set_notation.Rmd

---
title: "Set notation"
date: "`r Sys.Date()`"
output:
  workflowr::wflow_html:
    toc: true
---

```{r setup, include=FALSE}
library(tidyverse)
knitr::opts_chunk$set(echo = TRUE)
```

## Introduction

The **sample space**, $\Omega$, is the collection of possible outcomes of an
experiment, such as a die roll:

$$ \Omega = \{1, 2, 3, 4, 5, 6\} $$

An **event**, say $E$, is a subset of $\Omega$, such as the even dice rolls:

$$ E = \{2, 4, 6\} $$

An **elementary** or **simple** event is a particular result of an experiment,
such as the roll of 4 and is represented as a lowercase omega.

$$ \omega = 4 $$

A **null event** or the **empty set** is represented as:

$$ \emptyset $$.

## Interpretations of set operations

1. **Is an element of**

$$ \omega \in E $$

implies that $E$ occurs when $\omega$ occurs; for example if $\omega = 4$
occurs, then the event $E$ (even number) occurs.

2. **Not an element of**

$$ \omega \not \in E $$

implies that $E$ does not occur when $\omega$ occurs; for example if $\omega =
5$, $E$ (even numbers) does not occur.

3. **Subset**

$$ E \subset F $$

implies that the occurrence of $E$ implies the occurrence of $F$; for example if
$E = \{2, 4, 5\}$ and $F = \{2, 4, 5, 6\}$.

4. **Intersect**

$$ E \cap F $$

implies the event that both $E$ and $F$ occur; for example if $E = \{2, 4, 6\}$
(even numbers) and $F = \{2, 3, 5\}$ (prime numbers), then $E \cap F = 2$.

5. **Union**

$$ E \cup F $$

implies the event that at least one of $E$ or $F$ occur; for example if $E =
\{2, 4, 6\}$ (even numbers) and $F = \{2, 3, 5\}$ (prime numbers), then $E \cup
F = \{2, 3, 4, 5, 6\}$, which is that we get either an even number or a prime
number or **both**.

6. **If intersect == null event**

$$ E \cap F = \emptyset $$

implies that $E$ and $F$ are **mutually exclusive**, or cannot both
simultaneously occur; for example if $E = \{2, 4, 6\}$ (even numbers) and $F =
\{1, 3, 5\}$ (odd numbers).

7. **Complement**

$$ E^\complement $$

is the event that $E$ does not occur; for example if $E$ are the even numbers
then $E^\complement$ are the odd numbers.

## Set theory facts

**DeMorgan's laws**

$$ (A \cap B)^\complement = A^\complement \cup B^\complement$$

$$ (A \cup B)^\complement = A^\complement \cap B^\complement$$

To remember these laws, think of distributing the complements to $A$ and $B$ and
then inverting the cap into a cup or vice versa.

$$ (A^\complement)^\complement = A $$

The complement of $A$ complemented = $A$.

And lastly:

$$ (A \cup B) \cap C = (A \cap C) \cup (B \cap C) $$

## Probability and set notation

>Consider influenza epidemics for two parent heterosexual families. Suppose that
the probability is 17% that at least one of the parents has contracted the
disease. The probability that the father has contracted influenza is 12% while
the probability that both the mother and father have contracted the disease is
6%. What is the probability that the mother has contracted influenza?

Let $M$ be the mother and $F$ be the father.

Recall that the union ($\cup$) implies the event that at least one of $M$ or $F$
occurred, therefore:

$$ P(M \cup F) = 17\%. $$

The probability that the father contracted influenza is 12%:

$$ P(F) = 12\%. $$

An intersection ($\cap$) implies the event that both M and F occur, i.e. the
probability that both the mother and father have contracted the disease:

$$ P(M \cap F) = 6\%. $$

We know that:

$$ P(M \cup F) = P(M) + P(F) - P(M \cap F).$$

It is worth noting that if the events are mutually exclusive, $P(M \cap F) = 0$,
since they both can't occur together, but in our case the events can occur
together. From the above equation, we get:

$$
17 = P(M) + 12 - 6 \\
P(M) = 17 - 12 + 6 \\
P(M) = 11%.
$$

Therefore the probability of the mother contracting the disease is 11%.

You may be more familiar with the addition rule in probability, which is:

$$
P(M\ or\ F) = P(M) + P(F) - P(M\ and\ F) \\
17 = P(M) + 12 - 6 \\
P(M) = 17 - 12 + 6 \\
P(M) = 11
$$

## Conditional probability

The probability of getting a 1 when rolling a fair die is $\frac{1}{6}$. If we
are given the extra information that the die roll was an odd number (1, 3 or 5),
conditional on this new information, the probability of a 1 is now
$\frac{1}{3}$. Let's test this example out:

Let $B$ be an event so that $P(B) > 0$, i.e. it's an event that actually occurs.
Then the conditional probability of an event $A$ given that $B$ has occurred is:

$$ P(A | B) = \frac{P(A \cap B)}{P(B)}. $$

So using the example above:

$$ B = \{1, 3, 5\}\ and\ A = \{1\} $$

Therefore:

$$ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)} = \frac{\frac{1}{6}}{\frac{3}{6}} = \frac{1}{3} $$