Bayesian Updating: Discrete Priors: Spring 2014
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1 ian Updating: Discrete Priors: Spring January 1, / 16
2 Learning from experience Which treatment would you choose? 1. Treatment 1: cured 100% of patients in a trial. 2. Treatment 2: cured 95% of patients in a trial. 3. Treatment 3: cured 90% of patients in a trial. Which treatment would you choose? 1. Treatment 1: cured 3 out of 3 patients in a trial. 2. Treatment 2: cured 19 out of 20 patients treated in a trial. 3. Standard treatment: cured out of patients in clinical practice. January 1, / 16
3 Which die is it? I have a bag containing dice of two types: 4-sided and 10-sided. Suppose I pick a die at random and roll it. Based on what I rolled which type would you guess I picked? Suppose you find out that the bag contained one 4-sided die and one 10-sided die. Does this change your guess? Suppose you find out that the bag contained one 4-sided die and sided dice. Does this change your guess? January 1, / 16
4 Board Question: learning from data A certain disease has a prevalence of A screening test has 2% false positives an 1% false negatives. Suppose a patient is screened and has a positive test. 1 Represent this information with a tree and use theorem to compute the probabilities the patient does and doesn t have the disease Identify the data, hypotheses, likelihoods, prior probabilities and posterior probabilities. Make a full likelihood table containing all hypotheses and possible test data. Redo the computation using a ian update table. Match the terms in your table to the terms in your previous calculation. January 1, / 16
5 Board Question: Dice Five dice: 4-sided, 6-sided, 8-sided, 12-sided, 20-sided. Suppose I picked one at random and, without showing it to you, rolled it and reported a Make the full likelihood table (be smart about identical columns). 2. Make a ian update table and compute the posterior probabilities that the chosen die is each of the five dice. 3. Same question if I rolled a Same question if I rolled a 9. (Keep the tables for 5 and 9 handy! Do not erase!) January 1, / 16
6 Tabular solution D = rolled a 13 hypothesis prior likelihood numerator posterior H P(H) P(D H) P(D H)P(H) P(H D) H 4 1/ H 6 1/ H 8 1/ H 12 1/ H 20 1/5 1/20 1/100 1 total 1 1/100 1 January 1, / 16
7 Tabular solution D = rolled a 5 hypothesis prior likelihood numerator posterior H P(H) P(D H) P(D H)P(H) P(H D) H 4 1/ H 6 1/5 1/6 1/ H 8 1/5 1/8 1/ H 12 1/5 1/12 1/ H 20 1/5 1/20 1/ total January 1, / 16
8 Tabular solution D = rolled a 9 hypothesis prior likelihood numerator posterior H P(H) P(D H) P(D H)P(H) P(H D) H 4 1/ H 6 1/ H 8 1/ H 12 1/5 1/12 1/ H 20 1/5 1/20 1/ total January 1, / 16
9 Iterated Updates Suppose I rolled a 5 and then a 9. Update in two steps: First for the 5 Then update the update for the 9. January 1, / 16
10 Tabular solution D 1 = rolled a 5 D 2 = rolled a 9 numerator 1 = likelihood 1 prior. numerator 2 = likelihood 2 numerator 1 hyp. prior likel. 1 num. 1 likel. 2 num. 2 posterior H P(H) P(D 1 H) P(D 2 H) P(H D 1, D 2 ) H 4 1/ H 6 1/5 1/6 1/ H 8 1/5 1/8 1/ H 12 1/5 1/12 1/60 1/12 1/ H 20 1/5 1/20 1/100 1/20 1/ total January 1, / 16
11 Board Question Suppose I rolled a 9 and then a Do the ian update in two steps: First update for the 9. Then update the update for the Do the ian update in one step The data is D = 9 followed by 5 January 1, / 16
12 Tabular solution: two steps D 1 = rolled a 9 D 2 = rolled a 5 numerator 1 = likelihood 1 prior. numerator 2 = likelihood 2 numerator 1 hyp. prior likel. 1 num. 1 likel. 2 num. 2 posterior H P(H) P(D 1 H) P(D 2 H) P(H D 1, D 2 ) H 4 1/ H 6 1/ /6 0 0 H 8 1/ /8 0 0 H 12 1/5 1/12 1/60 1/12 1/ H 20 1/5 1/20 1/100 1/20 1/ total January 1, / 16
13 Tabular solution: one step D = rolled a 9 then a 5 hypothesis prior likelihood numerator posterior H P(H) P(D H) P(D H)P(H) P(H D) H 4 1/ H 6 1/ H 8 1/ H 12 1/5 1/144 1/ H 20 1/5 1/400 1/ total January 1, / 16
14 Board Question: probabilistic prediction Along with finding posterior probabilities of hypotheses. We might want to make posterior predictions about the next roll. With the same setup as before let: D 1 = result of first roll D 2 = result of second roll (a) Find P(D 1 = 5). (b) Find P(D 2 = 4 D 1 = 5). January 1, / 16
15 Solution D 1 = rolled a 5 D 2 = rolled a 4 hyp. prior likel. 1 num. 1 post. 1 likel. 2 post. 1 likel. 2 H P(H) P(D 1 H) P(H D 1 ) P(D 2 H, D 1 ) P(D 2 H, D 1 )P(H D 1 ) H 4 1/ H 6 1/5 1/6 1/ / /6 H 8 1/5 1/8 1/ / /40 H 12 1/5 1/12 1/ / /12 H 20 1/5 1/20 1/ / /20 total The law of total probability tells us P(D 1 ) is the sum of the numerator 1 column in the table: P(D 1 ) = The law of total probability tells us P(D 2 D 1 ) is the sum of the last column in the table: P(D 2 D 1 ) = January 1, / 16
16 MIT OpenCourseWare Introduction to Probability and Statistics Spring 2014 For information about citing these materials or our Terms of Use, visit:
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