
\(\overset {4}{S}\) |

2 |
F |
\(\overset {1}{S}\) |
F |
\(\overset {2}{S}\) |
\(\overset {3}{S}\) |
\(\overset {4}{S}\) |

3 |
F |
\(\overset {1}{S}\) |
\(\overset {2}{S}\) |
F |
\(\overset {3}{S}\) |
\(\overset {4}{S}\) |

4 |
F |
\(\overset {1}{S}\) |
\(\overset {2}{S}\) |
\(\overset {3}{S}\) |
F |
\(\overset {4}{S}\) |

5 |
\(\overset {1}{S}\) |
F |
F |
\(\overset {2}{S}\) |
\(\overset {3}{S}\) |
\(\overset {4}{S}\) |

6 |
\(\overset {1}{S}\) |
F |
\(\overset {2}{S}\) |
F |
\(\overset {3}{S}\) |
\(\overset {4}{S}\) |

7 |
\(\overset {1}{S}\) |
F |
\(\overset {2}{S}\) |
\(\overset {3}{S}\) |
F |
\(\overset {4}{S}\) |

8 |
\(\overset {1}{S}\) |
\(\overset {2}{S}\) |
F |
F |
\(\overset {3}{S}\) |
\(\overset {4}{S}\) |

9 |
\(\overset {1}{S}\) |
\(\overset {2}{S}\) |
F |
\(\overset {3}{S}\) |
F |
\(\overset {4}{S}\) |

10 |
\(\overset {1}{S}\) |
\(\overset {2}{S}\) |
\(\overset {3}{S}\) |
F |
F |
\(\overset {4}{S}\) |

If the probability Brian kicks a 35 yard field goal is p = 0:8, what is the probability it takes Brian exactly six tries to get his fourth successful kick? We can write this as

\[P \text {(it takes Brian six tries to make four field goals)}\]

\[= P \text {(Brian makes three of his first ve field goals, and he makes the sixth one)}\]

\[= P \text {(1st sequence OR 2nd sequence OR} \dots \text{OR 10th sequence)}\]

where the sequences are from Table 3.20. We can break down this last probability into the sum of ten disjoint possibilities:

\[P \text {(1st sequence OR 2nd sequence OR } \dots \text{ OR 10th sequence)}\]

\[= P(\text {1st sequence}) + P(\text {2nd sequence}) + \dots + P(\text {10th sequence})\]

The probability of the first sequence was identified in Exercise 3.57 as 0.0164, and each of the other sequences have the same probability. Since each of the ten sequence has the same probability, the total probability is ten times that of any individual sequence.

The way to compute this negative binomial probability is similar to how the binomial problems were solved in Section 3.4. The probability is broken into two pieces:

\[P\text {(it takes Brian six tries to make four field goals)}\]

\[= \text {[Number of possible sequences]} \times P \text {(Single sequence)}\]

Each part is examined separately, then we multiply to get the final result.

We first identify the probability of a single sequence. One particular case is to first observe all the failures (n - k of them) followed by the k successes:

\[\begin{align} P \text {(Single sequence)} \\[5pt] &= P \text {(n - k failures and then k successes)} \\[5pt] &= (1- p)^{n-k}p^k \end{align}\]

We must also identify the number of sequences for the general case. Above, ten sequences were identi ed where the fourth success came on the sixth attempt. These sequences were identified by fixing the last observation as a success and looking for all the ways to arrange the other observations. In other words, how many ways could we arrange \(k - 1\) successes in \(n - 1\) trials? This can be found using the n choose k coefficient but for n - 1 and k - 1 instead:

\[ \binom {n -1}{k - 1} = \frac {(n -1)!}{(k - 1)! ((n -1) - (k -1))!} = \frac {(n - 1)!}{(k - 1)! (n - k)!}\]

This is the number of different ways we can order k-1 successes and n-k failures in n-1 trials. If the factorial notation (the exclamation point) is unfamiliar, see page 138.

Negative binomial distribution

The negative binomial distribution describes the probability of observing the \(k^{th}\) success on the nth trial:

\[ P \text {(the k^{th} success on the nth trial)} = \binom {n - 1}{k - 1} p^k {(1 - p)}^{n - k} \label {3.58}\]

where p is the probability an individual trial is a success. All trials are assumed to be independent.

Example \(\PageIndex{3}\)

Show using Equation \ref{3.58} that the probability Brian kicks his fourth successful field goal on the sixth attempt is 0.164.

**Solution**

The probability of a single success is p = 0.8, the number of successes is k = 4, and the number of necessary attempts under this scenario is n = 6.

\[ \begin{align*} \binom {n -1}{k - 1} p^k {(1 - p)}^{n - k} &= \frac {(5)!}{(3)! (2)!} {(0.8)}^4{(0.2)}^2 \\[5pt] &= 10 \times 0.0164 \\[5pt] &= 0.164 \end{align*}\]

Exercise \(\PageIndex{3A}\)

The negative binomial distribution requires that each kick attempt by Brian is independent. Do you think it is reasonable to suggest that each of Brian's kick attempts are independent?

**Answer**-
Answers may vary. We cannot conclusively say they are or are not independent. However, many statistical reviews of athletic performance suggests such attempts are very nearly independent.

Exercise \(\PageIndex{3B}\)

Assume Brian's kick attempts are independent. What is the probability that Brian will kick his fourth field goal within 5 attempts?

**Answer**-
If his fourth field goal (k = 4) is within ve attempts, it either took him four or ve tries (n = 4 or n = 5). We have p = 0:8 from earlier. Use Equation \ref{3.58} to compute the probability of n = 4 tries and n = 5 tries, then add those probabilities together:

\[\begin{align*} P(n = 4 OR n = 5) &= P(n = 4) + P(n = 5) \\[5pt] &= \binom {4 - 1}{4 - 1} 0.8^4 + \binom {5 -1}{4 - 1} {(0.8)}^4 (1 - 0.8) = 1 \times 0.41 + 4 \times 0.082 = 0.41 + 0.33 = 0.71 \end{align*}\]

TIP: Binomial versus negative binomial

In the binomial case, we typically have a xed number of trials and instead consider the number of successes. In the negative binomial case, we examine how many trials it takes to observe a fixed number of successes and require that the last observation be a success.

Exercise \(\PageIndex{3C}\)

On 70% of days, a hospital admits at least one heart attack patient. On 30% of the days, no heart attack patients are admitted. Identify each case below as a binomial or negative binomial case, and compute the probability.^{45}

- What is the probability the hospital will admit a heart attack patient on exactly three days this week?
- What is the probability the second day with a heart attack patient will be the fourth day of the week?
- What is the probability the fth day of next month will be the first day with a heart attack patient?

**Answer**-
In each part, p = 0:7. (a) The number of days is xed, so this is binomial. The parameters are k = 3 and n = 7: 0.097. (b) The last "success" (admitting a heart attack patient) is xed to the last day, so we should apply the negative binomial distribution. The parameters are k = 2, n = 4: 0.132. (c) This problem is negative binomial with k = 1 and n = 5: 0.006. Note that the negative binomial case when k = 1 is the same as using the geometric distribution.

## The Poisson Distribution

Example \(\PageIndex{4}\)

There are about 8 million individuals in New York City. How many individuals might we expect to be hospitalized for acute myocardial infarction (AMI), i.e. a heart attack, each day? According to historical records, the average number is about 4.4 individuals. However, we would also like to know the approximate distribution of counts. What would a histogram of the number of AMI occurrences each day look like if we recorded the daily counts over an entire year?

**Solution**

A histogram of the number of occurrences of AMI on 365 days^{46} for NYC is shown in Figure \(\PageIndex{1}\). The sample mean (4.38) is similar to the historical average of 4.4. The sample standard deviation is about 2, and the histogram indicates that about 70% of the data fall between 2.4 and 6.4. The distribution's shape is unimodal and skewed to the right.

^{46}These data are simulated. In practice, we should check for an association between successive days.

The **Poisson distribution **is often useful for estimating the number of rare events in a large population over a unit of time. For instance, consider each of the following events, which are rare for any given individual:

- having a heart attack,
- getting married, and
- getting struck by lightning.

The Poisson distribution helps us describe the number of such events that will occur in a short unit of time for a fixed population if the individuals within the population are independent.

The histogram in Figure \(\PageIndex{1}\) approximates a Poisson distribution with rate equal to 4.4. The rate for a Poisson distribution is the average number of occurrences in a mostlyfixed population per unit of time. In Example 3.63, the time unit is a day, the population is all New York City residents, and the historical rate is 4.4. The parameter in the Poisson distribution is the rate - or how many rare events we expect to observe - and it is typically denoted by \(\lambda\) (the Greek letter lambda) or \(\mu\). Using the rate, we can describe the probability of observing exactly k rare events in a single unit of time.

Poisson distribution

Suppose we are watching for rare events and the number of observed events follows a Poisson distribution with rate \(\lambda\). Then

\[P\text {(observe k rare events)} = \frac {\lambda^k e^{-\lambda}}{k!}\]

where k may take a value 0, 1, 2, and so on, and k! represents k-factorial, as described on page 138. The letter \(e \approx 2:718\) is the base of the natural logarithm. The mean and standard deviation of this distribution are \(\lambda\) and \(\sqrt {\lambda}\), respectively.

We will leave a rigorous set of conditions for the Poisson distribution to a later course. However, we offer a few simple guidelines that can be used for an initial evaluation of whether the Poisson model would be appropriate.

TIP: Is it Poisson?

A random variable may follow a Poisson distribution if the event being considered is rare, the population is large, and the events occur independently of each other.

Even when rare events are not really independent - for instance, Saturdays and Sundays are especially popular for weddings - a Poisson model may sometimes still be reasonable if we allow it to have a different rate for different times. In the wedding example, the rate would be modeled as higher on weekends than on weekdays. The idea of modeling rates for a Poisson distribution against a second variable such as dayOfTheWeek forms the foundation of some more advanced methods that fall in the realm of **generalized linear models**. In Chapters 7 and 8, we will discuss a foundation of linear models.

## Contributors and Attributions

David M Diez (Google/YouTube), Christopher D Barr (Harvard School of Public Health), Mine Çetinkaya-Rundel (Duke University)