---
title: "Exponential and Gamma Distributions"
subtitle: "SD06 — Continuous Distributions"
author: "Samer El Kababji"
output:
  html_document:
    toc: true
format:
  html:
    code-links:
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        href: SD06-02-exponential-and-gamma-distributions.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, *SD06 — Continuous Distributions*.

The Poisson distribution counts events; the exponential and gamma
distributions measure how long we wait for those events.

## Exponential waiting time

For rate $\lambda>0$, $X\sim\operatorname{Exp}(\lambda)$ has density and CDF

$$
f(x)=\lambda e^{-\lambda x},\qquad
F(x)=1-e^{-\lambda x},\qquad x\geq0.
$$

Its survival probability is

$$
P(X>x)=e^{-\lambda x}.
$$

```{r exponential-density-cdf}
lambda <- 0.5
x <- c(0, 1, 2, 4)

data.frame(
  x,
  density = dexp(x, rate = lambda),
  CDF = pexp(x, rate = lambda),
  survival = pexp(x, rate = lambda, lower.tail = FALSE)
)
```

The exponential moments are

$$
E(X)=\frac1\lambda,
\qquad
\operatorname{Var}(X)=\frac1{\lambda^2}.
$$

```{r exponential-moments}
lambda <- 0.5
c(mean = 1 / lambda, variance = 1 / lambda^2, sd = 1 / lambda)
```

## Exponential examples

For calls arriving at rate 3 per hour, the probability the next call arrives
within an hour is

$$
P(X<1)=1-e^{-3}\approx0.9502.
$$

For a pipeline failure time with rate $0.5$ per year,

$$
P(X>2)=e^{-0.5(2)}=e^{-1}\approx0.3679.
$$

```{r exponential-examples}
c(
  call_within_one_hour = pexp(1, rate = 3),
  pipeline_survives_two_years = pexp(2, rate = 0.5, lower.tail = FALSE)
)
```

## Gamma waiting time

The gamma distribution measures the total wait to a later event. In the
shape-rate parameterization,

$$
f(x)=\frac{\lambda^\alpha}{\Gamma(\alpha)}
x^{\alpha-1}e^{-\lambda x},\qquad x\geq0.
$$

With scale $\beta=1/\lambda$, the same density is

$$
f(x)=\frac{1}{\Gamma(\alpha)\beta^\alpha}
x^{\alpha-1}e^{-x/\beta}.
$$

The moments are

$$
E(X)=\frac\alpha\lambda=\alpha\beta,
\qquad
\operatorname{Var}(X)=\frac\alpha{\lambda^2}=\alpha\beta^2.
$$

```{r gamma-parameterizations}
alpha <- 3
lambda <- 0.5
beta <- 1 / lambda
x <- seq(0, 15, length.out = 301)

max(abs(
  dgamma(x, shape = alpha, rate = lambda) -
    dgamma(x, shape = alpha, scale = beta)
))
c(mean = alpha / lambda, variance = alpha / lambda^2)
```

Do not supply both `rate` and `scale` to R. State which convention is being
used whenever gamma parameters are reported.

## Time to the third component failure

A machine fails after its third internal component failure. If failures arrive
at rate $0.5$ per hour, then $X\sim\operatorname{Gamma}(3,\text{rate}=0.5)$.
For integer shape,

$$
F(x)=1-e^{-\lambda x}
\sum_{j=0}^{\alpha-1}\frac{(\lambda x)^j}{j!}.
$$

At $x=4$ hours,

$$
P(X<4)=1-e^{-2}\left(1+2+\frac{2^2}{2!}\right)
\approx0.3233.
$$

```{r gamma-machine-example}
manual <- 1 - exp(-2) * sum(2^(0:2) / factorial(0:2))
built_in <- pgamma(4, shape = 3, rate = 0.5)

c(manual = manual, R = built_in)
```

## Relationship among the three distributions

- Poisson: number of events in a fixed interval.
- Exponential: wait to the first event.
- Gamma with integer shape $\alpha$: wait to the $\alpha$th event.

```{r gamma-simulation-extension}
set.seed(14270)
wait_to_third <- rgamma(100000, shape = 3, rate = 0.5)

c(
  simulated_mean = mean(wait_to_third),
  theoretical_mean = 6,
  simulated_P_less_than_4 = mean(wait_to_third < 4),
  exact_P_less_than_4 = pgamma(4, 3, rate = 0.5)
)
```

## Practice questions

1. If $X\sim\operatorname{Exp}(2)$, find $P(X>1.5)$ and $E(X)$.
2. If $Y\sim\operatorname{Gamma}(4,\text{rate}=2)$, find its mean and
   variance.
3. Find $P(Y\leq3)$.
4. Express the same gamma distribution using a scale parameter.

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

1. $P(X>1.5)=e^{-3}\approx0.0498$ and $E(X)=0.5$.
2. The mean is $4/2=2$ and the variance is $4/2^2=1$.
3. `pgamma(3, shape = 4, rate = 2)` gives approximately $0.8488$.
4. The scale is $\beta=1/2=0.5$, so use
   `Gamma(shape = 4, scale = 0.5)`.
:::