---
title: "Random Variables, PMFs, PDFs, and CDFs"
subtitle: "SD04 — Random Variables"
author: "Samer El Kababji"
output:
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    toc: true
format:
  html:
    code-links:
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        href: SD04-01-random-variables-pmf-pdf-and-cdf.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, *SD04 — Random Variables*.

This lesson develops the deck's distinction between discrete and continuous
random variables and connects probability mass, probability density, and
cumulative distribution functions. The practice questions and computational
checks extend the deck's examples.

## A random variable maps outcomes to numbers

A **random variable** is a function from a sample space $S$ to the real
numbers:

$$
X:S\longrightarrow\mathbb{R}.
$$

The randomness is in the outcome $\omega\in S$; once $\omega$ is known, the
value $X(\omega)$ is determined. For a phone call with outcomes Success and
Failure, one possible random variable is

$$
X(\text{Success})=1,
\qquad
X(\text{Failure})=0.
$$

```{r random-variable-map}
call_outcomes <- c("Success", "Failure")
X_call <- c(Success = 1, Failure = 0)

data.frame(
  outcome = call_outcomes,
  X = unname(X_call[call_outcomes])
)
```

Now test 9-volt batteries until the first acceptable battery, denoted $S$,
appears. With $F$ denoting failure,

$$
S=\{S,FS,FFS,FFFS,\ldots\},
$$

and the number tested is

$$
X(S)=1,\quad X(FS)=2,\quad X(FFS)=3,\quad\ldots.
$$

```{r battery-outcome-map}
battery_outcomes <- c("S", "FS", "FFS", "FFFS")
number_tested <- nchar(battery_outcomes)

data.frame(
  outcome = battery_outcomes,
  X = number_tested
)
```

## Discrete and continuous random variables

A **discrete** random variable has a finite or countably infinite set of
possible values. Counts such as customers per hour, students passing an exam,
or emails per day are discrete.

A **continuous** random variable can take values throughout an interval.
Completion time, temperature, weight, and rainfall amount are continuous.

For a continuous variable, a single point has zero probability even though an
interval can have positive probability:

$$
P(X=x)=0,
\qquad
P(a\leq X\leq b)>0\ \text{may hold}.
$$

The following draws illustrate the different supports. The simulations are
fictional extensions of the deck.

```{r discrete-versus-continuous}
set.seed(14270)

customers_per_hour <- rpois(8, lambda = 12)
completion_time <- rexp(8, rate = 0.5)

data.frame(
  discrete_count = customers_per_hour,
  continuous_time = round(completion_time, 3)
)
```

## Probability mass functions

For a discrete random variable $X$, its **probability mass function** (PMF)
is

$$
p(x)=P(X=x)
=P\bigl(\{\omega\in S:X(\omega)=x\}\bigr).
$$

A valid PMF satisfies

$$
p(x)\geq0
\quad\text{and}\quad
\sum_x p(x)=1.
$$

### Worked example: defects in a selected box

Six equally likely boxes contain respectively $0,2,0,1,2,0$ defective
components. If $X$ is the number of defectives in the selected box, then

$$
p(0)=\frac36=0.5,
\qquad
p(1)=\frac16,
\qquad
p(2)=\frac26=\frac13.
$$

```{r box-defects-pmf}
defects <- c(0, 2, 0, 1, 2, 0)
pmf_boxes <- prop.table(table(defects))

pmf_boxes
sum(pmf_boxes)
stopifnot(isTRUE(all.equal(sum(pmf_boxes), 1)))
```

The PMF groups sample points that produce the same value of $X$. Here three
different boxes all map to $X=0$.

## Probability density functions

For a continuous random variable $X$, a **probability density function**
(PDF) $f$ satisfies

$$
f(x)\geq0,
\qquad
\int_{-\infty}^{\infty}f(x)\,dx=1.
$$

Probabilities are areas under the density:

$$
P(a\leq X\leq b)=\int_a^b f(x)\,dx.
$$

Unlike a PMF value, $f(x)$ is not the probability of the point $x$. A density
may even exceed 1 as long as its total area is 1.

### Worked example: battery time to failure

The deck gives the density

$$
f(x)=
\begin{cases}
\dfrac{2}{(x+1)^3}, & x\geq0,\\[4pt]
0, & x<0.
\end{cases}
$$

It is normalized because

$$
\int_0^\infty\frac{2}{(x+1)^3}\,dx
=\left[-\frac{1}{(x+1)^2}\right]_0^\infty=1.
$$

For example,

$$
P(1\leq X\leq3)
=\int_1^3\frac{2}{(x+1)^3}\,dx
=\frac{3}{16}=0.1875.
$$

```{r battery-failure-pdf}
battery_pdf <- function(x) {
  ifelse(x >= 0, 2 / (x + 1)^3, 0)
}

total_area <- integrate(battery_pdf, lower = 0, upper = Inf)$value
P_between_1_and_3 <- integrate(battery_pdf, lower = 1, upper = 3)$value

c(total_area = total_area,
  P_between_1_and_3 = P_between_1_and_3)
stopifnot(isTRUE(all.equal(P_between_1_and_3, 3 / 16)))
```

```{r battery-failure-density-plot, fig.width=7, fig.height=4}
curve(
  battery_pdf,
  from = 0,
  to = 7,
  n = 501,
  lwd = 2,
  col = "steelblue",
  xlab = "x (hours)",
  ylab = "f(x)",
  main = "Battery time-to-failure density"
)
grid()
```

## The cumulative distribution function

The **cumulative distribution function** (CDF) works for every random
variable:

$$
F(x)=P(X\leq x).
$$

The standard definition includes equality. For a continuous random variable,
$P(X=x)=0$, so $P(X\leq x)=P(X<x)$. Every CDF is nondecreasing,
right-continuous, and satisfies

$$
\lim_{x\to-\infty}F(x)=0,
\qquad
\lim_{x\to\infty}F(x)=1.
$$

For a discrete random variable,

$$
F(x)=\sum_{y\leq x}p(y),
\qquad
p(x)=F(x)-F(x^-),
$$

where $F(x^-)$ is the value immediately to the left of $x$. For a continuous
random variable,

$$
F(x)=\int_{-\infty}^x f(u)\,du,
\qquad
f(x)=F'(x)
$$

where the derivative exists.

## Discrete CDF: flash-drive capacity

Let $Y$ be purchased memory capacity in GB, with PMF

| $y$ | 1 | 2 | 4 | 8 | 16 |
|---:|---:|---:|---:|---:|---:|
| $p(y)$ | 0.05 | 0.10 | 0.35 | 0.40 | 0.10 |

The cumulative probabilities are

$$
F(1)=0.05,\quad F(2)=0.15,\quad F(4)=0.50,
\quad F(8)=0.90,\quad F(16)=1.
$$

Between possible values, the CDF is constant. Thus

$$
F(2.7)=F(2)=0.15,
\qquad
F(7.999)=F(4)=0.50.
$$

```{r flash-drive-cdf}
capacity <- c(1, 2, 4, 8, 16)
capacity_pmf <- c(0.05, 0.10, 0.35, 0.40, 0.10)
capacity_cdf <- cumsum(capacity_pmf)

F_capacity <- function(q) {
  vapply(q, function(value) {
    sum(capacity_pmf[capacity <= value])
  }, numeric(1))
}

data.frame(capacity, capacity_pmf, capacity_cdf)
F_capacity(c(2.7, 7.999, 16))
```

```{r flash-drive-cdf-plot, fig.width=7, fig.height=4}
plot(
  stepfun(capacity, c(0, capacity_cdf)),
  verticals = FALSE,
  do.points = TRUE,
  pch = 19,
  xlab = "Memory capacity y (GB)",
  ylab = "F(y)",
  main = "CDF of purchased flash-drive capacity"
)
grid()
```

## Discrete CDF: waiting for the first success

Suppose independent trials have success probability $p$, and $X$ is the trial
number of the first success. Then

$$
P(X=x)=(1-p)^{x-1}p,
\qquad x=1,2,3,\ldots.
$$

Summing the geometric series gives

$$
F(x)=P(X\leq x)
=1-(1-p)^{\lfloor x\rfloor},
\qquad x\geq1,
$$

and $F(x)=0$ for $x<1$.

R's geometric distribution counts failures before the first success, so the
deck's $X$ equals the R value plus 1.

```{r geometric-pmf-and-cdf}
p_success <- 0.51
x <- 1:6

geometric_table <- data.frame(
  x = x,
  pmf_formula = (1 - p_success)^(x - 1) * p_success,
  pmf_R = dgeom(x - 1, prob = p_success),
  cdf_formula = 1 - (1 - p_success)^x,
  cdf_R = pgeom(x - 1, prob = p_success)
)

geometric_table
stopifnot(with(geometric_table,
               isTRUE(all.equal(pmf_formula, pmf_R)) &&
               isTRUE(all.equal(cdf_formula, cdf_R))))
```

## Continuous CDF: weekly gravel sales

Let $X$ be weekly gravel sales in tons with density

$$
f(x)=
\begin{cases}
\dfrac32(1-x^2), & 0\leq x\leq1,\\[4pt]
0, & \text{otherwise}.
\end{cases}
$$

For $0\leq x\leq1$,

$$
F(x)=\int_0^x\frac32(1-u^2)\,du
=\frac32\left(x-\frac{x^3}{3}\right).
$$

Therefore the complete CDF is

$$
F(x)=
\begin{cases}
0, & x<0,\\[2pt]
\dfrac32\left(x-\dfrac{x^3}{3}\right), & 0\leq x\leq1,\\[6pt]
1, & x>1.
\end{cases}
$$

```{r gravel-cdf}
gravel_pdf <- function(x) {
  ifelse(x >= 0 & x <= 1, 3 / 2 * (1 - x^2), 0)
}

gravel_cdf <- function(x) {
  ifelse(
    x < 0,
    0,
    ifelse(x <= 1, 3 / 2 * (x - x^3 / 3), 1)
  )
}

check_points <- c(0, 0.25, 0.5, 0.75, 1)
cbind(
  x = check_points,
  formula = gravel_cdf(check_points),
  integration = vapply(
    check_points,
    function(q) integrate(gravel_pdf, -Inf, q)$value,
    numeric(1)
  )
)
```

## Practice questions

1. Let $X$ be the number of heads in two fair coin tosses. Write its PMF and
   compute $F(1)$.
2. Verify from the flash-drive PMF that $P(2<Y\leq8)=F(8)-F(2)$.
3. For the battery time-to-failure density, derive $F(x)$ for $x\geq0$ and
   compute $P(X>2)$.
4. For the geometric model with $p=0.4$, compute $P(X=3)$ and $P(X\leq3)$.
5. For the gravel-sales distribution, compute $P(0.25\leq X\leq0.75)$ from
   the CDF.

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

1. $p(0)=1/4$, $p(1)=1/2$, and $p(2)=1/4$; hence
   $F(1)=P(X\leq1)=3/4$.
2. $F(8)-F(2)=0.90-0.15=0.75$, which is
   $P(Y=4)+P(Y=8)=0.35+0.40$.
3. $F(x)=1-(x+1)^{-2}$ for $x\geq0$, so
   $P(X>2)=1-F(2)=1/9$.
4. $P(X=3)=(0.6)^2(0.4)=0.144$ and
   $P(X\leq3)=1-(0.6)^3=0.784$.
5. $F(0.75)-F(0.25)=0.546875$.
:::