User:Saul/probability
Contents
Notations
Ω - The outcome space.
ω - An outcome.
∅ - A non event.
∪ - Union.
∩ - Intersection.
∈ - Is in.
∉ - Is not in.
⊂ - Subset of.
⊆ - Subset or equal to.
- Note: there is a small difference between ⊂ & ⊆ but sometimes they get used interchangeably.
⊄ - Is not a subset of.
⊃ - Superset of.
⊇ - Is not a superset of.
ℙ(A) - A set function of set A - Probability of A.
Ac - A compliment - The event A does not occur.
ω∈A - The outcome ω is in the event A
A∪B - The union of A and B - The set containing all the elements from A and B without duplicates.
A∩B - The intersection of A and B - The set containing all the common elements from A and B.
A∩B = ∅ - The sets A and B are disjoint.
ℙ(A∪B) = ℙ(A) + ℙ(B) - The sets A and B are disjoint.
#A - The number of elements A contains if A is finite.
ℙ(S) = 1 - The set S is a exhaustive set as it contains all the outcomes.
Possible outcome - may have a probability of 0.
Axioms
ℙ(A) ≥ 0 for all events A.
ℙ(Ω) = 1
(Countable additivity) For an infinite sequence of mutually exclusive events{A1,A2,A3,...}
Deducible Properties
ℙ(∅) = 0 Since Ω ∪ ∅ ∪ ∅ ∪ ... = Ω
Finite additivity.
ℙ(Ac) = 1 - ℙ(A) Since A ∪ Ac = Ω
If A ⊆ B Then ℙ(A) ≤ ℙ(B) Since A∪(Ac ∩ B) = B
ℙ(A) ≤ 1 Since A ⊆ Ω
ℙ(A ∪ B) = ℙ(A) + ℙ(B) - ℙ(A ∩ B)
Conditional Probability
ℙ(A | H) - Given H occurs what is the probability of A.
ℙ(A | H) = ℙ(A ∩ H) / ℙ(H)
- If ℙ(H) > 0
For example:
Two dice are rolled the outcomes being i and j.
let:
- A = {(i, j) : |i - j| ≤ 1}
- Note {} is the set, (i, j) is all the possible outcomes and the ":" means the set of those outcomes that satisfy the following condition.
and:
- H = {(i, j) : i + j = 7}
We could start by listing all the possible outcomes that are in set 'H:
i | j |
---|---|
1 | 6 |
2 | 5 |
3 | 4 |
4 | 3 |
5 | 2 |
6 | 1 |
And all the outcomes in set A
i | j |
---|---|
1 | 1 |
1 | 2 |
2 | 1 |
2 | 2 |
2 | 3 |
3 | 2 |
3 | 3 |
3 | 4 |
4 | 3 |
4 | 4 |
4 | 5 |
5 | 4 |
5 | 5 |
5 | 6 |
6 | 5 |
6 | 6 |
We can then find all the outcomes in A ∩ H (We could have just applied set H's condition straight to A):
i | j |
---|---|
3 | 4 |
4 | 3 |
We can say there are 6 * 6 = 36 total possible outcomes (Ω).
- Set H has 6 / 36 of these outcomes.
- Set A has 16 / 36 of these outcomes.
- Set A ∩ H has 2 / 36 of these outcomes.
We can now calculate the probability of A happening given H has happened.
- ℙ(A | H) = ℙ(2/36) / ℙ(6/36) = 1 / 3
And the probability of H happening given A has happened.
- ℙ(H | A) = ℙ(2/36) / ℙ(16/36) = 1 / 8
Positive/Negative Relationships
A positive' relationship means that given H has happened the chance of A has increased.
If this is true it also means the reverse is true: given A, H has an increased chance of happening.
- ℙ(A | H) > ℙ(A)
Which also means:
- ℙ(H | A) > ℙ(H)
The same goes for negative relationships, however it is flipped: given H, A has an decreased chance of happening.
- ℙ(A | H) < ℙ(A)
Which also means:
- ℙ(H | A) < ℙ(H)
Independent Events
If event H happening does not effect the chance of A then the two sets are said to be independent.
If:
- ℙ(A | H) = ℙ(A)
And/Or:
- ℙ(A ∩ H) = ℙ(A) * ℙ(H)
Then the two events(A and H) are independent.