---
title: "Weibull Distribution and Survival"
subtitle: "SD06 — Continuous Distributions"
author: "Samer El Kababji"
output:
  html_document:
    toc: true
format:
  html:
    code-links:
      - text: "Download R Markdown"
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        href: SD06-03-weibull-distribution-and-survival.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 Weibull distribution is a flexible model for lifetimes and survival. In
this lesson, $k$ is the shape and $\lambda$ is the **scale**, matching the
deck and base R.

## Density, CDF, and survival function

For $X\sim\operatorname{Weibull}(k,\lambda)$,

$$
f(x)=\frac{k}{\lambda}
\left(\frac{x}{\lambda}\right)^{k-1}
\exp\left[-\left(\frac{x}{\lambda}\right)^k\right],
\qquad x\geq0,
$$

and

$$
F(x)=1-\exp\left[-\left(\frac{x}{\lambda}\right)^k\right].
$$

Therefore the survival function is

$$
S(x)=P(X>x)=\exp\left[-\left(\frac{x}{\lambda}\right)^k\right].
$$

```{r weibull-functions}
shape <- 2
scale <- 100
x <- c(0, 40, 80, 120)

data.frame(
  x,
  density = dweibull(x, shape = shape, scale = scale),
  CDF = pweibull(x, shape = shape, scale = scale),
  survival = pweibull(
    x, shape = shape, scale = scale, lower.tail = FALSE
  )
)
```

## Component-lifetime example

For shape $k=2$ and scale $\lambda=100$,

$$
P(X<80)=1-\exp[-(80/100)^2]\approx0.4727.
$$

```{r weibull-component-example}
c(
  formula = 1 - exp(-(80 / 100)^2),
  R = pweibull(80, shape = 2, scale = 100)
)
```

## Mean and variance

The moments use the gamma function:

$$
E(X)=\lambda\Gamma\left(1+\frac1k\right),
$$

$$
\operatorname{Var}(X)=\lambda^2\left[
\Gamma\left(1+\frac2k\right)-
\Gamma\left(1+\frac1k\right)^2
\right].
$$

```{r weibull-moments}
k <- 2
lambda <- 100

mu <- lambda * gamma(1 + 1 / k)
variance <- lambda^2 * (
  gamma(1 + 2 / k) - gamma(1 + 1 / k)^2
)

c(mean = mu, variance = variance, sd = sqrt(variance))
```

## Shape and failure behavior

The Weibull hazard function is

$$
h(x)=\frac{f(x)}{S(x)}
=\frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1}.
$$

- $k<1$: decreasing hazard, often associated with early failures.
- $k=1$: constant hazard; the Weibull becomes exponential.
- $k>1$: increasing hazard, often associated with wear-out.

```{r weibull-hazard-plot, fig.width=7, fig.height=4}
x <- seq(0.05, 20, length.out = 500)
shapes <- c(0.6, 1, 2)
colors <- c("tomato", "darkorange", "steelblue")

matplot(
  x,
  sapply(shapes, function(k) (k / 10) * (x / 10)^(k - 1)),
  type = "l", lty = 1, lwd = 2, col = colors,
  xlab = "Time", ylab = "Hazard", main = "Weibull hazard shapes"
)
legend("topright", legend = paste("shape =", shapes), col = colors,
       lty = 1, lwd = 2, bty = "n")
grid()
```

## Scale changes the time axis

For fixed shape, a larger scale stretches lifetimes. Because

$$
F(\lambda)=1-e^{-1}\approx0.632,
$$

the scale is the time by which about 63.2% of the population has experienced
the event.

The deck's survival context uses shape 2 and scale 12 months.

```{r cancer-survival-context}
times <- c(6, 12, 18, 24)
data.frame(
  months = times,
  survival_probability = pweibull(
    times, shape = 2, scale = 12, lower.tail = FALSE
  )
)
```

These are model-based probabilities, not patient predictions.

## Practice questions

1. For shape 1.5 and scale 20, find $P(X\leq10)$.
2. Find $P(X>25)$.
3. Is the hazard increasing or decreasing?
4. What special distribution results when the shape is 1?

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

1. `pweibull(10, 1.5, 20)` gives approximately $0.2978$.
2. `pweibull(25, 1.5, 20, lower.tail = FALSE)` gives approximately $0.2472$.
3. It is increasing because $k=1.5>1$.
4. An exponential distribution with rate $1/\lambda$.
:::