Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’...
Transcript of Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’...
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Bayes’ Rule
Bayes’ Rule
Review
8.1
Lecture 8Bayes’ RuleText: A Course in Probability by Weiss 4.4
STAT 225 Introduction to Probability ModelsFebruary 2, 2014
Whitney HuangPurdue University
![Page 2: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/2.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.2
Agenda
1 Bayes’ Rule
2 Review
![Page 3: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/3.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.3
Motivating example
Example (The Monty Hall Problem)
There was an old television show called Let’s Make a Deal,whose original host was named Monty Hall. The set–up is asfollows. You are on a game show and you are given the choiceof three doors. Behind one door is a car, behind the others aregoats. You pick a door, and the host, who knows what isbehind the doors, opens another door (not your pick) which hasa goat behind it. Then he asks you if you want to change youroriginal pick. The question we ask you is, “Is it to youradvantage to switch your choice?"
![Page 4: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/4.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.4
The Monty Hall Problem
![Page 5: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/5.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.5
The Monty Hall Problem
Solution.
![Page 6: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/6.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.5
The Monty Hall Problem
Solution.
![Page 7: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/7.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.6
Bayes’ Rule
Let A1,A2, · · · ,Ak form a partition of the sample space. Thenfor every event B in the sample space,
P(Aj |B) =P(B|Aj )× P(Aj )∑ki=1 P(B|Ai )× P(Ai )
, j = 1,2, · · · , k
![Page 8: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/8.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.7
Example 24
Let us assume that a specific disease is only present in 5 out ofevery 1,000 people. Suppose that the test for the disease isaccurate 99% of the time a person has the disease and 95% ofthe time that a person lacks the disease. What is the probabilitythat the person has the disease given that they tested positive?
Solution.
P(D|+) = P(D∩+)P(+) = .005×.99
.005×.99+.995×.05 = .00495.0547 = .0905
The reason we get such a surprising result is because thedisease is so rare that the number of false positives greatlyoutnumbers the people who truly have the disease.
![Page 9: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/9.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.7
Example 24
Let us assume that a specific disease is only present in 5 out ofevery 1,000 people. Suppose that the test for the disease isaccurate 99% of the time a person has the disease and 95% ofthe time that a person lacks the disease. What is the probabilitythat the person has the disease given that they tested positive?
Solution.
P(D|+) = P(D∩+)P(+) = .005×.99
.005×.99+.995×.05 = .00495.0547 = .0905
The reason we get such a surprising result is because thedisease is so rare that the number of false positives greatlyoutnumbers the people who truly have the disease.
![Page 10: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/10.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.8
Basic Concepts
Random experiment: sample space, element, eventSet operations: union, intersectionSet relations: mutually exclusive, exhaust, partition
![Page 11: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/11.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.8
Basic Concepts
Random experiment: sample space, element, event
Set operations: union, intersectionSet relations: mutually exclusive, exhaust, partition
![Page 12: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/12.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.8
Basic Concepts
Random experiment: sample space, element, eventSet operations: union, intersection
Set relations: mutually exclusive, exhaust, partition
![Page 13: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/13.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.8
Basic Concepts
Random experiment: sample space, element, eventSet operations: union, intersectionSet relations: mutually exclusive, exhaust, partition
![Page 14: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/14.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.9
Probability Rules
0 ≤ P(E) ≤ 1 for any event E , P(∅) = 0, P(Ω) = 1Complement rule: P(A) = 1− P(Ac)
General addition rule: P(A ∪ B) = P(A) + P(B)− P(A ∩ B)
General multiplication rule:P(A ∩ B) = P(A|B)× P(B) = P(B|A)× P(A)
Conditional probability: P(A|B) = P(A∩B)P(B)
Law of total probability:P(B) =
∑ki=1 P(B ∩ Ai ) =
∑ki=1 P(B|Ai )× P(Ai )
Independence: if A and B are independent, thenP(A|B) = P(A), P(B|A) = P(B), and P(A ∩ B) = P(A)P(B)
![Page 15: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/15.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.9
Probability Rules
0 ≤ P(E) ≤ 1 for any event E , P(∅) = 0, P(Ω) = 1
Complement rule: P(A) = 1− P(Ac)
General addition rule: P(A ∪ B) = P(A) + P(B)− P(A ∩ B)
General multiplication rule:P(A ∩ B) = P(A|B)× P(B) = P(B|A)× P(A)
Conditional probability: P(A|B) = P(A∩B)P(B)
Law of total probability:P(B) =
∑ki=1 P(B ∩ Ai ) =
∑ki=1 P(B|Ai )× P(Ai )
Independence: if A and B are independent, thenP(A|B) = P(A), P(B|A) = P(B), and P(A ∩ B) = P(A)P(B)
![Page 16: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/16.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.9
Probability Rules
0 ≤ P(E) ≤ 1 for any event E , P(∅) = 0, P(Ω) = 1Complement rule: P(A) = 1− P(Ac)
General addition rule: P(A ∪ B) = P(A) + P(B)− P(A ∩ B)
General multiplication rule:P(A ∩ B) = P(A|B)× P(B) = P(B|A)× P(A)
Conditional probability: P(A|B) = P(A∩B)P(B)
Law of total probability:P(B) =
∑ki=1 P(B ∩ Ai ) =
∑ki=1 P(B|Ai )× P(Ai )
Independence: if A and B are independent, thenP(A|B) = P(A), P(B|A) = P(B), and P(A ∩ B) = P(A)P(B)
![Page 17: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/17.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.9
Probability Rules
0 ≤ P(E) ≤ 1 for any event E , P(∅) = 0, P(Ω) = 1Complement rule: P(A) = 1− P(Ac)
General addition rule: P(A ∪ B) = P(A) + P(B)− P(A ∩ B)
General multiplication rule:P(A ∩ B) = P(A|B)× P(B) = P(B|A)× P(A)
Conditional probability: P(A|B) = P(A∩B)P(B)
Law of total probability:P(B) =
∑ki=1 P(B ∩ Ai ) =
∑ki=1 P(B|Ai )× P(Ai )
Independence: if A and B are independent, thenP(A|B) = P(A), P(B|A) = P(B), and P(A ∩ B) = P(A)P(B)
![Page 18: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/18.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.9
Probability Rules
0 ≤ P(E) ≤ 1 for any event E , P(∅) = 0, P(Ω) = 1Complement rule: P(A) = 1− P(Ac)
General addition rule: P(A ∪ B) = P(A) + P(B)− P(A ∩ B)
General multiplication rule:P(A ∩ B) = P(A|B)× P(B) = P(B|A)× P(A)
Conditional probability: P(A|B) = P(A∩B)P(B)
Law of total probability:P(B) =
∑ki=1 P(B ∩ Ai ) =
∑ki=1 P(B|Ai )× P(Ai )
Independence: if A and B are independent, thenP(A|B) = P(A), P(B|A) = P(B), and P(A ∩ B) = P(A)P(B)
![Page 19: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/19.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.9
Probability Rules
0 ≤ P(E) ≤ 1 for any event E , P(∅) = 0, P(Ω) = 1Complement rule: P(A) = 1− P(Ac)
General addition rule: P(A ∪ B) = P(A) + P(B)− P(A ∩ B)
General multiplication rule:P(A ∩ B) = P(A|B)× P(B) = P(B|A)× P(A)
Conditional probability: P(A|B) = P(A∩B)P(B)
Law of total probability:P(B) =
∑ki=1 P(B ∩ Ai ) =
∑ki=1 P(B|Ai )× P(Ai )
Independence: if A and B are independent, thenP(A|B) = P(A), P(B|A) = P(B), and P(A ∩ B) = P(A)P(B)
![Page 20: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/20.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.9
Probability Rules
0 ≤ P(E) ≤ 1 for any event E , P(∅) = 0, P(Ω) = 1Complement rule: P(A) = 1− P(Ac)
General addition rule: P(A ∪ B) = P(A) + P(B)− P(A ∩ B)
General multiplication rule:P(A ∩ B) = P(A|B)× P(B) = P(B|A)× P(A)
Conditional probability: P(A|B) = P(A∩B)P(B)
Law of total probability:P(B) =
∑ki=1 P(B ∩ Ai ) =
∑ki=1 P(B|Ai )× P(Ai )
Independence: if A and B are independent, thenP(A|B) = P(A), P(B|A) = P(B), and P(A ∩ B) = P(A)P(B)
![Page 21: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/21.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.9
Probability Rules
0 ≤ P(E) ≤ 1 for any event E , P(∅) = 0, P(Ω) = 1Complement rule: P(A) = 1− P(Ac)
General addition rule: P(A ∪ B) = P(A) + P(B)− P(A ∩ B)
General multiplication rule:P(A ∩ B) = P(A|B)× P(B) = P(B|A)× P(A)
Conditional probability: P(A|B) = P(A∩B)P(B)
Law of total probability:P(B) =
∑ki=1 P(B ∩ Ai ) =
∑ki=1 P(B|Ai )× P(Ai )
Independence: if A and B are independent, thenP(A|B) = P(A), P(B|A) = P(B), and P(A ∩ B) = P(A)P(B)
![Page 22: Bayes’ Rule Lecture 8 - Purdue University › ~huang251 › slides8.pdf · Lecture 8 Bayes’ Rule Text: A Course in Probability by Weiss 4:4 STAT 225 Introduction to Probability](https://reader033.fdocuments.us/reader033/viewer/2022060405/5f0f26e47e708231d442be5b/html5/thumbnails/22.jpg)
Bayes’ Rule
Bayes’ Rule
Review
8.10
Counting Problem
Basic counting rulePermutationCombinationOrdered partition