18.05 S22 Class 12: Bayesian Updating: Odds

Bayesian Updating: Odds Class 12, 18.05 1 Learning Goals 1. Be able to convert between odds and probability. 2. Be able to update prior odds to posterior odds using Bayes factors. 3. Understand how Bayes factors measure the extent to which data provides evidence for or against a hypothesis. 2 Odds When comparing two events, it common to phrase probability statements in terms of odds. Definition The odds of event 𝐸 versus event 𝐸 ′ are the ratio of their probabilities 𝑃 (𝐸)/𝑃 (𝐸 ′ ). If unspecified, the second event is assumed to be the complement 𝐸 𝑐 . So the odds of 𝐸 are: 𝑂(𝐸) = 𝑃 (𝐸) . 𝑃 (𝐸 𝑐 ) For example, 𝑂(rain) = 2 means that the probability of rain is twice the probability of no rain (2/3 versus 1/3). We might say ‘the odds of rain are 2 to 1.’ Example. For a fair coin, 𝑂(heads) = 1/2 = 1. We might say the odds of heads are 1 to 1/2 1 or fifty-fifty. Example. For a standard die, the odds of rolling a 4 are 1/6 1 = . We might say the odds 5/6 5 are ‘1 to 5 for’ or ‘5 to 1 against’ rolling a 4. Example. The probability of a pair in a five card poker hand is 0.42257. So the odds of a pair are 0.42257/(1-0.42257) = 0.73181. We can go back and forth between probability and odds as follows. 𝑝 𝑞 Conversion formulas: if 𝑃 (𝐸) = 𝑝 then 𝑂(𝐸) = . If 𝑂(𝐸) = 𝑞 then 𝑃 (𝐸) = . 1−𝑝 1+𝑞 Notes: 1. The second formula simply solves 𝑞 = 𝑝/(1 − 𝑝) for 𝑝. 2. Probabilities are between 0 and 1, while odds are between 0 to ∞. 3. The property 𝑃 (𝐸 𝑐 ) = 1 − 𝑃 (𝐸) becomes 𝑂(𝐸 𝑐 ) = 1/𝑂(𝐸). Example. Let 𝐹 be the event that a five card poker hand is a full house. Then 𝑃 (𝐹 ) = 0.00145214 so 𝑂(𝐹 ) = 0.0014521/(1 − 0.0014521) = 0.0014542. The odds not having a full house are 𝑂(𝐹 𝑐 ) = (1 − 0.0014521)/0.0014521 = 687 = 1/𝑂(𝐹 ). 1 18.05 Class 12, Bayesian Updating: Odds, Spring 2022 2 4. If 𝑃 (𝐸) or 𝑂(𝐸) is small then 𝑂(𝐸) ≈ 𝑃 (𝐸). This follows from the conversion formulas. Example. In the poker example where 𝐹 = ‘full house’ we saw that 𝑃 (𝐹 ) and 𝑂(𝐹 ) differ only in the fourth significant digit. 3 3.1 Updating odds Introduction In Bayesian updating, we used the likelihood of data to update prior probabilities of hypotheses to posterior probabilities. In the language of odds, we will update prior odds to posterior odds. One of our key points will be that the data can provide evidence supporting or negating a hypothesis depending on whether its posterior odds are greater or less than its prior odds. We’ll begin by returning to our familiar example of a screening test for a disease. Example 1. Briefly, a screening test for a disease is both sensitive and specific. Assume the true positive rate is 99% and the false positive rate is 2%. Suppose the prevalence of the disease in the general population is 0.5%. For a randomly chosen person, what are the prior odds that they have the disease? Suppose they test positive, now what are the posterior odds that they have the disease? By what factor have the odds changed as a result of the test? Solution: We’ll use our, by now, standard notation: ℋ+ = have disease, ℋ− = do not have disease, 𝒯+ = test positive, 𝒯− = test negative. To start with the prior odds that they have the disease are 𝑂(ℋ+ ) = 𝑃 (ℋ+ ) 0.005 ≈ 0.005 = 𝑃 (ℋ− ) 0.995 For the posterior odds, we’ll do the computation with trees and then repeat it with tables. Here is the tree describing the scenario. 0.005 0.995 ℋ+ ℋ− 0.99 0.01 0.02 0.98 𝒯+ 𝒯− 𝒯+ 𝒯− Bayes’ theorem yields 𝑂(ℋ+ |𝒯+ ) = 𝑃 (𝒯+ |ℋ+ )𝑃 (ℋ+ )/𝑃 (𝒯+ ) 𝑃 (𝒯+ |ℋ+ )𝑃 (ℋ+ ) 𝑃 (ℋ+ |𝒯+ ) = = 𝑃 (ℋ− |𝒯+ ) 𝑃 (𝒯+ |ℋ− )𝑃 (ℋ− )/𝑃 (𝒯+ ) 𝑃 (𝒯+ |ℋ− )𝑃 (ℋ− ) This is great! For the odds, the total probability 𝑃 (𝒯+ ) cancels and does not need to be computed. Putting in numbers we see the posterior odds are 𝑂(ℋ+ |𝒯+ ) = 0.99 ⋅ 0.005 ≈ 50 ⋅ 0.005 = 1/4. 0.02 ⋅ 0.995 18.05 Class 12, Bayesian Updating: Odds, Spring 2022 3 We’ve structured the presentation so you can easily see that the posterior odds are only one in four. However, the posterior odds are about a factor of 50 greater than the prior odds. Redoing this calculation using a Bayesian update table: hypothesis ℋ ℋ+ ℋ− total prior 𝑃 (ℋ) 0.005 0.995 1 likelihood 𝑃 (𝒯+ |ℋ) 0.99 0.02 NO SUM Bayes numerator 𝑃 (𝒯+ |ℋ)𝑃 (ℋ) 0.00495 0.01990 𝑃 (𝒯+ ) = 0.02485 posterior 𝑃 (ℋ|𝒯+ ) 0.19920 0.80080 1 The prior odds are computed using the prior column of the table. As above, they are 𝑃 (ℋ+ ) 0.005 = . 𝑃 (ℋ− ) 0.995 The posterior odds are computed using either the posterior or Bayes numerator columns of the table. We can use either column, because, they only differ by the normalazing factor of 𝑃 (𝒯+ ) in the denominator of the posteriors. We get the same answer as above: 𝑂(ℋ+ |𝒯+ ) = 0.00495 5 ≈ . 0.01990 20 You should see that these odds come by multiplying the prior odds by the ratio of the likelihoods. 3.2 Example: Marfan syndrome Marfan syndrome is a genetic disease of connective tissue that occurs in 1 of every 15000 people. The main ocular features of Marfan syndrome include bilateral ectopia lentis (lens dislocation), myopia and retinal detachment. About 70% of people with Marfan syndrome have a least one of these ocular features; only 7% of people without Marfan syndrome do. (We don’t guarantee the accuracy of these numbers, but they will work perfectly well for our example.) If a person has at least one of these ocular features, what are the odds that they have Marfan syndrome? Solution: This is a standard Bayesian updating problem. Our hypotheses are: 𝑀 = ‘the person has Marfan syndrome’ 𝑀 𝑐 = ‘the person does not have Marfan syndrome’ The data is: 𝐹 = ‘the person has at least one ocular feature’. We are given the prior probability of 𝑀 and the likelihoods of 𝐹 given 𝑀 or 𝑀 𝑐 : 𝑃 (𝑀 ) = 1/15000, 𝑃 (𝐹 |𝑀 ) = 0.7, 𝑃 (𝐹 |𝑀 𝑐 ) = 0.07. As before, we can compute the posterior probabilities using a table: 18.05 Class 12, Bayesian Updating: Odds, Spring 2022 hypothesis 𝐻 𝑀 𝑀𝑐 total prior 𝑃 (𝐻) 0.000067 0.999933 1 likelihood 𝑃 (𝐹 |𝐻) 0.7 0.07 no sum Bayes numerator 𝑃 (𝐹 |𝐻)𝑃 (𝐻) 0.0000467 0.069995 𝑃 (𝐹 ) = 0.07004 4 posterior 𝑃 (𝐻|𝐹 ) 0.00066 0.99933 1 First we find the prior odds: 𝑂(𝑀 ) = 𝑃 (𝑀 ) 1/15000 1 = = ≈ 0.000067. 𝑐 14999/15000 𝑃 (𝑀 ) 14999 The posterior odds are given by the ratio of the posterior probabilities or the Bayes numerators, since the normalizing factor will be the same in both numerator and denominator. 𝑂(𝑀 |𝐹 ) = 𝑃 (𝑀 |𝐹 ) 𝑃 (𝐹 |𝑀 )𝑃 (𝑀 ) = = 0.000667. 𝑐 𝑃 (𝐹 |𝑀 𝑐 )𝑃 (𝑀 𝑐 ) 𝑃 (𝑀 |𝐹 ) The posterior odds are a factor of 10 larger than the prior odds. In that sense, having an ocular feature is strong evidence in favor of the hypothesis 𝑀 . However, because the prior odds are so small, it is still highly unlikely the person has Marfan syndrome. 4 Bayes factors and strength of evidence The factor of 10 in the previous example is called a Bayes factor or a likelihood ratio. The exact definition is the following. Definition: For a hypothesis 𝐻 and data 𝐷, the Bayes factor is the ratio of the likelihoods: Bayes factor = 𝑃 (𝐷|𝐻) . 𝑃 (𝐷|𝐻 𝑐 ) This is also called the likelihood ratio. Let’s see exactly where the Bayes factor arises in updating odds. We have 𝑃 (𝐻|𝐷) 𝑃 (𝐻 𝑐 |𝐷) 𝑃 (𝐷|𝐻) 𝑃 (𝐻) = 𝑃 (𝐷|𝐻 𝑐 )𝑃 (𝐻 𝑐 ) 𝑃 (𝐷|𝐻) 𝑃 (𝐻) = ⋅ 𝑐 𝑃 (𝐷|𝐻 ) 𝑃 (𝐻 𝑐 ) 𝑃 (𝐷|𝐻) = ⋅ 𝑂(𝐻) 𝑃 (𝐷|𝐻 𝑐 ) 𝑂(𝐻|𝐷) = posterior odds = Bayes factor × prior odds From this formula, we see that the Bayes’ factor (𝐵𝐹 ) tells us whether the data provides evidence for or against the hypothesis. 18.05 Class 12, Bayesian Updating: Odds, Spring 2022 5 • If 𝐵𝐹 > 1 then the posterior odds are greater than the prior odds. So the data provides evidence for the hypothesis. • If 𝐵𝐹 < 1 then the posterior odds are less than the prior odds. So the data provides evidence against the hypothesis. • If 𝐵𝐹 = 1 then the prior and posterior odds are equal. So the data provides no evidence either way. The following example is taken from the textbook Information Theory, Inference, and Learning Algorithms by David J. C. Mackay, who has this to say regarding trial evidence. In my view, a jury’s task should generally be to multiply together carefully evaluated likelihood ratios from each independent piece of admissible evidence with an equally carefully reasoned prior probability. This view is shared by many statisticians but learned British appeal judges recently disagreed and actually overturned the verdict of a trial because the jurors had been taught to use Bayes’ theorem to handle complicated DNA evidence. Example 2. Two people have left traces of their own blood at the scene of a crime. A suspect , Oliver, is tested and found to have type ‘O’ blood. The blood groups of the two traces are found to be of type ‘O’ (a common type in the local population, having frequency 60%) and type ‘AB’ (a rare type, with frequency 1%). Does this data (type ‘O’ and ‘AB’ blood were found at the scene) give evidence in favor of the proposition that Oliver was one of the two people present at the scene of the crime?” Solution: There are two hypotheses: 𝑆 = ‘Oliver and another unknown person were at the scene of the crime 𝑆 𝑐 = ‘two unknown people were at the scene of the crime’ The data is: 𝐷 = ‘type ‘O’ and ‘AB’ blood were found’ The Bayes factor for Oliver’s presence is 𝐵𝐹Oliver = 𝑃 (𝐷|𝑆) . We compute the numerator 𝑃 (𝐷|𝑆 𝑐 ) and denominator of this separately. The data says that both type O and type AB blood were found. If Oliver was at the scene then ‘type O’ blood would be there. So 𝑃 (𝐷|𝑆) is the probability that the other person had type AB blood. We are told this is 0.01, so 𝑃 (𝐷|𝑆) = 0.01. If Oliver was not at the scene then there were two random people one with type O and one with type AB blood. The probability of this is 2 ⋅ 0.6 ⋅ 0.01.* Thus the Bayes factor for Oliver’s presence is 𝑃 (𝐷|𝑆) 0.01 𝐵𝐹Oliver = = 0.83. = 𝑐 𝑃 (𝐷|𝑆 ) 2 ⋅ 0.6 ⋅ 0.01 Since 𝐵𝐹Oliver < 1, the data provides (weak) evidence against Oliver being at the scene. *The factor of 2 is, perhaps surprising. The following careful counting will explain it. Suppose there are 𝑁 people in the population, 𝑁𝑂 have type O blood and 𝑁𝐴𝐵 have type AB. So 𝑁𝑂 /𝑁 = 0.6 18.05 Class 12, Bayesian Updating: Odds, Spring 2022 6 and 𝑁𝐴𝐵 /𝑁 = 0.01. We want the probability that a random choice of 2 people will pick one of type O and one of type AB. This is clearly (𝑁1𝑂 )(𝑁𝐴𝐵 𝑁𝑂 𝑁𝐴𝐵 𝑁 𝑁 1 ) = = 2 ⋅ 𝑂 ⋅ 𝐴𝐵 ≈ 2 ⋅ 0.6 ⋅ 0.01. 𝑁 (𝑁 − 1)/2 𝑁 𝑁 −1 (𝑁2 ) In the last approximation, we assumed that 𝑁 is large enough the 𝑁𝐴𝐵 /(𝑁 − 1) ≈ 𝑁𝐴𝐵 /𝑁 . Example 3. Another suspect Alberto is found to have type ‘AB’ blood. Do the same data give evidence in favor of the proposition that Alberto was one of the two people present at the crime? Solution: Reusing the above notation with Alberto in place of Oliver we have: 𝐵𝐹Alberto = 𝑃 (𝐷|𝑆) 0.6 = = 50. 𝑐 𝑃 (𝐷|𝑆 ) 2 ⋅ 0.6 ⋅ 0.01 Since 𝐵𝐹Alberto ≫ 1, the data provides strong evidence in favor of Alberto being at the scene. Notes: 1. In both examples, we have only computed the Bayes factor, not the posterior odds. To compute the latter, we would need to know the prior odds that Oliver (or Alberto) was at the scene based on other evidence. 2. Note that if 50% of the population had type O blood instead of 60%, then the Oliver’s Bayes factor would be 1 (neither for nor against). More generally, the break-even point for blood type evidence is when the proportion of the suspect’s blood type in the general population equals the proportion of the suspect’s blood type among those who left blood at the scene. 4.1 Updating again and again Suppose we collect data in two stages, first 𝐷1 , then 𝐷2 . We have seen in our dice and coin examples that the final posterior can be computed all at once or in two stages where we first update the prior using the likelihoods for 𝐷1 and then update the resulting posterior using the likelihoods for 𝐷2 . The latter approach works whenever likelihoods multiply: 𝑃 (𝐷1 , 𝐷2 |𝐻) = 𝑃 (𝐷1 |𝐻)𝑃 (𝐷2 |𝐻). Since likelihoods are conditioned on hypotheses, we say that 𝐷1 and 𝐷2 are conditionally independent if the above equation holds for every hypothesis 𝐻. Example. There are five dice in a drawer, with 4, 6, 8, 12, and 20 sides (these are the hypotheses). I pick a die at random and roll it twice. The first roll gives 7. The second roll gives 11. Are these results conditionally independent? Are they independent? Solution: These results are conditionally independent. For example, for the hypothesis of the 8-sided die we have: 𝑃 (7 on roll 1 | 8-sided die) = 1/8 𝑃 (11 on roll 2 | 8-sided die) = 0 𝑃 (7 on roll 1, 11 on roll 2 | 8-sided die) = 0 18.05 Class 12, Bayesian Updating: Odds, Spring 2022 7 For the hypothesis of the 20-sided die we have: 𝑃 (7 on roll 1 | 20-sided die) = 1/20 𝑃 (11 on roll 2 | 20-sided die) = 1/20 𝑃 (7 on roll 1, 11 on roll 2 | 20-sided die) = (1/20)2 However, the results of the rolls are not independent. That is: 𝑃 (7 on roll 1, 11 on roll 2) ≠ 𝑃 (7 on roll 1)𝑃 (11 on roll 2). Intuitively, this is because a 7 on the roll 1 allows us to rule out the 4- and 6-sided dice, making an 11 on roll 2 more likely. Let’s check this intuition by computing both sides precisely. On the righthand side we have: 1 5 1 𝑃 (11 on roll 2) = 5 𝑃 (7 on roll 1) = 1 1 1 1 1 31 + ⋅ + ⋅ = 8 5 12 5 20 600 1 1 1 2 ⋅ + ⋅ = 12 5 20 75 ⋅ On the lefthand side we have: 𝑃 (7 on roll 1, 11 on roll 2) = 𝑃 (11 on roll 2 | 7 on roll 1)𝑃 (7 on roll 1) 30 1 6 1 31 ⋅ )⋅ =( ⋅ + 93 12 31 20 600 17 31 17 = ⋅ = 465 600 9000 6 Here 30 93 and 31 are the posterior probabilities of the 12- and 20-sided dice given a 7 on roll 1. We conclude that, without conditioning on hypotheses, the rolls are not independent. Returning the to general setup, if 𝐷1 and 𝐷2 are conditionally independent for 𝐻 and 𝐻 𝑐 then it makes sense to consider each Bayes factor independently: 𝐵𝐹𝑖 = 𝑃 (𝐷𝑖 |𝐻) . 𝑃 (𝐷𝑖 |𝐻 𝑐 ) The prior odds of 𝐻 are 𝑂(𝐻). The posterior odds after 𝐷1 are 𝑂(𝐻|𝐷1 ) = 𝐵𝐹1 ⋅ 𝑂(𝐻). And the posterior odds after 𝐷1 and 𝐷2 are 𝑂(𝐻|𝐷1 , 𝐷2 ) = 𝐵𝐹2 ⋅ 𝑂(𝐻|𝐷1 ) = 𝐵𝐹2 ⋅ 𝐵𝐹1 ⋅ 𝑂(𝐻) We have the beautifully simple notion that updating with new data just amounts to multiplying the current posterior odds by the Bayes factor of the new data. Example 4. Other symptoms of Marfan Syndrome Recall from the earlier example that the Bayes factor for a least one ocular feature (𝐹 ) is 𝐵𝐹𝐹 = 𝑃 (𝐹 |𝑀 ) 0.7 = = 10. 𝑐 𝑃 (𝐹 |𝑀 ) 0.07 18.05 Class 12, Bayesian Updating: Odds, Spring 2022 8 The wrist sign (𝑊 ) is the ability to wrap one hand around your other wrist to cover your pinky nail with your thumb. Assume 10% of the population have the wrist sign, while 90% of people with Marfan’s have it. Therefore the Bayes factor for the wrist sign is 𝐵𝐹𝑊 = 𝑃 (𝑊 |𝑀 ) 0.9 = = 9. 𝑐 𝑃 (𝑊 |𝑀 ) 0.1 We will assume that 𝐹 and 𝑊 are conditionally independent symptoms. That is, among people with Marfan syndrome, ocular features and the wrist sign are independent, and among people without Marfan syndrome, ocular features and the wrist sign are independent. Given this assumption, the posterior odds of Marfan syndrome for someone with both an ocular feature and the wrist sign are 𝑂(𝑀 |𝐹 , 𝑊 ) = 𝐵𝐹𝑊 ⋅ 𝐵𝐹𝐹 ⋅ 𝑂(𝑀 ) = 9 ⋅ 10 ⋅ 1 6 ≈ . 14999 1000 We can convert the posterior odds back to probability, but since the odds are so small the result is nearly the same: 𝑃 (𝑀 |𝐹 , 𝑊 ) ≈ 6 ≈ 0.596%. 1000 + 6 So ocular features and the wrist sign are both strong evidence in favor of the hypothesis 𝑀 , and taken together they are very strong evidence. Again, because the prior odds are so small, it is still unlikely that the person has Marfan syndrome, but at this point it might be worth undergoing further testing given potentially fatal consequences of the disease (such as aortic aneurysm or dissection). Note also that if a person has exactly one of the two symptoms, then the product of the Bayes factors is near 1 (either 9/10 or 10/9). So the two pieces of data essentially cancel each other out with regard to the evidence they provide for Marfan’s syndrome. 5 Log odds In practice, people often find it convenient to work with the natural log of the odds in place of odds. Naturally enough these are called the log odds. The Bayesian update formula 𝑂(𝐻|𝐷1 , 𝐷2 ) = 𝐵𝐹2 ⋅ 𝐵𝐹1 ⋅ 𝑂(𝐻) becomes ln(𝑂(𝐻|𝐷1 , 𝐷2 )) = ln(𝐵𝐹2 ) + ln(𝐵𝐹1 ) + ln(𝑂(𝐻)). We can interpret the above formula for the posterior log odds as the sum of the prior log odds and all the evidence ln(𝐵𝐹𝑖 ) provided by the data. Note that by taking logs, evidence in favor (𝐵𝐹𝑖 > 1) is positive and evidence against (𝐵𝐹𝑖 < 1) is negative. To avoid lengthier computations, we will work with odds rather than log odds in this course. Log odds are nice because sums are often more intuitive then products. Log odds also play a central role in logistic regression, an important statistical model related to linear regression.