A4 - independent events

independent events

two events are said to be independent if knowing one happened does not change how likely the other is
events $E$ and $F$ are independent, ie. $E ⊥ F$ , if:

P (E \cap F) = P (E) P (F)

P (E | F) = P (E)

this implies that $P (E)$ is unaffected by whether $F$ is true or not
if they are mutually exclusive then $P (E | F) = 0 \neq P (E)$ , so they cannot be independent
independent events:
- a coin flip and a die roll
- the day being a tuesday, and a card drawn being an ace
dependent events:
- drawing aces without replacement
- rain and people carrying umbrellas
consider rolling a fair die twice
$P (sum = 7) = 1 / 6$ and given that the second roll is $4$ , $P (second = 4) = 1 / 6$ , $P (sum = 7 | second = 4) = 1 / 6$ , thus they are independent
now, if given that the second roll is $4$ and $P (sum = 8) = 5 / 36$ , $P (sum = 8 | second) = 1 / 6$ , which is not the same, therefore these events are dependent
generalising for more than two events

independence

events $A_{1}, A_{2}, \dots, A_{n}$ are independent if, for every subset $S$ of $1, 2, \dots, n :$

P (\cap_{i \in S}) = \prod_{i \in S} P (A_{i})

pair-wise independence

$\forall i \neq j, P (A_{i} \cap A_{j}) = P (A_{i}) P (A_{j})$

pair-wise independence is a weaker version of independence, but still admits some nice properties

conditional independence

two events $E$ and $F$ are said to be conditionally independent given event $G$ , where $P (G) > 0$ , if

P (E \cap F | G) = P (E | G) P (F | G)