---
title: "The Poisson Distribution"
subtitle: "SD05 — Discrete Distributions"
author: "Samer El Kababji"
output:
  html_document:
    toc: true
format:
  html:
    code-links:
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        href: SD05-04-poisson-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 Poisson distribution models counts of events in a fixed interval when the
events occur independently at a stable average rate.

## Probability mass function

If $X\sim\operatorname{Poisson}(\mu)$, then

$$
P(X=x)=e^{-\mu}\frac{\mu^x}{x!},
\qquad x=0,1,2,\ldots,
$$

where $\mu>0$ is the expected count per interval.

```{r poisson-pmf}
mu <- 2.4
x <- 0:10

formula_pmf <- exp(-mu) * mu^x / factorial(x)
R_pmf <- dpois(x, lambda = mu)

data.frame(x, formula_pmf, R_pmf)
```

## Tile-defect example

If a tile has an average of $\mu=2.4$ defects, then

$$
P(X=0)=e^{-2.4}\approx0.0907.
$$

The probability of at least four defects is easiest through the complement:

$$
P(X\geq4)=1-P(X\leq3)\approx0.2213.
$$

```{r tile-defect-example}
c(
  no_defects = dpois(0, lambda = 2.4),
  at_least_four = ppois(3, lambda = 2.4, lower.tail = FALSE),
  complement_check = 1 - ppois(3, lambda = 2.4)
)
```

Using `lower.tail = FALSE` avoids manually subtracting from 1 and can be more
accurate for very small upper-tail probabilities.

## Mean and variance

The Poisson parameter controls both center and spread:

$$
E(X)=\mu,\qquad
\operatorname{Var}(X)=\mu,\qquad
\operatorname{SD}(X)=\sqrt{\mu}.
$$

```{r poisson-moments}
mu <- 2.4
c(mean = mu, variance = mu, sd = sqrt(mu))
```

Equality of the mean and variance is a distinctive Poisson property. Strong
overdispersion or underdispersion in observed counts can signal that a simple
Poisson model is inadequate.

## Changing the interval

If the rate is 2.4 events per hour, the expected count in $t$ hours is
$\mu_t=2.4t$.

```{r poisson-time-scaling}
rate_per_hour <- 2.4
hours <- c(0.5, 1, 3)
mu_by_interval <- rate_per_hour * hours

data.frame(
  hours,
  expected_count = mu_by_interval,
  P_no_events = dpois(0, lambda = mu_by_interval)
)
```

## Poisson approximation to the binomial

For $X\sim\operatorname{Bin}(n,p)$, if $n$ is large and $p$ is small while
$np\to\mu$, then

$$
X\ \dot\sim\ \operatorname{Poisson}(\mu),\qquad \mu=np.
$$

The deck's practical rule is $n>50$ and $np<5$.

```{r binomial-poisson-approximation}
n <- 200
p <- 0.01
mu <- n * p
k <- 0:8

comparison <- data.frame(
  k,
  binomial = dbinom(k, size = n, prob = p),
  poisson = dpois(k, lambda = mu)
)
comparison$absolute_difference <- abs(comparison$binomial - comparison$poisson)
comparison
```

## Simulation check (fictional extension)

```{r poisson-simulation}
set.seed(14270)
simulated_counts <- rpois(100000, lambda = 2.4)

c(
  simulated_mean = mean(simulated_counts),
  simulated_variance = var(simulated_counts),
  theoretical_value = 2.4
)
```

## Practice questions

1. A help desk averages 3.5 calls per hour. Find $P(X=2)$ in one hour.
2. Find $P(X\geq5)$ in one hour.
3. Find the probability of no calls in 30 minutes.
4. Approximate $P(X=1)$ for $X\sim\operatorname{Bin}(100,0.02)$ with a
   Poisson distribution.

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

1. `dpois(2, 3.5)` gives approximately $0.1850$.
2. `ppois(4, 3.5, lower.tail = FALSE)` gives approximately $0.2746$.
3. The half-hour mean is $1.75$, so the probability is $e^{-1.75}\approx0.1738$.
4. Use $\mu=np=2$: `dpois(1, 2)` gives approximately $0.2707$ (the exact
   binomial value is about $0.2707$).
:::