1. P(B)P(B)P(B ) Bayes’ Formula Exactly how does one event A affect the probability of another event B? 1 AP(B)P(B) prior probability posterior probability P(B ∩ A)P(A)P(B ∩ A)P(A) P(A | B) P(B) P(A) ??? Recall……

2. P(A | B) P(B) P(A) P(B | A) = P(B)P(B)P(B ) Bayes’ Formula Exactly how does one event A affect the probability of another event B? 2 AP(B)P(B) prior probability posterior probability Recall… Recall… … Bayes’ Law (short form)

3. BBCBC A P(A ⋂ B ) P(A ⋂ B C ) P(A)P(A) ACAC P(A C ⋂ B ) P(A C ⋂ B C ) P(AC)P(AC) P(B)P(B) P(BC)P(BC) 1.0 A ⋂ BA ⋂ B A ⋂ BcA ⋂ Bc Ac ⋂ BAc ⋂ B “A only” “B only” Ac ⋂ BcAc ⋂ Bc “Neither A nor B” “A and B” A B BcBc P(A | B) P(B) P(A) P(B | A) = Bayes’ Law (short form) Recall… ??? For any two events A and B, there are 4 disjoint intersections: Also recall…… Probability Table “LAW OF TOTAL PROBABILITY” Bayes’ Law (full form, n = 2)

4. B1B1 B 2 (= B 1 c ) A AcAc.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities P(A) “LAW OF TOTAL PROBABILITY” posterior probabilities

5. B1B1 B 2 (= B 1 c ) A AcAc.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities P(A) “LAW OF TOTAL PROBABILITY” posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4

6. B1B1 B 2 (= B 1 c ) A(.2)(.9)(.6)(.1) AcAc.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities P(A) “LAW OF TOTAL PROBABILITY” posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4

7. B1B1 B 2 (= B 1 c ) A.18.06 AcAc.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities P(A) “LAW OF TOTAL PROBABILITY” posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4

8. B1B1 B 2 (= B 1 c ) A.18.06.24 AcAc.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities P(A) “LAW OF TOTAL PROBABILITY” posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4

9. B1B1 B 2 (= B 1 c ) A.18.06.24 AcAc.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 In the highlands… B 1 decreases from a prior of 90% to a posterior of 75% B 2 increases from a prior of 10% to a posterior of 25%

10. B1B1 B 2 (= B 1 c ) A.18.06.24 AcAc.72.04.76.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 In the lowlands… B 1 increases from a prior of 90% to a posterior of 94.7% B 2 decreases from a prior of 10% to a posterior of 5.3%

11. Exercise: ??? Species B 1 – with a prior prob of ??? – ?????? ??? to a posterior of ??? in the highlands ?????? ??? to a posterior of ??? in the lowlands. ??? Species B 2 – with a prior prob of ??? – ?????? ??? to a posterior of ??? in the highlands ?????? ??? to a posterior of ??? in the lowlands. Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4

12. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. “10% of pop is B 1 -deficient (only), 20% is B 2 -deficient (only), and 30% is B 3 -deficient (only). The remaining 40% is not B-deficient.” Given:

13. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40 A = Alcoholic A c = Not Alcoholic Given: P(A ∩ B 1 ) To find these intersection probabilities, we need more information! Prior probs 1.00 P(A ∩ B 2 )P(A ∩ B 3 ) P(A ∩ B 4 ) P(A c ∩ B 1 ) P(A c ∩ B 2 )P(A c ∩ B 3 )P(A c ∩ B 4 )

14. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40 Also given… “Alcoholics comprise 35%, 30%, 25%, and 20% of the B 1, B 2, B 3, B 4 groups, respectively.” Given: A = Alcoholic A c = Not Alcoholic Prior probs 1.00 P(A ∩ B 1 )P(A ∩ B 2 )P(A ∩ B 3 ) P(A ∩ B 4 ) P(A c ∩ B 1 ) P(A c ∩ B 2 )P(A c ∩ B 3 )P(A c ∩ B 4 )

15. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40 Also given… P(A | B 1 ) =.35P(A | B 2 ) =.30P(A | B 3 ) =.25P(A | B 4 ) =.20 Prior probs 1.00 Given: A = Alcoholic A c = Not Alcoholic P(A ∩ B 1 )P(A ∩ B 2 )P(A ∩ B 3 ) P(A ∩ B 4 ) P(A c ∩ B 1 ) P(A c ∩ B 2 )P(A c ∩ B 3 )P(A c ∩ B 4 )

16. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. Also given… Prior probs 1.00 Given: A = Alcoholic A c = Not Alcoholic P(A ∩ B 1 )P(A ∩ B 2 )P(A ∩ B 3 ) P(A ∩ B 4 ) P(A c ∩ B 1 ) P(A c ∩ B 2 )P(A c ∩ B 3 )P(A c ∩ B 4 ) P(A | B 1 ) =.35P(A | B 2 ) =.30P(A | B 3 ) =.25P(A | B 4 ) =.20 P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40 P(A ∩ B) = P(A | B) P(B) Recall:

17. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. Also given… Prior probs 1.00 Given: A = Alcoholic A c = Not Alcoholic P(A c ∩ B 1 ) P(A c ∩ B 2 )P(A c ∩ B 3 )P(A c ∩ B 4 ) P(A ∩ B) = P(A | B) P(B) Recall: P(A | B 1 ) =.35P(A | B 2 ) =.30P(A | B 3 ) =.25P(A | B 4 ) =.20.10 .35.20 .30.30 .25.40 .20.035.060.075.080 P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40

18. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. Prior probs 1.00 Given: A = Alcoholic A c = Not Alcoholic.10 .35.20 .30.30 .25.40 .20.035.060.075.080.065.140.225.320 P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40 P(B 1 | A) = ?P(B 2 | A) = ?P(B 3 | A) = ? P(B 4 | A) = ? Posterior probabilities

19. B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Example: Vitamin B-complex deficiency among general population Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40 P(B 1 | A) = ?P(B 2 | A) = ?P(B 3 | A) = ? P(B 4 | A) = ?.035.060.075.080.065.140.225.320 P(A) =.25 P(A c ) =.75.035 1.00 P(B 1 ∩ A) P(A) Prior probs Given: A = Alcoholic A c = Not Alcoholic Posterior probabilities

20. B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Example: Vitamin B-complex deficiency among general population Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 1 | A) =P(B 2 | A) =P(B 3 | A) = P(B 4 | A) =.035.060.075.080.065.140.225.320 P(A) =.25 P(A c ) =.75.035.060.075.080 Prior probs Given: A = Alcoholic A c = Not Alcoholic Posterior probabilities 1.00 P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40

21. B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Example: Vitamin B-complex deficiency among general population Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 1 | A) =.14P(B 2 | A) =.24P(B 3 | A) =.30 P(B 4 | A) =.32.035.060.075.080.065.140.225.320 P(A) =.25 P(A c ) =.75 1.00 INCREASE DECREASE NO CHANGE; A and B 3 are independent! Prior probs Given: A = Alcoholic A c = Not Alcoholic Posterior probabilities P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40

22. B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Example: Vitamin B-complex deficiency among general population Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 1 | A c ) = ??P(B 2 | A c ) = ??P(B 3 | A c ) = ?? P(B 4 | A c ) = ??.035.060.075.080.065.140.225.320 P(A) =.25 P(A c ) =.75 1.00 Exercise: Prior probs Given: A = Alcoholic A c = Not Alcoholic Posterior probabilities P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40

23. Example: Vitamin B-complex deficiency among general population Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. A (Yes) A c (No) Alcoholic etc. Non- deficient Thiamine- deficient Riboflavin- deficient Niacin- deficient C1C1 C2C2 C5C5 C6C6 C4C4 C3C3 C7C7 C8C8 C 1 C 2 C 3 C 4 C 5 C 6 C 7 C 8

24. Prior probabilities: BAYES’ FORMULA Assume B 1, B 2, …, B n “partition” the population, i.e., they are disjoint and exhaustive. AAcAAc B 1 B 2 B 3 ……etc……. B n Given… P(B 1 ) P(B 2 ) P(B 3 ) ……etc……. P(B n ) Conditional probabilities: P(A|B 1 ) P(A|B 2 ) P(A|B 3 ) ……etc……. P(A|B n ) 1 Then… Posterior probabilities: P(B 1 |A) P(B 2 |A) P(B 3 |A) ……etc……. P(B n |A) are computed via P(B i | A) = P(B i ∩ A) P(A) “LAW OF TOTAL PROBABILITY” P(A | B i ) P(B i ) P(A | B 1 ) P(B 1 ) + P(A | B 2 ) P(B 2 ) + …+ P(A | B n ) P(B n ) = P(A) = P(A | B 1 ) P(B 1 ) + P(A | B 2 ) P(B 2 ) + …+ P(A | B n ) P(B n ) for i = 1, 2, 3,…, n P(A ∩ B 1 )P(A ∩ B 2 )P(A ∩ B 3 )P(A ∩ B n ) ……etc……. P(A)P(A) P(A c ∩ B 1 ) P(Ac ∩B2)P(Ac ∩B2)P(A c ∩ B 3 ) ……etc……. P(A c ∩ B n ) P(Ac)P(Ac)

25. Prior probabilities: BAYES’ FORMULA Assume B 1, B 2, …, B n “partition” the population, i.e., they are disjoint and exhaustive. AAcAAc B 1 B 2 B 3 ……etc……. B n Given… P(B 1 ) P(B 2 ) P(B 3 ) ……etc……. P(B n ) 1 Then… Posterior probabilities: P(B 1 |A) P(B 2 |A) P(B 3 |A) ……etc……. P(B n |A) are computed via P(B i | A) = P(B i ∩ A) P(A) P(A | B i ) P(B i ) P(A | B 1 ) P(B 1 ) + P(A | B 2 ) P(B 2 ) + …+ P(A | B n ) P(B n ) = for i = 1, 2, 3,…, n P(A ∩ B 1 )P(A ∩ B 2 )P(A ∩ B 3 )P(A ∩ B n ) ……etc……. P(A)P(A) P(A c ∩ B 1 ) P(Ac ∩B2)P(Ac ∩B2)P(A c ∩ B 3 ) ……etc……. P(A c ∩ B n ) P(Ac)P(Ac) ……etc……

26. P(D–) =.90P(D+) =.10 Example: Screening Tests: D = Disease (+ or –), T = Test (+ or –) “prevalence” T– ∩ D+ T– ∩ D– T+ ∩ D+ T+ ∩ D– prior probabilities Highly sensitive and highly specific, but expensive. Cost-effective for adults 50+ Ex: Colorectal Cancer True Negative rate, “specificity” P(T– | D–) =.98 False Positive rate P(T+ | D–) =.02 False Negative rate P(T– | D+) =.15 True Positive rate, “sensitivity” P(T+ | D+) =.85

27. Example: Screening Tests: D = Disease (+ or –), T = Test (+ or –) “prevalence” T– ∩ D+T– ∩ D– T+ ∩ D+T+ ∩ D– P(D+) =.10P(D–) =.90 prior probabilities Highly sensitive and highly specific, but expensive. Cost-effective for adults 50+ Ex: Colorectal Cancer Fecal Occult Blood Test (FOBT) Cheap, fast, easy (… but disgusting) True Negative rate, “specificity” P(T– | D–) =.98 False Positive rate P(T+ | D–) =.02 False Negative rate P(T– | D+) =.15 True Positive rate, “sensitivity” P(T+ | D+) =.85 True Negative rate, “specificity” P(T– | D–) = ??? False Positive rate P(T+ | D–) = ??? False Negative rate P(T– | D+) = ??? True Positive rate, “sensitivity” P(T+ | D+) = ???

28. True Negative rate, “specificity” P(T– | D–) = ??? False Positive rate P(T+ | D–) = ??? False Negative rate P(T– | D+) = ??? True Positive rate, “sensitivity” P(T+ | D+) = ??? D+D+D–D– T+11436150 T– 644 50 12080200 D+D+D–D– T+.57.18.75 T–.03.22.25.60.401 Example: Screening Tests: D = Disease (+ or –), T = Test (+ or –) “prevalence” T– ∩ D+T– ∩ D– T+ ∩ D+T+ ∩ D– P(D+) =.10P(D–) =.90 prior probabilities Highly sensitive and highly specific, but expensive. Cost-effective for adults 50+ Ex: Colorectal Cancer Fecal Occult Blood Test (FOBT) Cheap, fast, easy (… but disgusting) First determine D  via gold standard on a clinical sample…

29. True Negative rate, “specificity” P(T– | D–) = ??? False Positive rate P(T+ | D–) = ??? False Negative rate P(T– | D+) = ??? True Positive rate, “sensitivity” P(T+ | D+) = ??? Example: Screening Tests: D = Disease (+ or –), T = Test (+ or –) “prevalence” T– ∩ D+T– ∩ D– T+ ∩ D+T+ ∩ D– P(D+) =.10P(D–) =.90 prior probabilities Highly sensitive and highly specific, but expensive. Cost-effective for adults 50+ Ex: Colorectal Cancer Fecal Occult Blood Test (FOBT) Cheap, fast, easy (… but disgusting) D+D+D–D– T+.57.18.75 T–.03.22.25.60.401 First determine D  via gold standard on a clinical sample…

30. True Negative rate, “specificity” P(T– | D–) = ??? False Positive rate P(T+ | D–) = ??? False Negative rate P(T– | D+) = ??? True Positive rate, “sensitivity” P(T+ | D+) =.95 Example: Screening Tests: D = Disease (+ or –), T = Test (+ or –) “prevalence” T– ∩ D+T– ∩ D– T+ ∩ D+T+ ∩ D– P(D+) =.10P(D–) =.90 prior probabilities Highly sensitive and highly specific, but expensive. Cost-effective for adults 50+ Ex: Colorectal Cancer Fecal Occult Blood Test (FOBT) Cheap, fast, easy (… but disgusting) D+D+D–D– T+.57.18.75 T–.03.22.25.60.401 First determine D  via gold standard on a clinical sample…

31. True Negative rate, “specificity” P(T– | D–) = ??? False Positive rate P(T+ | D–) = ??? False Negative rate P(T– | D+) =.05 True Positive rate, “sensitivity” P(T+ | D+) =.95 Example: Screening Tests: D = Disease (+ or –), T = Test (+ or –) “prevalence” T– ∩ D+T– ∩ D– T+ ∩ D+T+ ∩ D– P(D+) =.10P(D–) =.90 prior probabilities Highly sensitive and highly specific, but expensive. Cost-effective for adults 50+ Ex: Colorectal Cancer Fecal Occult Blood Test (FOBT) Cheap, fast, easy (… but disgusting) D+D+D–D– T+.57.18.75 T–.03.22.25.60.401 First determine D  via gold standard on a clinical sample…

32. True Negative rate, “specificity” P(T– | D–) = ??? False Positive rate P(T+ | D–) =.45 False Negative rate P(T– | D+) =.05 True Positive rate, “sensitivity” P(T+ | D+) =.95 Example: Screening Tests: D = Disease (+ or –), T = Test (+ or –) “prevalence” T– ∩ D+T– ∩ D– T+ ∩ D+T+ ∩ D– P(D+) =.10P(D–) =.90 prior probabilities Highly sensitive and highly specific, but expensive. Cost-effective for adults 50+ Ex: Colorectal Cancer Fecal Occult Blood Test (FOBT) Cheap, fast, easy (… but disgusting) D+D+D–D– T+.57.18.75 T–.03.22.25.60.401 First determine D  via gold standard on a clinical sample…

33. True Negative rate, “specificity” P(T– | D–) =.55 False Positive rate P(T+ | D–) =.45 False Negative rate P(T– | D+) =.05 True Positive rate, “sensitivity” P(T+ | D+) =.95 Example: Screening Tests: D = Disease (+ or –), T = Test (+ or –) “prevalence” T– ∩ D+T– ∩ D– T+ ∩ D+T+ ∩ D– P(D+) =.10P(D–) =.90 prior probabilities Highly sensitive and highly specific, but expensive. Cost-effective for adults 50+ Ex: Colorectal Cancer Fecal Occult Blood Test (FOBT) Cheap, fast, easy (… but disgusting) D+D+D–D– T+.57.18.75 T–.03.22.25.60.401 First determine D  via gold standard on a clinical sample…

34. “Predictive Value” (PV) for the general population D+D+D–D– T+.57.18.75 T–.03.22.25.60.401 Example: Screening Tests: D = Disease (+ or –), T = Test (+ or –) “prevalence” T– ∩ D+T– ∩ D– T+ ∩ D+T+ ∩ D– P(D+) =.10P(D–) =.90 prior probabilities Highly sensitive and highly specific, but expensive. Cost-effective for adults 50+ Ex: Colorectal Cancer Fecal Occult Blood Test (FOBT) Cheap, fast, easy (… but disgusting) of a positive test: First determine D  via gold standard on a clinical sample… of a negative test: Then calculate PV+ and PV– from the table. posterior probabilities

35. D+D+D–D– T+.57.18.75 T–.03.22.25.60.401 Example: Screening Tests: D = Disease (+ or –), T = Test (+ or –) “prevalence” T– ∩ D+T– ∩ D– T+ ∩ D+T+ ∩ D– P(D+) =.10P(D–) =.90 prior probabilities Highly sensitive and highly specific, but expensive. Cost-effective for adults 50+ Ex: Colorectal Cancer Fecal Occult Blood Test (FOBT) Cheap, fast, easy (… but disgusting) of a positive test: Then calculate PV+ and PV– from the table. First determine D  via gold standard on a clinical sample… of a negative test: “Predictive Value” (PV) for the general population posterior probabilities

36. D+D+D–D– T+.57.18.75 T–.03.22.25.60.401 Example: Screening Tests: D = Disease (+ or –), T = Test (+ or –) “prevalence” T– ∩ D+T– ∩ D– T+ ∩ D+T+ ∩ D– P(D+) =.10P(D–) =.90 prior probabilities Highly sensitive and highly specific, but expensive. Cost-effective for adults 50+ Ex: Colorectal Cancer Fecal Occult Blood Test (FOBT) Cheap, fast, easy (… but disgusting) of a positive test: of a negative test: Then calculate PV+ and PV– from the table. First determine D  via gold standard on a clinical sample… CRUDE gross overestimate of disease prevalence in population “Predictive Value” (PV) for the general population

37. Example: Screening Tests: D = Disease (+ or –), T = Test (+ or –) “prevalence” T– ∩ D+T– ∩ D– T+ ∩ D+T+ ∩ D– P(D+) =.10P(D–) =.90 prior probabilities Highly sensitive and highly specific, but expensive. Cost-effective for adults 50+ Ex: Colorectal Cancer Fecal Occult Blood Test (FOBT) Cheap, fast, easy (… but disgusting) posterior probabilities of a positive test: of a negative test: “Predictive Value” (PV) for the general population

38. .99 True Negative rate, “specificity” P(T– | D–) =.55 False Positive rate P(T+ | D–) =.45 False Negative rate P(T– | D+) =.05 Example: Screening Tests: D = Disease (+ or –), T = Test (+ or –) “prevalence” T– ∩ D+T– ∩ D– T+ ∩ D+T+ ∩ D– prior probabilities Highly sensitive and highly specific, but expensive. Cost-effective for adults 50+ Ex: Colorectal Cancer Fecal Occult Blood Test (FOBT) Cheap, fast, easy (… but disgusting) posterior probabilities P(D+ | T+) of a positive test: P(D– | T–) of a negative test: “Predictive Value” (PV) True Positive rate, “sensitivity” P(T+ | D+) =.95 P(D–) =.90P(D+) =.10 ???.19 ???????????

39. 39 Frequentists Forever Bayes’ Rules! The Debate Continues…

40. “Receiver Operating Characteristic” (ROC) Curves

41. AUC (Area Under Curve) = () P(Test is correct in a random (D+, D–) pair). “Receiver Operating Characteristic” (ROC) Curves

42. AUC (Area Under Curve) = () P(Test is correct in a random (D+, D–) pair). as a function of D “Receiver Operating Characteristic” (ROC) Curves

43. 43 EXAM 1 ENDS HERE