---
title: "Uniform Distribution and Monte Carlo Simulation"
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: SD06-01-uniform-distribution-and-monte-carlo.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*.

This lesson connects the continuous uniform distribution to simulation and
Monte Carlo estimation. Computational experiments extend the deck and are
labeled as such.

## The continuous uniform distribution

For $X\sim U(a,b)$, every equal-length subinterval inside $[a,b]$ has the
same probability. Its density is

$$
f(x)=
\begin{cases}
\dfrac{1}{b-a}, & a\leq x\leq b,\\[4pt]
0, & \text{otherwise}.
\end{cases}
$$

```{r uniform-density}
a <- 2
b <- 6
x <- seq(0, 8, length.out = 401)

plot(
  x, dunif(x, min = a, max = b), type = "l", lwd = 2,
  xlab = "x", ylab = "f(x)", main = "Uniform density on [2, 6]"
)
grid()
```

## CDF and interval probabilities

Inside the support,

$$
F(x)=P(X\leq x)=\frac{x-a}{b-a}.
$$

For $a\leq c\leq d\leq b$,

$$
P(c\leq X\leq d)=\frac{d-c}{b-a}.
$$

Thus, for $X\sim U(2,6)$,

$$
P(3\leq X\leq5)=\frac{5-3}{6-2}=0.5.
$$

```{r uniform-interval-probability}
geometric_probability <- (5 - 3) / (6 - 2)
cdf_probability <- punif(5, 2, 6) - punif(3, 2, 6)

c(geometry = geometric_probability, R = cdf_probability)
```

For continuous variables, including or excluding endpoints does not change
the probability because $P(X=x)=0$.

## Mean and variance

The center and spread are

$$
E(X)=\frac{a+b}{2},\qquad
\operatorname{Var}(X)=\frac{(b-a)^2}{12}.
$$

```{r uniform-moments}
a <- 2
b <- 6

c(
  mean = (a + b) / 2,
  variance = (b - a)^2 / 12,
  sd = (b - a) / sqrt(12)
)
```

## Random-number generation

`runif()` generates pseudorandom uniform values. The following is a fictional
simulation extending the deck.

```{r uniform-simulation}
set.seed(14270)
draws <- runif(100000, min = 2, max = 6)

c(
  simulated_mean = mean(draws),
  theoretical_mean = 4,
  simulated_variance = var(draws),
  theoretical_variance = 16 / 12,
  simulated_P_3_to_5 = mean(draws >= 3 & draws <= 5)
)
```

## Monte Carlo estimation of pi

Draw a point $(X,Y)$ uniformly from the square $[-1,1]^2$. The square has
area 4, while the unit circle inside it has area $\pi$. Therefore,

$$
P(X^2+Y^2\leq1)=\frac{\pi}{4}
$$

and a Monte Carlo estimator is

$$
\widehat\pi=4\left(\frac{\text{points inside the circle}}{n}\right).
$$

```{r monte-carlo-pi}
set.seed(14270)
n <- 100000
x <- runif(n, -1, 1)
y <- runif(n, -1, 1)
inside <- x^2 + y^2 <= 1

pi_hat <- 4 * mean(inside)
c(estimate = pi_hat, truth = pi, error = pi_hat - pi)
```

```{r monte-carlo-plot, fig.width=6, fig.height=6}
keep <- seq_len(2500)
plot(
  x[keep], y[keep], pch = 19, cex = 0.45,
  col = ifelse(inside[keep], "steelblue", "tomato"),
  asp = 1, xlab = "X", ylab = "Y", main = "Monte Carlo points"
)
symbols(0, 0, circles = 1, inches = FALSE, add = TRUE, lwd = 2)
```

## Practice questions

1. If $X\sim U(10,18)$, find $P(12\leq X\leq15)$.
2. Find $E(X)$ and $\operatorname{SD}(X)$.
3. Use `qunif()` to find the 90th percentile.
4. Why is the Monte Carlo estimate of $\pi$ not exactly the same on every run
   unless the random seed is fixed?

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

1. The probability is $(15-12)/(18-10)=3/8=0.375$.
2. The mean is 14 and the standard deviation is $8/\sqrt{12}\approx2.309$.
3. `qunif(0.90, 10, 18)` gives $17.2$.
4. It is computed from a finite random sample, so the sampled proportion has
   simulation variability.
:::