1. Statistics Workshop Bayes Theorem J-Term 2009 Bert Kritzer

2. Bayesian Inference A method of using prior information about the probability of some event combined with conditional probabilities about consequences of that event to obtain post hoc probabilities using the actual observed consequences. We know the percentage of the population having some medical condition and we also know the probabilities of observing a symptom among those who do and those who do not have that condition. What is the probability that someone has the condition if we observe the symptom?

3. Addition & Multiplication Rules Revisted General case Special case If E and F are statistically independent If E and F are mutually exclusive As a single eventAs two events occurring together

4. Buses and Pedestrians 95% of buses are Metro Transit (MT) 80% of the time client correctly identify type P(B) = Probability of client identifying as MT P(A) = Prior Probability of bus being MT (.95) P(B|A) = Conditional probability of client saying it was MT give that it was in fact MT (.8) P(A|B) = Post hoc probability that it was in fact MT given that client says it was MT P(AB) = Joint probability that client says it was MT and it was in fact MT

5. Combined Probabilities ConditionalJoint P(A|B) P(B|A) P(AB) P(AB) A A B B A bus is Metro B client says bus is Metro

6. Multiplication Rule Probability of saying it was MT:

7. Summary P(buses that are Metro) P(iding as Metro) Joint P that bus is Metro Was MTP(A)=.95P(B|A)=.8P(AB)=.76 Not MTP(A)=.05P(B|A)=.2P(AB)=.01

8. Bayes Theorem

9. P(MT given client says its MT) P(A) =.95P of any bus being MT P(A)=.05P of any bus not MT P(B|A) =.80P of saying MT if it was MT P(B|A)=.20P of saying MT if not MT

10. Thinking Pictorially P(AB)=.01 P(AB)=.19 P(AB)=.04 P(AB)=.76 A A B B

11. P(not MT given client says not MT) P(A) =.95P of any bus being MT P(A)=.05P of any bus not MT P(B|A)=.80P of saying not MT if not MT P(B|A)=.20P of saying not MT if it is MT

12. Client Says It Is Not MT P(AB)=.01 P(AB)=.19 P(AB)=.04 P(AB)=.76 A A B B

13. Tree Trimming MT ID-MT MT.95.05.8.76.01.2 ID-MT.8.04 ID-MT.19.2

14. Online Bayes Calculator http://statpages.org/bayes.html

15. Diagnosing Sexual Abuse P(B) Probability of a child being diagnosed as having been abused P(A|B) = ? Probability of having been abused given positive diagnosis P(A) =.10 Probability of any random child having been abused P(A) =.90 Probability of any random child not having been abused P(B|A) =.90 Probability of child who was abused being correctly diagnosed as abused P(B|A) =.10 Probability of child who was not abused being incorrectly diagnosed as abused

16. 95% Accurate Diagnosis P(B) Probability of a child being diagnosed as having been abused P(A|B) = ? Probability of having been abused given positive diagnosis P(A) =.10 Probability of any random child having been abused P(A) =.90 Probability of any random child not having been abused P(B|A) =.95 Probability of child who was abused being correctly diagnosed as abused P(B|A) =.05 Probability of child who was not abused being incorrectly diagnosed as abused

17. 20% Abuse Rate P(B) Probability of a child being diagnosed as having been abused P(A|B) = ? Probability of having been abused given positive diagnosis P(A) =.20 Probability of any random child having been abused P(A) =.80 Probability of any random child not having been abused P(B|A) =.95 Probability of child who was abused being correctly diagnosed as abused P(B|A) =.05 Probability of child who was not abused being incorrectly diagnosed as abused

18. Different Error Rates P(B) Probability of a child being diagnosed as having been abused P(A|B) = ? Probability of having been abused given positive diagnosis P(A) =.10 Probability of any random child having been abused P(A) =.90 Probability of any random child not having been abused P(B|A) =.90 Probability of child who was abused being correctly diagnosed as abused (10% error rate) P(B|A) =.20 Probability of child who was not abused being incorrectly diagnosed as abused (80% accurate)