---
title: "The Negative Binomial Distribution"
subtitle: "SD05 — Discrete Distributions"
author: "Samer El Kababji"
output:
  html_document:
    toc: true
format:
  html:
    code-links:
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        href: SD05-03-negative-binomial-distribution.Rmd?download=1
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE, message = FALSE, warning = TRUE)
```

**Course:** Statistics and Probability for Data Science (DS14270).  
**Source deck:** Samer El Kababji, *SD05 — Discrete Distributions*.

The negative binomial distribution reverses the binomial stopping rule: fix
the number of successes and count how many failures occur first.

## Counting failures before the rth success

In independent Bernoulli trials with constant success probability $p$, let
$X$ be the number of failures before the $r$th success. Then

$$
P(X=x)=\binom{x+r-1}{r-1}p^r(1-p)^x,
\qquad x=0,1,2,\ldots.
$$

The last trial must be the $r$th success. Among the preceding $x+r-1$ trials,
choose the positions of the other $r-1$ successes.

```{r negative-binomial-pmf}
r <- 5
p <- 0.20
x <- 0:20

formula_pmf <- choose(x + r - 1, r - 1) * p^r * (1 - p)^x
R_pmf <- dnbinom(x, size = r, prob = p)

head(data.frame(failures = x, formula_pmf, R_pmf), 8)
sum(R_pmf)
```

The displayed sum is less than 1 only because the infinite support was
truncated at 20 failures.

## Pediatrician recruitment example

A pediatrician asks couples to participate until 5 agree. Each couple agrees
with probability $p=0.20$. If exactly 15 couples are asked, there were 10
failures before the fifth success:

$$
P(X=10)=\binom{14}{4}(0.2)^5(0.8)^{10}\approx0.034.
$$

```{r pediatrician-example}
r <- 5
p <- 0.20
failures <- 15 - r

c(
  formula = choose(14, 4) * p^5 * (1 - p)^10,
  R = dnbinom(failures, size = r, prob = p),
  at_most_15_couples = pnbinom(failures, size = r, prob = p)
)
```

The probability that the fifth agreement arrives by the 15th couple is
$P(X\leq10)\approx0.164$.

## Mean and variance

For this failures-before-successes convention,

$$
E(X)=\frac{r(1-p)}p,
\qquad
\operatorname{Var}(X)=\frac{r(1-p)}{p^2}.
$$

If instead $T=X+r$ counts the **total number of trials** through the $r$th
success, then $E(T)=r/p$ and $\operatorname{Var}(T)=\operatorname{Var}(X)$.

```{r negative-binomial-moments}
r <- 5
p <- 0.20

mean_failures <- r * (1 - p) / p
variance_failures <- r * (1 - p) / p^2

c(
  mean_failures = mean_failures,
  mean_total_trials = mean_failures + r,
  variance = variance_failures,
  sd = sqrt(variance_failures)
)
```

## Quantiles and a simulation check (fictional extension)

```{r negative-binomial-quantiles-simulation}
# Failures needed so there is at least 90% probability of having r successes
q90_failures <- qnbinom(0.90, size = 5, prob = 0.20)
q90_total_trials <- q90_failures + 5

set.seed(14270)
simulated_failures <- rnbinom(100000, size = 5, prob = 0.20)

c(
  q90_failures = q90_failures,
  q90_total_trials = q90_total_trials,
  simulated_mean = mean(simulated_failures),
  theoretical_mean = 5 * 0.8 / 0.2
)
```

## Do not confuse the counting conventions

Base R's `dnbinom()`, `pnbinom()`, `qnbinom()`, and `rnbinom()` count failures
before `size` successes. If a question gives the total trial number $t$,
convert it with $x=t-r$.

## Practice questions

1. A sale occurs with probability $0.25$ on each independent call. Find the
   probability of exactly 8 failures before the third sale.
2. Find the probability that the third sale occurs by the 11th call.
3. Find the expected number and variance of failures before the third sale.
4. Find the expected total number of calls through the third sale.

::: {.callout-note collapse="true"}
### Check your answers

1. `dnbinom(8, size = 3, prob = 0.25)` gives approximately $0.0704$.
2. The condition is $X\leq8$, so `pnbinom(8, 3, 0.25)` gives approximately
   $0.5448$.
3. The mean is $3(0.75)/0.25=9$ and the variance is
   $3(0.75)/0.25^2=36$.
4. Add the three successes: $E(T)=9+3=12$ calls.
:::