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IntroductionProbability
Math 141Introduction to Probability and Statistics
Albyn Jones
Mathematics DepartmentLibrary 304
jones@reed.eduwww.people.reed.edu/∼jones/courses/141
September 3, 2014
Albyn Jones Math 141
IntroductionProbability
MotivationHow likely is an eruption at Mount Rainier in the next 25 years?
Albyn Jones Math 141
IntroductionProbability
Data!Post ice-age eruptions
−8000 −6000 −4000 −2000 0
Year
Mount Rainier vs a Poisson Point Process
Mount Rainier
Poisson
Albyn Jones Math 141
IntroductionProbability
Two ModelsDon’t worry about the details!
Poisson Process: Events uniformly distributed in time.Roughly 50 events in the last 12000 years: one every 240years.
Prediction: roughly a 10% chance of an eruption in thenext 25 years, regardless of the elapsed time since the lasteruption.Hidden Markov Model: Two (unobservable) states withdifferent rates. Given the last eruption was roughly 1050years ago, we think we are in a low rate regime: roughlyone eruption every 650 years.Prediction: roughly a 3.7% chance of an eruption in thenext 25 years.
Albyn Jones Math 141
IntroductionProbability
Two ModelsDon’t worry about the details!
Poisson Process: Events uniformly distributed in time.Roughly 50 events in the last 12000 years: one every 240years.Prediction: roughly a 10% chance of an eruption in thenext 25 years, regardless of the elapsed time since the lasteruption.
Hidden Markov Model: Two (unobservable) states withdifferent rates. Given the last eruption was roughly 1050years ago, we think we are in a low rate regime: roughlyone eruption every 650 years.Prediction: roughly a 3.7% chance of an eruption in thenext 25 years.
Albyn Jones Math 141
IntroductionProbability
Two ModelsDon’t worry about the details!
Poisson Process: Events uniformly distributed in time.Roughly 50 events in the last 12000 years: one every 240years.Prediction: roughly a 10% chance of an eruption in thenext 25 years, regardless of the elapsed time since the lasteruption.Hidden Markov Model: Two (unobservable) states withdifferent rates. Given the last eruption was roughly 1050years ago, we think we are in a low rate regime: roughlyone eruption every 650 years.
Prediction: roughly a 3.7% chance of an eruption in thenext 25 years.
Albyn Jones Math 141
IntroductionProbability
Two ModelsDon’t worry about the details!
Poisson Process: Events uniformly distributed in time.Roughly 50 events in the last 12000 years: one every 240years.Prediction: roughly a 10% chance of an eruption in thenext 25 years, regardless of the elapsed time since the lasteruption.Hidden Markov Model: Two (unobservable) states withdifferent rates. Given the last eruption was roughly 1050years ago, we think we are in a low rate regime: roughlyone eruption every 650 years.Prediction: roughly a 3.7% chance of an eruption in thenext 25 years.
Albyn Jones Math 141
IntroductionProbability
QuestionsHint: statistical analysis!
Which of those predictions is more reliable?In other words: which is the better model?
How do we produce estimates for those models?How accurate or trustworthy are those estimates?How do we validate the models?
Albyn Jones Math 141
IntroductionProbability
QuestionsHint: statistical analysis!
Which of those predictions is more reliable?In other words: which is the better model?How do we produce estimates for those models?
How accurate or trustworthy are those estimates?How do we validate the models?
Albyn Jones Math 141
IntroductionProbability
QuestionsHint: statistical analysis!
Which of those predictions is more reliable?In other words: which is the better model?How do we produce estimates for those models?How accurate or trustworthy are those estimates?
How do we validate the models?
Albyn Jones Math 141
IntroductionProbability
QuestionsHint: statistical analysis!
Which of those predictions is more reliable?In other words: which is the better model?How do we produce estimates for those models?How accurate or trustworthy are those estimates?How do we validate the models?
Albyn Jones Math 141
IntroductionProbability
Statisticswhat is it all about?
Formal inference: estimates, confidence intervals andhypothesis tests; quantification of uncertainty.
Tools: probability theory, computational engines like R.
Informal inference: judgements about statistical models —model choice, model validationTools: graphical methods, computational engines like R.
Note the computational theme!
Albyn Jones Math 141
IntroductionProbability
Statisticswhat is it all about?
Formal inference: estimates, confidence intervals andhypothesis tests; quantification of uncertainty.Tools: probability theory, computational engines like R.
Informal inference: judgements about statistical models —model choice, model validationTools: graphical methods, computational engines like R.
Note the computational theme!
Albyn Jones Math 141
IntroductionProbability
Statisticswhat is it all about?
Formal inference: estimates, confidence intervals andhypothesis tests; quantification of uncertainty.Tools: probability theory, computational engines like R.
Informal inference: judgements about statistical models —model choice, model validation
Tools: graphical methods, computational engines like R.
Note the computational theme!
Albyn Jones Math 141
IntroductionProbability
Statisticswhat is it all about?
Formal inference: estimates, confidence intervals andhypothesis tests; quantification of uncertainty.Tools: probability theory, computational engines like R.
Informal inference: judgements about statistical models —model choice, model validationTools: graphical methods, computational engines like R.
Note the computational theme!
Albyn Jones Math 141
IntroductionProbability
Statisticswhat is it all about?
Formal inference: estimates, confidence intervals andhypothesis tests; quantification of uncertainty.Tools: probability theory, computational engines like R.
Informal inference: judgements about statistical models —model choice, model validationTools: graphical methods, computational engines like R.
Note the computational theme!
Albyn Jones Math 141
IntroductionProbability
A Little Probability TheoryThe mathematics we need to quantify uncertainty
A little History: gambling! dice! cards!
A little Philosophy: epistemology and subjective probability,positivism and ‘objective’ probability.
Albyn Jones Math 141
IntroductionProbability
Example
Toss a fair coin 3 times. What is the probability that two of thethree tosses yield heads?
We need some terminology and notation:Sample Space: the set of possible outcomes.
Event: a subset of the sample space.Probability: a function assigning real numbers to events.
Albyn Jones Math 141
IntroductionProbability
Example
Toss a fair coin 3 times. What is the probability that two of thethree tosses yield heads?
We need some terminology and notation:Sample Space: the set of possible outcomes.Event: a subset of the sample space.
Probability: a function assigning real numbers to events.
Albyn Jones Math 141
IntroductionProbability
Example
Toss a fair coin 3 times. What is the probability that two of thethree tosses yield heads?
We need some terminology and notation:Sample Space: the set of possible outcomes.Event: a subset of the sample space.Probability: a function assigning real numbers to events.
Albyn Jones Math 141
IntroductionProbability
Sample Space: Ω
Toss a fair coin 3 times. What are the possible outcomes?
HHHHHT, HTH, THHHTT, THT, TTH
TTT
These are the events in our sample space.
Albyn Jones Math 141
IntroductionProbability
NotationA little set theory
Let A and B be events (subsets of the sample space Ω).
Term Notation Interpretation
Union A ∪ B A or B occurs (or both!)
Intersection A ∩ B A and B both occur
Complement Ac , !A, (Ω \ A) A does not occur
Disjoint Events A ∩ B = ∅ A and B can not both occur
Albyn Jones Math 141
IntroductionProbability
Notation: Examplesthree coin tosses again
‘two heads’: a union of three events
HHT ∪ HTH ∪ THH
‘at least one head’: the complement of ‘no heads’
TTTc = TTH ∪ THT ∪ . . . ∪ HHH
an impossible event!
TTT ∩ HHH
Albyn Jones Math 141
IntroductionProbability
Notation: Examplesthree coin tosses again
‘two heads’: a union of three events
HHT ∪ HTH ∪ THH
‘at least one head’: the complement of ‘no heads’
TTTc = TTH ∪ THT ∪ . . . ∪ HHH
an impossible event!
TTT ∩ HHH
Albyn Jones Math 141
IntroductionProbability
Notation: Examplesthree coin tosses again
‘two heads’: a union of three events
HHT ∪ HTH ∪ THH
‘at least one head’: the complement of ‘no heads’
TTTc = TTH ∪ THT ∪ . . . ∪ HHH
an impossible event!
TTT ∩ HHH
Albyn Jones Math 141
IntroductionProbability
Probabilitythree coin tosses again
Let’s assign probabilities to our 8 events, giving each event in Ωthe same probability (why?).
PHHH = PHHT = . . .PTTT =18
Note the probabilities of the 8 events add up to 1.
Albyn Jones Math 141
IntroductionProbability
More Probability!three coin tosses again
Now, what is the probability of getting two heads in threetosses?
PHHT = PHTH = PTHH =18
The probabilities of these 3 events add up to 3/8. Is that thecorrect value for the probability of getting two heads?
We need some rules for computing probabilities!
Albyn Jones Math 141
IntroductionProbability
Rules for Probabilityaka AXIOMS
Let Ω be a sample space, and E1, E2, E3, . . . be events.P : Events → R according to the following three rules:
1 For any event E :0 ≤ P(E) ≤ 1
2 P(Ω) = 13 If E1, E2, E3, . . . are disjoint events, then
P(E1 ∪ E2 ∪ . . .) =∑
P(Ei) = P(E1) + P(E2) + . . .
Albyn Jones Math 141
IntroductionProbability
Rules for Probabilityaka AXIOMS
Let Ω be a sample space, and E1, E2, E3, . . . be events.P : Events → R according to the following three rules:
1 For any event E :0 ≤ P(E) ≤ 1
2 P(Ω) = 13 If E1, E2, E3, . . . are disjoint events, then
P(E1 ∪ E2 ∪ . . .) =∑
P(Ei) = P(E1) + P(E2) + . . .
Albyn Jones Math 141
IntroductionProbability
Rules for Probabilityaka AXIOMS
Let Ω be a sample space, and E1, E2, E3, . . . be events.P : Events → R according to the following three rules:
1 For any event E :0 ≤ P(E) ≤ 1
2 P(Ω) = 1
3 If E1, E2, E3, . . . are disjoint events, then
P(E1 ∪ E2 ∪ . . .) =∑
P(Ei) = P(E1) + P(E2) + . . .
Albyn Jones Math 141
IntroductionProbability
Rules for Probabilityaka AXIOMS
Let Ω be a sample space, and E1, E2, E3, . . . be events.P : Events → R according to the following three rules:
1 For any event E :0 ≤ P(E) ≤ 1
2 P(Ω) = 13 If E1, E2, E3, . . . are disjoint events, then
P(E1 ∪ E2 ∪ . . .) =∑
P(Ei) = P(E1) + P(E2) + . . .
Albyn Jones Math 141
IntroductionProbability
Examplethree coin tosses again
Now, what is the probability of getting two heads in threetosses, given our assignment of equal probability to each:
P(HHT) = P(HTH) = P(THH) =18
Are these events disjoint?yes!Therefore, by axiom 3,
P(HHT ∪ HTH ∪ THH)
= P(HHT) + P(HTH) + P(THH) =38
Albyn Jones Math 141
IntroductionProbability
Examplethree coin tosses again
Now, what is the probability of getting two heads in threetosses, given our assignment of equal probability to each:
P(HHT) = P(HTH) = P(THH) =18
Are these events disjoint?
yes!Therefore, by axiom 3,
P(HHT ∪ HTH ∪ THH)
= P(HHT) + P(HTH) + P(THH) =38
Albyn Jones Math 141
IntroductionProbability
Examplethree coin tosses again
Now, what is the probability of getting two heads in threetosses, given our assignment of equal probability to each:
P(HHT) = P(HTH) = P(THH) =18
Are these events disjoint?yes!
Therefore, by axiom 3,
P(HHT ∪ HTH ∪ THH)
= P(HHT) + P(HTH) + P(THH) =38
Albyn Jones Math 141
IntroductionProbability
Examplethree coin tosses again
Now, what is the probability of getting two heads in threetosses, given our assignment of equal probability to each:
P(HHT) = P(HTH) = P(THH) =18
Are these events disjoint?yes!Therefore, by axiom 3,
P(HHT ∪ HTH ∪ THH)
= P(HHT) + P(HTH) + P(THH) =38
Albyn Jones Math 141
IntroductionProbability
Complementary Events
What do we know about the events E and Ec?
What is E ∩ Ec?
What is E ∪ Ec?
Albyn Jones Math 141
IntroductionProbability
More on Complementary Events
For any event E , E ∩ Ec = ∅, so E and Ec are disjoint.
For any event E , E ∪ Ec = Ω.
Putting these facts together with our axioms:
1 = P(Ω) = P(E ∪ Ec) = P(E) + P(Ec)
ThusP(Ec) = 1− P(E)
Albyn Jones Math 141
IntroductionProbability
Example: Complementary Events
What is the probability of at least one head in three tosses?
What is the complement of ‘at least one head’?No heads! (All tails.)Using the last result we have
P(at least one head) = P(TTTc)
= 1− P(TTT) = 1− 18
=78
Albyn Jones Math 141
IntroductionProbability
Example: Complementary Events
What is the probability of at least one head in three tosses?What is the complement of ‘at least one head’?
No heads! (All tails.)Using the last result we have
P(at least one head) = P(TTTc)
= 1− P(TTT) = 1− 18
=78
Albyn Jones Math 141
IntroductionProbability
Example: Complementary Events
What is the probability of at least one head in three tosses?What is the complement of ‘at least one head’?No heads! (All tails.)
Using the last result we have
P(at least one head) = P(TTTc)
= 1− P(TTT) = 1− 18
=78
Albyn Jones Math 141
IntroductionProbability
Example: Complementary Events
What is the probability of at least one head in three tosses?What is the complement of ‘at least one head’?No heads! (All tails.)Using the last result we have
P(at least one head) = P(TTTc)
= 1− P(TTT) = 1− 18
=78
Albyn Jones Math 141
IntroductionProbability
A probability inequality
If A ⊂ B, then P(A) ≤ P(B), proof by picture:
B
A
Albyn Jones Math 141
IntroductionProbability
A General Addition Formula: Inclusion/Exclusion
P(A ∪ B) = P(A) + P(B)− P(A ∩ B)
AB
Albyn Jones Math 141
IntroductionProbability
Summary
1 Definitions: Sample Space, Events, Disjoint Events2 Axioms or Rules of Probability3 P(E) = 1− P(Ec)
4 Addition formula:
P(A ∪ B) = P(A) + P(B)− P(A ∩ B)
Albyn Jones Math 141
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