Decision Theory III (MATH 3071) Lecture 1 2017/18 Useful - - PowerPoint PPT Presentation
Decision Theory III (MATH 3071) Lecture 1 2017/18 Useful Information Lecturers: Dr Camila Caiado (Michaelmas), Prof Frank Coolen (Epiphany) Office: CM321 Email: c.c.d.s.caiado@durham.ac.uk Lecture Times: Wednesdays and Fridays 9h
2017/18
◮ Lecturers: Dr Camila Caiado (Michaelmas), Prof Frank
Coolen (Epiphany)
◮ Office: CM321 ◮ Email: c.c.d.s.caiado@durham.ac.uk ◮ Lecture Times: Wednesdays and Fridays 9h CG85 ◮ Problem Classes: Mondays 13h CG85, Weeks: 3, 5, 7, 9, 12,
14, 16, 18
◮ DUO page: Course ID 73497
DUO: Summaries, problem sheets, solutions, extras Recommended books:
◮ M. Peterson, An Introduction to Decision Theory, Cambridge
University Press, 2009
◮ J. Q. Smith, Decision Analysis - A Bayesian Approach,
Chapman and Hall, 1988
◮ D. V. Lindley, Making Decision, 2nd Ed, Wiley, 1985 ◮ S. French, Decision Theory: An Introduction to the
Mathematics of Rationality, Ellis, Horwood, Chichester, 1986
◮ M. H. DeGroot, Optiomal Statistical Decisions, McGraw-Hill,
2004
◮ R. T. Clemen, Making Hard Decisions, 2nd Ed, Duxbury Press,
1995
◮ Homework: 4 per term ◮ Office Hours: Friday 10h-12h00 (tentative) ◮ Contact: Questions by e-mail are also welcome, include a
picture/scan of your attempt when possible. Answers to the most common questions will be posted on DUO.
Definition: Suppose Ω is a sample space. A probability distribution
satisfying the following axioms:
◮ A1. P(A) ≥ 0 for every A ⊆ Ω ◮ A2. P(Ω) = 1 ◮ A3. if A and B are incompatible or mutually exclusive events
then P(A ∪ B) = P(A) + P(B).
Mutually Exclusive
Black Red
Definition: Suppose Ω is a sample space. A probability distribution
satisfying the following axioms:
◮ A1. P(A) ≥ 0 for every A ⊆ Ω ◮ A2. P(Ω) = 1 ◮ A3. if A and B are incompatible or mutually exclusive events
then P(A ∪ B) = P(A) + P(B). From the axioms, we can derive:
◮ C1. For any events A and B, P(B − A) = P(B) − P(A ∩ B). ◮ C2. For any event A, P(AC) = 1 − P(A). ◮ C3. P(∅) = 0. ◮ C4. For any two events A and B,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
◮ C5. If A ⊆ B then P(A) ≤ P(B). ◮ C6. For any two events A and B,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
Conditional Probability: The conditional probability of an event A given B, P(A|B), is defined as: P(A|B) = P(A ∩ B) P(B) , P(B) = 0. Independent Events: Two events A and B are independent if and
P(A ∩ B) = P(A)P(B).
(Partition Theorem) Let A1, . . . , An be a partition of Ω, i.e. A1, . . . , An are mutually disjoint and A1 ∪ . . . ∪ An = Ω. If P(Aj) = 0, j = 1, . . . , n then, for an event B ⊆ Ω, we have P(B) =
n
P (B|Aj) P (Aj) . (Bayes’ Theorem) Let A1, . . . , An be a partition of Ω. For B ∈ Ω, P(B) = 0, we have P(Aj|B) = P(B|Aj)P(Aj)
n
k=1 P(B|Ak)P(Ak), j = 1, . . . , n.
In a survey it is found that 40% of the population like dogs, 60% like cats, and 70% of those who like dogs also like cats.
population likes both dogs and cats.
population likes dogs or cats but not both.
Suppose 0.001 of the population have a certain disease. A diagnostic test is carried out for the disease: the outcome of the test is either positive or negative. It is known from past experience that the test is 90% reliable, i.e. a person with the disease will test (correctly) positive with probability 0.9, while a person without the disease will test (incorrectly) positive with probability 0.1.
they test positive?
they have the disease?
A plane is missing, and it is presumed that it was equally likely to have gone down in any of three possible regions. The probability that the plane will be found upon a search of region 1, when the plane actually is in that region, is judged to be 0.9. The corresponding figures for regions 2 and 3 are 0.95 and 0.85 respectively. What is the conditional probability that the plane is in region 3 given that searches in regions 1 and 2 have been unsuccessful?