Difference between revisions of "User:Saul/probability"
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'''⊃''' - Superset of.<br> | '''⊃''' - Superset of.<br> | ||
'''⊇''' - Is not a superset of.<br> | '''⊇''' - Is not a superset of.<br> | ||
| − | '''ℙ(A)''' - Probability of '''A'''.<br> | + | '''ℙ(A)''' - A ''set'' function of set '''A''' - Probability of '''A'''.<br> |
'''A<sup>c</sup>''' - A compliment - The event '''A''' does not occur.<br> | '''A<sup>c</sup>''' - A compliment - The event '''A''' does not occur.<br> | ||
'''ω∈A''' - The outcome '''ω''' is in the event '''A'''<br> | '''ω∈A''' - The outcome '''ω''' is in the event '''A'''<br> | ||
| Line 21: | Line 21: | ||
'''ℙ(A∪B) = ℙ(A) + ℙ(B)''' - The sets '''A''' and '''B''' are disjoint.<br> | '''ℙ(A∪B) = ℙ(A) + ℙ(B)''' - The sets '''A''' and '''B''' are disjoint.<br> | ||
'''#A''' - The number of elements '''A''' contains if '''A''' is finite.<br> | '''#A''' - The number of elements '''A''' contains if '''A''' is finite.<br> | ||
| + | '''ℙ(S) = 1''' - The set '''S''' is a ''exhaustive set'' as it contains all the outcomes.<br> | ||
| + | ''Possible outcome'' - may have a probability of '''0'''.<br> | ||
== Axioms == | == Axioms == | ||
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''Finite additivity.''<br> | ''Finite additivity.''<br> | ||
'''ℙ(A<sup>c</sup>) = 1 - ℙ(A)''' ''Since'' '''A ∪ A<sup>c</sup> = Ω'''<br> | '''ℙ(A<sup>c</sup>) = 1 - ℙ(A)''' ''Since'' '''A ∪ A<sup>c</sup> = Ω'''<br> | ||
| − | ''If'' '''A⊆B''' ''Then'' '''ℙ(A) ≤ ℙ(B)''' ''Since'' '''A∪(A<sup>c</sup>∩B) = B'''<br> | + | ''If'' '''A ⊆ B''' ''Then'' '''ℙ(A) ≤ ℙ(B)''' ''Since'' '''A∪(A<sup>c</sup> ∩ B) = B'''<br> |
'''ℙ(A) ≤ 1''' ''Since'' '''A ⊆ Ω'''<br> | '''ℙ(A) ≤ 1''' ''Since'' '''A ⊆ Ω'''<br> | ||
| − | '''ℙ(A∪B) = ℙ(A) + ℙ(B) - ℙ(A∩B)'''<br> | + | '''ℙ(A ∪ B) = ℙ(A) + ℙ(B) - ℙ(A ∩ B)'''<br> |
| + | |||
| + | == Conditional Probability == | ||
| + | '''ℙ(A | H)''' - Given '''H'' occurs what is the probability of '''A''.<br> | ||
| + | '''ℙ(A | H) = ℙ(A ∩ H) / ℙ(H)''' | ||
| + | : ''If'' '''ℙ(H) > 0'''<br> | ||
| + | <br> | ||
| + | For example:<br> | ||
| + | Two dice are rolled the outcomes being '''i''' and '''j'''.<br> | ||
| + | ''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}''' | ||
| + | <br> | ||
| + | We could start by listing all the possible outcomes that are in set '''H'':<br> | ||
| + | {| class="wikitable" | ||
| + | |- | ||
| + | ! '''i''' !! '''j''' | ||
| + | |- | ||
| + | | 1 || 6 | ||
| + | |- | ||
| + | | 2 || 5 | ||
| + | |- | ||
| + | | 3 || 4 | ||
| + | |- | ||
| + | | 4 || 3 | ||
| + | |- | ||
| + | | 5 || 2 | ||
| + | |- | ||
| + | | 6 || 1 | ||
| + | |} | ||
| + | And all the outcomes in set '''A'''<br> | ||
| + | {| class="wikitable" | ||
| + | |- | ||
| + | ! '''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'''): | ||
| + | {| class="wikitable" | ||
| + | |- | ||
| + | ! '''i''' !! '''j''' | ||
| + | |- | ||
| + | | 3 || 4 | ||
| + | |- | ||
| + | | 4 || 3 | ||
| + | |} | ||
| + | We can say there are '''6 * 6 = 36''' total possible outcomes ('''Ω''').<br> | ||
| + | : 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'''.<br> | ||
| + | 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)''' | ||
| + | <br> | ||
| + | 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.<br> | ||
| + | ''If:'' | ||
| + | : '''ℙ(A | H) = ℙ(A)''' | ||
| + | ''And/Or:'' | ||
| + | : '''ℙ(A ∩ H) = ℙ(A) * ℙ(H) ''' | ||
| + | Then the two events('''A''' and '''H''') are independent. | ||
Latest revision as of 00:28, 7 March 2020
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.
