And the Bayesians and the frequentists shall lie down together...
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1 And the Bayesians and the frequentists shall lie down together... Keith Winstein MIT CSAIL October 16, 2013
2 Axioms of Probability (1933) S: a finite set (the sample space) A: any subset of S (an event) P(A): the probability of A satisfies P(A) R P(A) 0 P(S) = 1 P(A B) = P(A)+P(B) if A B = If S infinite, axiom becomes: for an infinite sequence of disjoint subsets A 1,A 2,..., ( ) P A i = P(A i ) i=1 i=1
3 Some Theorems P(A) = 1 P(A) P( ) = 0 P(A) P(B) if A B P(A) 1 P(A B) = P(A)+P(B) P(A B) P(A B) P(A)+P(B)
4 Joint & Conditional Probability For A,B S, P(A B) is joint probability of A and B. The conditional probability of A given B in: P(A B) = P(A B) P(B) A and B are independent iff P(A B) = P(A)P(B). A,B independent P(A B) = P(A).
5 Bayes Theorem We have: P(A B) = P(A B) P(B) P(B A) = P(A B) P(A) Therefore: P(A B) = P(A B)P(B) = P(B A)P(A) Bayes Theorem: P(A B) = P(B A)P(A) P(B)
6 On the islands of Ste. Frequentiste and Bayesienne...
7 On the islands of Ste. Frequentiste and Bayesienne... The king has been poisoned!
8 A letter goes out... Dear Governor: Attached is a blood test for proximity to the poison. It has a 0% rate of false negative and a 1% rate of false positive. Jail those responsible. But remember the nationwide law: You must be 95% certain to send a citizen to jail.
9 On Ste. Frequentiste: Test has a 0% rate of false negative and a 1% rate of false positive. You must be 95% certain to send a citizen to jail. P(Positive Guilty) = 1 P(Negative Guilty) = 0 P(Positive Innocent) = 0.01 P(Negative Innocent) = 0.99 How to interpret the law? We must be 95% certain
10 On Ste. Frequentiste: Test has a 0% rate of false negative and a 1% rate of false positive. You must be 95% certain to send a citizen to jail. P(Positive Guilty) = 1 P(Negative Guilty) = 0 P(Positive Innocent) = 0.01 P(Negative Innocent) = 0.99 How to interpret the law? We must be 95% certain P(Jail Innocent) 0.05
11 On Ste. Frequentiste: Test has a 0% rate of false negative and a 1% rate of false positive. You must be 95% certain to send a citizen to jail. P(Positive Guilty) = 1 P(Negative Guilty) = 0 P(Positive Innocent) = 0.01 P(Negative Innocent) = 0.99 How to interpret the law? We must be 95% certain P(Jail Innocent) 0.05 Can Positive Jail? Yes.
12 On Isle Bayesienne: Test has a 0% rate of false negative and a 1% rate of false positive. You must be 95% certain to send a citizen to jail. P(Positive Guilty) = 1 P(Negative Guilty) = 0 P(Positive Innocent) = 0.01 P(Negative Innocent) = 0.99 How to interpret the law? We must be 95% certain
13 On Isle Bayesienne: Test has a 0% rate of false negative and a 1% rate of false positive. You must be 95% certain to send a citizen to jail. P(Positive Guilty) = 1 P(Negative Guilty) = 0 P(Positive Innocent) = 0.01 P(Negative Innocent) = 0.99 How to interpret the law? We must be 95% certain P(Innocent Jail) 0.05
14 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Positive ) = P(Positive Innocent) P(Innocent) P(Positive)
15 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Positive ) = P(Positive Innocent) P(Innocent) P(Positive)
16 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = P(Positive Innocent) P(Innocent) P(Positive)
17 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = P(Positive Innocent) P(Innocent) P(Positive)
18 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = P(Positive Innocent) P(Innocent) P(Positive)
19 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = (0.01) P(Innocent) P(Positive)
20 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = (0.01) P(Innocent) P(Positive)
21 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = (0.01) P(Innocent) P(Positive)
22 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = (0.01) P(Innocent) P(Positive Innocent) P(Innocent) + P(Positive Guilty) P(Guilty)
23 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = (0.01) P(Innocent) P(Positive Innocent) P(Innocent) + P(Positive Guilty) P(Guilty)
24 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = (0.01) P(Innocent) P(Positive Innocent) P(Innocent) + P(Positive Guilty) P(Guilty)
25 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = (0.01) P(Innocent) 1 - (0.99) P(Innocent)
26 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = (0.01) P(Innocent) 1 - (0.99) P(Innocent)
27 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = (0.01) P(Innocent) 1 - (0.99) P(Innocent)
28 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = (0.01) P(Innocent) 1 - (0.99) P(Innocent) P(Innocent) =???
29 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = (0.01) P(Innocent) 1 - (0.99) P(Innocent) P(Innocent) = 0.9
30 Isle Bayesienne: the need for prior assumptions We must be 95% certain P(Innocent Jail) 0.05 Can Positive Jail? Apply Bayes theorem P(Innocent Jail ) = (0.01) P(Innocent) 1 - (0.99) P(Innocent) P(Innocent) = 0.9 P(Innocent Jail) 0.08!!
31 On the islands of Ste. Frequentiste and Bayesienne... More than 1% of Ste. Frequentiste goes to jail. On Isle Bayesienne, 10% are assumed guilty, but nobody goes to jail. The disagreement wasn t about math or how to interpret P(). What was it about?
32 The islanders concerns Frequentist cares about the rate of jailings among innocent people. Concern: overall rate of false positive Bayesian cares about the rate of innocence among jail inmates. Concern: rate of error among positives The Bayesian had to make an assumption about the overall probability of innocence.
33 Quantifying uncertainty Underlying parameters
34 Quantifying uncertainty the truth Underlying parameters
35 Quantifying uncertainty the truth Experiment Underlying parameters
36 Quantifying uncertainty
37 Quantifying uncertainty
38 Quantifying uncertainty
39 Quantifying uncertainty
40 Jewel s Cookies Cookie jars A, B, C, D have 100 cookies each, but different numbers of chocolate chips per cookie: P( chips jar ) A B C D total 100% 100% 100% 100%
41 Quantifying cookie jar uncertainty
42 Quantifying cookie jar uncertainty
43 Quantifying cookie jar uncertainty
44 Quantifying cookie jar uncertainty
45 Quantifying cookie jar uncertainty
46 Quantifying cookie jar uncertainty
47 Frequentist inference A 70% confidence interval method includes the correct jar with at least 70% probability in the worst case, no matter what.
48 Making 70% confidence intervals Cookie jars A, B, C, D have 100 cookies each, but different numbers of chocolate chips per cookie: P( chips jar ) A B C D coverage
49 Making 70% confidence intervals Cookie jars A, B, C, D have 100 cookies each, but different numbers of chocolate chips per cookie: P( chips jar ) A B C D coverage 70%
50 Making 70% confidence intervals Cookie jars A, B, C, D have 100 cookies each, but different numbers of chocolate chips per cookie: P( chips jar ) A B C D coverage 70% 25%
51 Making 70% confidence intervals Cookie jars A, B, C, D have 100 cookies each, but different numbers of chocolate chips per cookie: P( chips jar ) A B C D coverage 70% 49%
52 Making 70% confidence intervals Cookie jars A, B, C, D have 100 cookies each, but different numbers of chocolate chips per cookie: P( chips jar ) A B C D coverage 70% 69%
53 Making 70% confidence intervals Cookie jars A, B, C, D have 100 cookies each, but different numbers of chocolate chips per cookie: P( chips jar ) A B C D coverage 70% 88%
54 Making 70% confidence intervals Cookie jars A, B, C, D have 100 cookies each, but different numbers of chocolate chips per cookie: P( chips jar ) A B C D coverage 70% 88% 67%
55 Making 70% confidence intervals Cookie jars A, B, C, D have 100 cookies each, but different numbers of chocolate chips per cookie: P( chips jar ) A B C D coverage 70% 88% 87%
56 Making 70% confidence intervals Cookie jars A, B, C, D have 100 cookies each, but different numbers of chocolate chips per cookie: P( chips jar ) A B C D coverage 70% 88% 87% 70%
57 Bayesian inference A 70% credible interval has at least 70% conditional probability of including the correct jar, given the observation and the prior assumptions.
58 Uniform prior Our prior assumption: jars A, B, C, and D have equal probability.
59 Conditional probabilities P( chips jar ) A B C D total 100% 100% 100% 100%
60 Conditional probabilities P( chips jar ) A B C D total 100% 100% 100% 100%
61 Joint probabilities under uniform prior P( chips jar ) A B C D P(chips) 0 1/4 12/4 13/4 27/ % 1 1/4 19/4 20/4 70/ % 2 70/4 24/4 0/4 1/ % 3 28/4 20/4 0/4 1/ % 4 0/4 25/4 67/4 1/ % total P(jar) 25% 25% 25% 25% 100%
62 Joint probabilities under uniform prior P( chips jar ) A B C D P(chips) 0 1/4 12/4 13/4 27/ % 1 1/4 19/4 20/4 70/ % 2 70/4 24/4 0/4 1/ % 3 28/4 20/4 0/4 1/ % 4 0/4 25/4 67/4 1/ % total P(jar) 25% 25% 25% 25% 100%
63 Posterior probabilities under uniform prior A B C D P(chips) 0 1/4 12/4 13/4 27/ % 1 1/4 19/4 20/4 70/ % 2 70/4 24/4 0/4 1/ % 3 28/4 20/4 0/4 1/ % 4 0/4 25/4 67/4 1/ %
64 Posterior probabilities under uniform prior A B C D P(chips) % 1 1/4 19/4 20/4 70/ % 2 70/4 24/4 0/4 1/ % 3 28/4 20/4 0/4 1/ % 4 0/4 25/4 67/4 1/ %
65 Posterior probabilities under uniform prior A B C D P(chips) % 1 1/4 19/4 20/4 70/ % 2 70/4 24/4 0/4 1/ % 3 28/4 20/4 0/4 1/ % 4 0/4 25/4 67/4 1/ %
66 Posterior probabilities under uniform prior A B C D P(chips) % % 2 70/4 24/4 0/4 1/ % 3 28/4 20/4 0/4 1/ % 4 0/4 25/4 67/4 1/ %
67 Posterior probabilities under uniform prior A B C D P(chips) % % 2 70/4 24/4 0/4 1/ % 3 28/4 20/4 0/4 1/ % 4 0/4 25/4 67/4 1/ %
68 Posterior probabilities under uniform prior A B C D P(chips) % % % 3 28/4 20/4 0/4 1/ % 4 0/4 25/4 67/4 1/ %
69 Posterior probabilities under uniform prior A B C D P(chips) % % % 3 28/4 20/4 0/4 1/ % 4 0/4 25/4 67/4 1/ %
70 Posterior probabilities under uniform prior A B C D P(chips) % % % % 4 0/4 25/4 67/4 1/ %
71 Posterior probabilities under uniform prior A B C D P(chips) % % % % 4 0/4 25/4 67/4 1/ %
72 Posterior probabilities under uniform prior A B C D P(chips) % % % % %
73 Posterior probabilities under uniform prior P( jar chips ) A B C D P(chips) % % % % %
74 70% credible intervals P( jar chips ) A B C D credibility
75 70% credible intervals P( jar chips ) A B C D credibility %
76 70% credible intervals P( jar chips ) A B C D credibility %
77 70% credible intervals P( jar chips ) A B C D credibility % %
78 70% credible intervals P( jar chips ) A B C D credibility % %
79 70% credible intervals P( jar chips ) A B C D credibility % % %
80 70% credible intervals P( jar chips ) A B C D credibility % % % %
81 70% credible intervals P( jar chips ) A B C D credibility % % % %
82 70% credible intervals P( jar chips ) A B C D credibility % % % % %
83 Confidence & credible intervals together confidence A B C D credibility % % % % % coverage 70% 88% 87% 70% credible A B C D credibility % % % % % coverage 98% 20% 100% 97%
84 Correlation of error
85 Correlation of error
86 Correlation of error
87 Criticism of frequentist style Why Most Published Research Findings Are False, Ioannidis JPA, PLoS Medicine Vol. 2, No. 8, e124 doi: /journal.pmed
88 Criticism of Bayesian style
89 Disagreement in the real world Avandia: world s #1 diabetes drug, approved in Sold by GlaxoSmithKline PLC Sales: $3 billion in 2006 In 2004, GSK releases results of many small studies.
90 GSK releases 42 small studies Study Avandia heart attacks Control heart attacks /391 1/ /110 2/114 DREAM 15/2635 9/ /561 2/ /706 0/325...
91 In 2007, Dr. Nissen crashes the party
92 Frequentist inference
93
94 GlaxoSmithKline loses $12 billion GBP (millions) Avandia worldwide sales
95 Bayesian inference disagrees, for risk ratio 4 P.D.F. on Avandia s risk ratio for heart attack (Joint work with Joshua Mandel, Children s Hospital Informatics Program)
96 Or does it? Here, risk difference model 800 P.D.F. on Avandia s risk difference for heart attack (Joint work with Joshua Mandel, Children s Hospital Informatics Program)
97 The TAXUS ATLAS Experiment Boston Scientific proposed to show that new heart stent was not inferior to old heart stent, with 95% confidence. Inferior means three percentage points more bad events. Control 7% vs. Treatment 10.5% inferior Control 7% vs. Treatment 9.5% non-inferior.
98 ATLAS Results (May 2006) May 16, 2006 NATICK, Mass. and PARIS, May 16 /PRNewswire-FirstCall/ Boston Scientific Corporation today announced nine-month data from its TAXUS ATLAS clinical trial. [...] The trial met its primary endpoint of nine-month target vessel revascularization (TVR), a measure of the effectiveness of a coronary stent in reducing the need for a repeat procedure.
99 ATLAS Results (April 2007) Turco et al., Polymer-Based, Paclitaxel-Eluting TAXUS Liberté Stent in De Novo Lesions, Journal of the American College of Cardiology, Vol. 49, No. 16, Results: The primary non-inferiority end point was met with the 1-sided 95% confidence bound of 2.98% less than the pre-specified non-inferiority margin of 3% (p = ). Statistical methodology. Student t test was used to compare independent continuous variables, while chi-square or Fisher exact test was used to compare proportions.
100 p-value p < % confidence interval excludes inferiority
101 The problem Event No event Total Control Treatment Total
102 The problem Event No event Total Control Treatment Total With uniform prior on rates, Pr(inferior data) Posterior probability of non-inferiority is less than 95%.
103 ATLAS trial solution Confidence interval: approximate each binomial separately with a normal distribution. Known as Wald interval. p = ( ) 0.03 N i m j n, i(m i) + j(n j) m 3 n 3 p < 0.05 success
104 The ultimate close call Wald s area ( p) with (m,n) = (855,956) TVR (Liberte) TVR (Express)
105 confidence A B C D coverage 70% 88% 87% 70%
106 confidence A B C D coverage 70% 88% 87% 70% false positive rate 30% 12% 13% 30%
107 The Wald interval undercovers False Positive Rate of ATLAS non-inferiority test along critical line 5.4 False Positive Rate (%) (2,5) (4,7) (6,9) (8,11) (10,13) (12,15) (14,17) (16,19) (18,21) (20,23) (π TVRe,π TVRl ) (%)
108 Better approximation: score interval False Positive Rate of maximum-likelihood z-test along critical line 5.4 False Positive Rate (%) (2,5) (4,7) (6,9) (8,11) (10,13) (12,15) (14,17) (16,19) (18,21) (20,23) (π TVRe,π TVRl ) (%)
109 Other methods all yield failure Method p-value or confidence bound Result Wald interval p = Pass z-test, constrained max likelihood standard error p = Fail
110 Other methods all yield failure Method p-value or confidence bound Result Wald interval p = Pass z-test, constrained max likelihood standard error p = Fail z-test with Yates continuity correction c = Fail Agresti-Caffo I 4 interval p = Fail Wilson score c = Fail Wilson score with continuity correction c = Fail Farrington & Manning score p = Fail Miettinen & Nurminen score p = Fail Gart & Nam score p = Fail NCSS s bootstrap method c = Fail NCSS s quasi-exact Chen c = Fail NCSS s exact double-binomial test p = Fail StatXact s approximate unconditional test of non-inferiority p = Fail StatXact s exact unconditional test of non-inferiority p = Fail StatXact s exact CI based on difference of observed rates c = Fail StatXact s approximate CI from inverted 2-sided test c = Fail StatXact s exact CI from inverted 2-sided test c = Fail
111 Nerdiest chart contender?
112
113 The statistician says... np > 5, therefore, the Central Limit Theorem applies and a Gaussian approximation is appropriate. We had even more data points than we powered the study for, so there was adequate safety margin. Exact tests are too conservative.
114 StatXact calculates an exact test Other statistical applications often rely on large-scale assumptions for inferences, risking incorrect conclusions from data sets not normally distributed. StatXact utilizes Cytel s own powerful algorithms to make exact inferences...
115 StatXact calculates an exact test Other statistical applications often rely on large-scale assumptions for inferences, risking incorrect conclusions from data sets not normally distributed. StatXact utilizes Cytel s own powerful algorithms to make exact inferences...
116 Graphing the coverage Problem: hard to calculate a million exact p-values (StatXact: about 10 minutes each) Contribution: method for calculating whole tableau Calculates all p-values in time for hardest square Trick: calculate in the right order, cache partial results
117 StatXact calculates an exact test Other statistical applications often rely on large-scale assumptions for inferences, risking incorrect conclusions from data sets not normally distributed. StatXact utilizes Cytel s own powerful algorithms to make exact inferences...
118 StatXact calculates an exact test Other statistical applications often rely on large-scale assumptions for inferences, risking incorrect conclusions from data sets not normally distributed. StatXact utilizes Cytel s own powerful algorithms to make exact inferences... Type I rate of StatXAct 8 non-inferiority test (Berger Boos-adjusted Chan) 5.4 False Positive Rate (%) (2,5) (4,7) (6,9) (8,11) (10,13) (12,15) (14,17) (16,19) (18,21) (20,23) (π TVRe,π TVRl ) (%)
119 Animation
120 Barnard s test for N =
121 Assume a prior hypothesis...
122 Frequentists can benefit from priors too!
123 Final thoughts Bayesian and frequentist schools have much in common. If stark disagreement between Bayesian and frequentist methods, probably sign of bigger problems! What s important: say what we re trying to infer, how we get there, what we care about.
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