Constraints on Credence Conditional Probability Conditional Probability Question
Constraints on Credence
Conditional Probability
Conditional Probability Question: What is the probability of rolling a prime number (2, 3, 5) when one rolls a 6 sided die?
Conditional Probability Answer: since all the sides are equal: 3/6, or 1/2.
Conditional Probability POSSIBLE ROLLS 1 2 3 4 5 6
Conditional Probability Question: What is the probability of rolling a prime number (2, 3, 5), given that you know the number is even?
STEP 1: Ignore Odd Rolls POSSIBLE ROLLS 1 2 3 4 5 6
STEP 2: Find the Prime Rolls POSSIBLE ROLLS 1 2 3 4 5 6
STEP 3: Count the Prime-and-Even Rolls as a Proportion of the Even Rolls POSSIBLE ROLLS 1 2 3 4 5 6
Conditional Probability Answer: 1/3 This is the probability of prime conditional on rolling even.
Definition of Conditional Probability Pr( p & q ) Pr( p / q ) = Pr( q )
Exercise 1 (from the book) Suppose: Pr( wind ) = 0. 6 Pr( rain ) = 0. 5 Pr( wind & rain ) = 0. 4 What is Pr( wind / rain) and Pr( rain / wind) ?
Pr( wind / rain ) = ? Pr( p & q ) Pr( p / q ) = Pr( q )
Pr( wind / rain ) = ? Pr( wind & rain ) Pr( wind / rain ) = Pr( rain ) Pr( wind ) = 0. 6 Pr( rain ) = 0. 5 Pr( wind & rain ) = 0. 4
Pr( wind / rain ) = ? Pr( wind / rain ) = 0. 4 Pr( rain ) Pr( wind ) = 0. 6 Pr( rain ) = 0. 5 Pr( wind & rain ) = 0. 4
Pr( wind / rain ) = ? Pr( wind / rain ) = 0. 4 0. 5 Pr( wind ) = 0. 6 Pr( rain ) = 0. 5 Pr( wind & rain ) = 0. 4
Pr( wind / rain ) = ? Pr( wind / rain ) = 0. 4 0. 5 = 4/5 Pr( wind ) = 0. 6 Pr( rain ) = 0. 5 Pr( wind & rain ) = 0. 4
Pr( rain / wind ) = ? Pr( p & q ) Pr( p / q ) = Pr( q )
Pr( rain / wind ) = ? Pr( rain & wind ) Pr( rain / wind ) = Pr( wind ) = 0. 6 Pr( rain ) = 0. 5 Pr( wind & rain ) = 0. 4
Pr( rain / wind ) = ? Pr( rain & wind ) Pr( rain / wind ) = Pr( wind ) = 0. 6 Pr( rain ) = 0. 5 Pr( wind & rain ) = 0. 4
Pr( rain / wind ) = ? Pr( rain / wind ) = 0. 4 Pr( wind ) = 0. 6 Pr( rain ) = 0. 5 Pr( wind & rain ) = 0. 4
Pr( rain / wind ) = ? Pr( rain / wind ) = 0. 4 0. 6 Pr( wind ) = 0. 6 Pr( rain ) = 0. 5 Pr( wind & rain ) = 0. 4
Pr( rain / wind ) = ? Pr( rain / wind ) = 0. 4 0. 6 = 2/3 Pr( wind ) = 0. 6 Pr( rain ) = 0. 5 Pr( wind & rain ) = 0. 4
Bayes’ Theorem
Reverend Thomas Bayes • Relatively obscure British mathematician. • Proved a (more specific) instance of theorem that bears his name
Bayes’ Theorem Pr( q / p ) x Pr( p ) Pr( p / q ) = Pr( q )
Proof 1. Pr( a / b ) = Pr( a & b ) ÷ Pr( b ) by Definition of Conditional Probability 2. Pr( a / b ) x Pr( b ) = Pr( a & b ) multiplying both sides by Pr( b ) 3. Pr( a & b ) = Pr( a / b ) x Pr( b ) symmetry of = 4. Pr( q & p ) = Pr( q / p ) x Pr( p ) substitution
Formula from Last Time Area(Z & B) = Area(Z) x Area(B/ Z) This is the proportion of Z that is taken up by B
Proof Part 2 1. Pr( p / q ) = Pr( p & q ) ÷ Pr( q ) by Definition of Conditional Probability 2. p & q = q & p by logic 3. Pr( p / q ) = Pr( q & p ) ÷ Pr( q ) by 1, 2, and substitution 4. Pr( p / q ) = [ Pr( q / p ) x Pr( p ) ] ÷ Pr( q ) substitution from last slide
Base Rate Fallacy
Elements of Bayes’ Theorem “likelihood” “prior probability” Pr( q / p ) x Pr( p ) Pr( p / q ) = Pr( q )
Bayes’ Theorem Baye’s theorem lets us calculate the probability of A conditional on B when we have the probability of B conditional on A.
Base Rate Fallacy • There are ½ million people in Russia are affected by HIV/ AIDS. • There are 150 million people in Russia.
Base Rate Fallacy Imagine that the government decides this is bad and that they should test everyone for HIV/ AIDS.
The Test If someone has HIV/ AIDS, then : • 95% of the time the test will be positive (correct) • 5% of the time will it be negative (incorrect)
The Test If someone does not have HIV/ AIDS, then: • 95% of the time the test will be negative (correct) • 5% of the time will it be positive (incorrect)
Suppose you test positive. We’re interested in the conditional probability: what is the probability you have HIV assuming that you test positive. We’re interested in Pr(HIV = yes/ test = pos)
Known: Pr(sick) = 1/300 Known: Pr(positive/ sick) = 95% Known: Pr(positive/ not-sick) = 5% Unknown: Pr(positive) Unknown: Pr(sick/ positive)
Pr(positive) = True positives + false positives = [Pr(positive/ sick) x Pr(sick)] + [Pr(positive/ not-sick) x Pr(not-sick)] = [95% x 1/300] + [5% x 299/300] = 5. 3% Known: Pr(sick) = 1/300 Known: Pr(positive/ sick) = 95% Known: Pr(positive/ not-sick) = 5%
Known: Pr(sick) = 1/300 Known: Pr(positive/ sick) = 95% Known: Pr(positive) = 5. 3% Unknown: Pr(sick/ positive)
Pr(sick/ positive) Pr(A/ B) = [Pr(B/ A) x Pr(A)] ÷ Pr(B) Pr(sick/ positive) = [Pr(positive/ sick) x Pr(sick)] ÷ Pr(positive) = [95% x 1/300] ÷ 5. 3% = 5. 975% Known: Pr(sick) = 1/300 Known: Pr(positive/ sick) = 95% Known: Pr(positive) = 5. 3%
Conditionalization
Confirming Hypotheses Suppose there is a certain hypothesis H for which you are collecting evidence E. For example, the hypothesis could be that some person is or is not HIVpositive.
Elements of Bayes’ Theorem “likelihood” “prior probability” Pr( E / H ) x Pr( H ) Pr( H / E ) = Pr( E )
Elements of Bayes’ Theorem Pr( H ) is the prior probability of your hypothesis H being true– that is, prior to collecting any evidence. Pr( E / H ) is the likelihood that you will observe evidence E, on the assumption that your hypothesis H is true.
Confirming Hypotheses Now suppose you collect the evidence and you know that E is true. Your degree of belief in your hypothesis H should now change (higher if E supports it, lower if it supports not-H).
Bayesianism, i. e. Conditionalization Pr. NEW( H ) = Pr. OLD( H / E )
Diachronic Constraint The axioms of probability are synchronic constraints on your degrees of belief: they are how those degrees should relate to one another at any moment. A bunch of different probability functions all satisfy these axioms.
Diachronic Constraint Therefore, for all the axioms care, Pr. NEW(H) can be anything at all, so long as, for example Pr. NEW(not-H) is 1 – Pr. NEW(H), etc. The axioms don’t tell you how your degrees of belief should relate to each other over time.
Conditionalization Does! Pr. NEW( H ) = Pr. OLD( H / E )
Exercise 3 (from the book) Suppose you have good reason to believe that Pr( H ) = 0. 1 Pr( E ) = 0. 2 Pr( E / H ) = 0. 8 Then you learn E. What probability should you now attach to H?
Start with Bayes’ Theorem Pr( E / H ) x Pr( H ) Pr( H / E ) = Pr( E ) Pr( H ) = 0. 1 Pr( E ) = 0. 2 Pr( E / H ) = 0. 8
Fill in What You Know Pr( H / E ) = 0. 8 x 0. 1 0. 2 Pr( H ) = 0. 1 Pr( E ) = 0. 2 Pr( E / H ) = 0. 8
Calculate! Pr( H / E ) = 0. 8 x 0. 1 0. 2 = 0. 4 Pr( H ) = 0. 1 Pr( E ) = 0. 2 Pr( E / H ) = 0. 8
Become a Bayesian! Pr. NEW( H ) = Pr. OLD( H / E ) = 0. 8 x 0. 1 0. 2 = 0. 4 Pr( H ) = 0. 1 Pr( E ) = 0. 2 Pr( E / H ) = 0. 8
Problem #4 & Theorem of Total Probability
Problem #4 Warning: NOT ON THE TEST! You have a 10% degree of belief that a coin is not fair but has a 75% bias in favor of heads. You toss it twice and see two heads. What should be your degree of belief that it is fair?
Problem #4 Warning: NOT ON THE TEST! You have a 10% degree of belief that a coin is not fair but has a 75% bias in favor of heads. You toss it twice and see two heads. What should be your degree of belief that it is fair? ASSUMPTION: You have 90% degree of belief that it is fair.
What Do We Know? Hypothesis H: Coin is biased 75% in favor of heads. Pr( H ) = 10%
What Do We Know? Evidence E = Coin lands heads twice in a row. Pr( E ) = ? ? ?
What Do We Know What is Pr( E / H )? Well, if the coin is biased 75% towards heads, then there is a 75% probability that it lands heads on the first toss, and a 75% probability that it lands heads on the second toss. Since the tosses are independent: Pr( both heads ) = Pr ( heads first ) x Pr (heads second) = 3/4 x 3/4 = 9/16.
How Do We Solve? we know this Pr( E / H ) x Pr( H ) Pr( H / E ) = Pr( E ) we don’t know this
Logic to the Rescue Theorem: P is equivalent to [ ( P & Q ) or ( P & not-Q) ] Proof part 1: Suppose P. Now either Q is true or it is false. But you know P is true. So you are either living in a world where P and Q are both true, or you are living in a world where P is true but Q is false. Proof part 2: Suppose that one of the following two possibilities is true (a) P and Q (b) P and not-Q. If the first possibility is true, then P. If the second possibility is true, also P. So either way, P.
Using the Theorem Pr( E ) = Pr[ ( E & H ) or ( E & not-H ) ] = Pr( E & H ) + Pr( E & not-H ) because ‘E & H’ and ‘E & not-H’ are exclusive = Pr( E / H ) x Pr( H ) + Pr( E / not-H) x Pr( not-H ) by the Definition of Conditional Probability
Definition of Conditional Probability Pr( E & H ) Pr( E / H ) = Pr( H )
Using the Theorem Pr( E ) = Pr[ ( E & H ) or ( E & not-H ) ] = Pr( E & H ) + Pr( E & not-H ) because ‘E & H’ and ‘E & not-H’ are exclusive = Pr( E / H ) x Pr( H ) + Pr( E / not-H) x Pr( not-H ) by the Definition of Conditional Probability = 9/16 x 1/10 + Pr( E / not-H ) x 9/10 earlier calculations = 9/16 x 1/10 + 1/4 x 9/10 = 9/32
How Do We Solve? we know this Pr( E / H ) x Pr( H ) Pr( H / E ) = Pr( E ) now we know this
How Do We Solve? we know this Pr( H / E ) = we know this 9/16 x 1/10 9/32 now we know this
How Do We Solve? we know this Pr( H / E ) = we know this 9/16 x 1/10 9/32 = 0. 2 now we know this
The Principal Principle
Cr( p / Pr(p) = x
- Slides: 71