math 141 - introduction to probability and statistics

IntroductionProbability

Math 141Introduction to Probability and Statistics

Albyn Jones

Mathematics DepartmentLibrary 304

[email protected]/∼jones/courses/141

September 3, 2014

Albyn Jones Math 141


MotivationHow likely is an eruption at Mount Rainier in the next 25 years?



Data!Post ice-age eruptions

−8000 −6000 −4000 −2000 0

Year

Mount Rainier vs a Poisson Point Process

Mount Rainier

Poisson



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.




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.




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.




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.



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?




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?




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?




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?



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!




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.




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.



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.



Example


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.



Example


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.



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.



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



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



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.



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!



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) + . . .






2 P(Ω) = 1

3 If E1, E2, E3, . . . are disjoint events, then

P(E1 ∪ E2 ∪ . . .) =∑

P(Ei) = P(E1) + P(E2) + . . .






2 P(Ω) = 13 If E1, E2, E3, . . . are disjoint events, then

P(E1 ∪ E2 ∪ . . .) =∑

P(Ei) = P(E1) + P(E2) + . . .



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






Are these events disjoint?

yes!Therefore, by axiom 3,








Are these events disjoint?yes!

Therefore, by axiom 3,








Are these events disjoint?yes!Therefore, by axiom 3,





Complementary Events

What do we know about the events E and Ec?

What is E ∩ Ec?

What is E ∪ Ec?



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)



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




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


= 1− P(TTT) = 1− 18

=78




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


= 1− P(TTT) = 1− 18

=78




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


= 1− P(TTT) = 1− 18

=78



A probability inequality

If A ⊂ B, then P(A) ≤ P(B), proof by picture:

B

A



A General Addition Formula: Inclusion/Exclusion

P(A ∪ B) = P(A) + P(B)− P(A ∩ B)

AB



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)



Assignment!

Read Chapter 2.


math 141 - introduction to probability and statistics

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