reasoning with probs

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Reasoning with Probs How does evidence lead to conclusions in situations of uncertainty? Bayes Theorem Data fusion, use of techniques that combine data from multiple sources and gather that information in order to achieve inferences, which will be more efficient and potentially more accurate than if they were achieved by means of a single source. Spam Cancer Screening Law Exams 1 ST2004 Week 7

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Reasoning with Probs. How does evidence lead to conclusions in situations of uncertainty? Bayes Theorem - PowerPoint PPT Presentation

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Page 1: Reasoning with  Probs

ST2004 Week 7 1

Reasoning with Probs

How does evidence lead to conclusions in situations of uncertainty? Bayes Theorem

Data fusion, use of techniques that combine data from multiple sources and gather that information in order to achieve inferences, which will be more efficient and potentially more accurate than if they were achieved by means of a single source.

SpamCancer ScreeningLawExams

Page 2: Reasoning with  Probs

ST2004 Week 7 2

Probability RulesConditional Prob and Independence

Important special Pr( ) Pr( | ) Pr( )

Pr( ) Pr( ) Prcase

( )

Aand B A B B

Aand B A B when independent

Multiplication Rule

All computed probs: depend on real world knowledgeassumptions about real world

Sometimes Useful to be explicitWhat - if

Page 3: Reasoning with  Probs

ST2004 Week 7 3

Exam Q from 2010

Page 4: Reasoning with  Probs

ST2004 Week 7 4

Brain Teasers• Two regular dice are rolled.

One is a 6. What’s the Pr (Other is 6) ? (Tijms, 8.1)

• Who is the murderer? (Tijms, Ch 1 Q6)?

Murder committed; know either X or Y – equally likely.Evidence: actual perp has blood group A

10% of people group A; X is group A Seek Pr( X is perp | evidence)

Page 5: Reasoning with  Probs

ST2004 Week 7 5

Brain Teasers

• Monty Hall Game Show (Tijms, Ch 1, Q11)

One car, behind one of three doors.Player selects one: say Door 1 Before opening this door

host opens one of two others: say Door 2 GOAT!host offers chance to change selection.

Issue Is there any point changing?

Page 6: Reasoning with  Probs

ST2004 Week 7 6

Life Expectancy in Ireland

Average age at death = 75

Average age at death = 80

Average age at death,given survival to 60, = 79

http://understandinguncertainty.org/node/272

Page 7: Reasoning with  Probs

ST2004 Week 7 7

Light

Metro Wed 10 Nov 2010Teens at risk from hyper-texting

Teenagers who send more than 100 text messages per day are more likely to have had sex, tried drugs, research has revealed.

4200 students at 20 schools; hyper-texting 19.2%Such teens 43% more likely to have tried alcohol.

Page 8: Reasoning with  Probs

ST2004 Week 7 8

Serious

Sally Clarke - Sudden Infant Death SID The case was widely criticised because of the way statistical evidence was misrepresented in the original trial, particularly by Meadow. He stated in evidence as an expert witness that "one sudden infant death in a family is a tragedy, two is suspicious and three is murder unless proven otherwise" (Meadow's law).

He claimed that, for an affluent non-smoking family like the Clarks, the probability of a single cot death was 1 in 8,543, so the probability of two cot deaths in the same family was around "1 in 73 million" (8543 × 8543).

Page 9: Reasoning with  Probs

ST2004 Week 7 10

Cond Prob for LifetimesKnowledge of current age impacts uncertainty on age at death

Probability DistributionPoss LiveTimes 1 2 3 4 5 6Corresp Probs 0.1 0.2 0.3 0.3 0.05 0.05

Pr(death at end day 3, given alive at start day 3)

Page 10: Reasoning with  Probs

ST2004 Week 7 12

Conditional Prob: Chain Rule

Extension Chain Rule

Pr Pr( ) | ) Pr( )

Pr(

Pr( )Pr( ) P

| ) Pr( | ) Pr( | )

Pr Pr( | ) Pr( | ) Pr(

r( | ) Pr( ) Pr( | )Pr( )

Pr( , , ) Pr( | , ) Pr( | ) P (

)

r )

A ANDB ANDC A ANDB C C

But A ANDB C A B AND

Aand BAand B A B B A BB

A

C B C

Thus A ANDB ANDC A B ANDC

B C

B

A B C B C C

Sp

C C

Pr( ) Pr( ) Pr( )ecial case if events probabilistically iA B C ndep

Page 11: Reasoning with  Probs

ST2004 Week 7 13

Conditional Prob: Chain Rule

th = Ace on i drawSeek Pr (3A in 3 Draws)

Event Identity

Prob EquationPr 3A in

Defi

3 d

n

ws

e

ra

iA

Page 12: Reasoning with  Probs

ST2004 Week 7 17

Decomposition via Conditional Probs

1 2

1 2

1 2

( )

Pr( )

................

................

...

Pr( )

n

n

n

A A AND B OR B A AND B OR A ANDB

A

B OR B OR B

A A AND AAND B OR B OR B

A ANDB OR A ANDB OR A ANDB

A

Page 13: Reasoning with  Probs

ST2004 Week 7 18

Decomposition via Conditional Probs

nd

Regular pack of cards; no replacement

Pr( 2 card is Q)

Chance Tree

Page 14: Reasoning with  Probs

ST2004 Week 7 21

Decomposition via Conditional Probs

Define Score on die; no heads in tossesS F S

Chance Tree

Page 15: Reasoning with  Probs

ST2004 Week 7 23

Brain TeasersMonty Hall Game Show (Tijms, Ch 1, Q11)One car, behind one of three doors.

Player selects one: say Door 1 Before opening this door

host opens one of two others: say Door 2 GOAT!host offers chance to change selection.

Issue Is there any point changing?

Page 16: Reasoning with  Probs

ST2004 Week 7 24

Decomposition by Conditional Sim

Contestant always chooses Door 1Car behind random door – equal probsHost actions Don’t Switch Do Switch PrizeCar behind

123

Pseudo Code

Page 17: Reasoning with  Probs

ST2004 Week 7 25

Decomposition by Conditional SimContestant Always Chooses Door 1Car behind Random DoorGiven Car behind Door

1 2 3 Car behind randomly chosen door2

Pr Host opens Host Opens Door 31 0 0.5 0.52 0 0 1 Relevant Cum probs3 0 1 0 1 2 3

0 0 0 1Cum Probs

1 0 0 0.5 12 0 0 0 1 Win3 0 0 1 1 Don't switch 1 Goat

Switch to door 2 Car

Final Choice Reps

1 Goat Car2 Car Goat3 Car Goat4 Goat Car 645 3555 Goat Car6 Goat Car

Don't switch

Do Switch

Don't switch

Do Switch

Number of Times win Car

Page 18: Reasoning with  Probs

ST2004 Week 7 26

Decomposition via Conditional ProbsChance Tree

2

1

1 2

3

3

2

3

Contestantchooses door1, for example

3

2

2

3

Car

Car

Goat

Goat

Contestantdoes not switch;

Car

Car

Goat

Goat

Contestantswitches

Prob wins =

Carbehind

Hostopens

Page 19: Reasoning with  Probs

ST2004 Week 7 29

Bayes Rule

Pr( ) Pr( | ) Pr( ) Pr( ) Pr( | ) Pr( )Pr( )

Pr( | ) Pr( | )Pr( )

Aand B A B B Band A B A AB

B A A BA

Inverting the Conditioning

Multiple Possibilities

1 2

1 1

................Pr( | )Pr( )Pr( | )

Pr( | )Pr( ) ....... Pr( | )Pr( )

n

i ii

n n

B B OR B OR BA B BB A

A B B A B B

Page 20: Reasoning with  Probs

ST2004 Week 7 30

Brain Teasers• Who is the murderer? (Tijms, Ch 1 Q6)?

Murder committed; know either X or Y – equally likely.Evidence: actual perp has blood group A

10% of people group A; X is group A Seek Pr( X is perp | evidence)

Page 21: Reasoning with  Probs

ST2004 Week 7 34

Brain Teaser

Roll regular die: note score NToss fair coin N times

Observe no heads.What now Pr(scored 2)?Pr(scored 2|no heads)?

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ST2004 Week 7 35

Bayes Rule

Odds Rule Form for EvidencePr( ) Pr( )Pr( | ) Pr( | ) ; Pr( | ) Pr( | )Pr( ) Pr( )

Pr( | ) Pr( | ) Pr( )Pr( | ) Pr( | ) Pr( )

B AA E E A A E E AA E

A E E A AA E E A A

Evidence Fusion

1 1

1 1

Pr( | ..... ) Pr( | )Pr( | ) Pr( )....Pr( | ..... ) Pr( | ) Pr( | ) Pr( )

if evidence indep given ,

n n

n n

A E AND AND E E AE A AA E AND AND E E A E A A

A A

Page 23: Reasoning with  Probs

ST2004 Week 7 37

Serious

Sally Clarke - Sudden Infant Death SID He claimed that, for an affluent non-smoking family like the Clarks, the

probability of a single cot death was 1 in 8,543, so the probability of two cot deaths in the same family was around "1 in 73 million" (8543 × 8543).

218543Pr(2 Cot Deaths | Normal Family) = Pr(Normal Family | 2 Cot Deaths)

Pr(Normal | 2 Deaths) Pr(Normal ) Pr(2 Deaths | Normal)= Pr(Not normal | 2 Deaths) Pr(Not normal ) Pr(2 Deaths | Not normal)

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ST2004 Week 7 39

HomeWork• 2010 Exam Q• Discuss Metro• Tijms

– Q1, Ch 1 22 players + ref Friend bets €10 at least one common birthday

What is fair price of the bet

– Q12, Ch 1Told family has two children; one is a daughterProb other is daughter?Told family has two children; ring bell – girl opensProb other also a girl?

Page 25: Reasoning with  Probs

ST2004 Week 7 43

Spam Exam QA simple spam filter is used on a single incoming message. You know only a 1% chance of such messages are spam. You also know that the filter is imperfect - with false positive (ie positive for spam given not spam) and false negative rates of 5% and 2% respectively.

Defining events F± and S in a natural way, restate this information in terms

Pr( ),Pr( | ),Pr( | )S F S F S

Define Reported as spamPr( ) 0.01Pr( | ) Pr( | ) Pr(False Negative) = 0.02

Pr( | ) Pr( | ) Pr(False Positive) = 0.05

RSR S F S

R S F S