Bayes’ Theorem Special Type of Conditional Probability
Recall- Conditional Probability • P(YTC|S) will be used to calculate P(S|YTC) • P(YTC|F) will be used to calculate P(F|YTC) • HOW????? • We will learn in the next lesson? • BAYES THEOREM
Definition of Partition Let the events B1, B2, , Bn be non-empty subsets of a sample space S for an experiment. The Bi’s are a partitionof S if the intersection of any two of them is empty, and if their union is S. This may be stated symbolically in the following way. 1. Bi Bj = , unless i = j. 2. B1 B2 Bn = S.
Partition Example S B1 B2 B3
Example 1 Your retail business is considering holding a sidewalk sale promotion next Saturday. Past experience indicates that the probability of a successful sale is 60%, if it does not rain. This drops to 30% if it does rain on Saturday. A phone call to the weather bureau finds an estimated probability of 20% for rain. What is the probability that you have a successful sale?
Example 1 Events R- rains next Saturday N -does not rain next Saturday. A -sale is successful U- sale is unsuccessful. Given P(A|N) = 0.6 and P(A|R) = 0.3. P(R) = 0.2. In addition we know R and N are complementary events P(N)=1-P(R)=0.8 Our goal is to compute P(A).
Using Venn diagram –Method1 Event A is the disjoint union of event R A & event N A S=RN A R N • P(A) = P(R A) + P(N A)
P(A)- Probability that you have a Successful Sale We need P(R A) and P(N A) Recall from conditional probability P(R A)= P(R )* P(A|R)=0.2*0.3=0.06 Similarly P(N A)= P(N )* P(A|N)=0.8*0.6=0.48 Using P(A) = P(R A) + P(N A) =0.06+0.48=0.54
S=RN A R N Let us examine P(A|R) • Consider P(A|R) • The conditional probability that sale is successful given that it rains • Using conditional probability formula
Conditional Probability Probability Event Probability 0.3 AR A 0.20.3 = 0.06 R 0.2 0.7 UR U 0.20.7 = 0.14 Saturday 0.6 AN A 0.80.6 = 0.48 0.8 N UN U 0.80.4 = 0.32 0.4 Bayes’, Partitions Tree Diagram-Method 2 P(R ) P(A|R) P(N ) P(A|N) *Each Branch of the tree represents the intersection of two events *The four branches represent Mutually Exclusive events
Method 2-Tree Diagram Using P(A) = P(R A) + P(N A) =0.06+0.48=0.54
Extension of Example1 Consider P(R|A) The conditional probability that it rains given that sale is successful the How do we calculate? Using conditional probability formula = = 0.1111 *show slide 7
Example 2 • In a recent New York Times article, it was reported that light trucks, which include SUV’s, pick-up trucks and minivans, accounted for 40% of all personal vehicles on the road in 2002. Assume the rest are cars. Of every 100,000 car accidents, 20 involve a fatality; of every 100,000 light truck accidents, 25 involve a fatality. If a fatal accident is chosen at random, what is the probability the accident involved a light truck?
Example 2 Events C- Cars T –Light truck F –Fatal Accident N- Not a Fatal Accident Given P(F|C) = 20/10000 and P(F|T) = 25/100000 P(T) = 0.4 In addition we know C and Tare complementary events P(C)=1-P(T)=0.6 Our goal is to compute the conditional probability of a Light truck accident given that it is fatalP(T|F).
Goal P(T|F) S=CT Consider P(T|F) Conditional probability of a Light truck accident given that it is fatal Using conditional probability formula F C T
P(T|F)-Method1 Consider P(T|F) Conditional probability of a Light truck accident given that it is fatal How do we calculate? Using conditional probability formula = = 0.4545
Conditional Probability Probability Event Probability 0.0002 FC F 0.6 0.0002 = .00012 C 0.6 0.9998 NC N 0.6 0.9998 = 0.59988 Vehicle 0.00025 FT F 0.40.00025= 0.0001 0.4 T NT N 0.40.99975= .3999 0.99975 Tree Diagram- Method2
Tree Diagram- Method2 = = 0.4545
S A B1 B2 B3 Partition
Law of Total Probability Let the events B1, B2, , Bn partition the finite discrete sample space S for an experiment and let A be an event defined on S.
Bayes’ Theorem • Suppose that the events B1, B2, B3, . . . , Bn partition the sample space S for some experiment and that A is an event defined on S. For any integer, k, such that we have
Focus on the Project Recall • P(YTC|S) will be used to calculate P(S|YTC) • P(YTC|F) will be used to calculate P(F|YTC)
How can Bayes’ Theorem help us with thedecision on whether or not to attempt a loan work out? Partitions • Event S • Event F Given P(YTC|S) P(YTC|F) Need P(S|YTC) P(F|YTC)
P(S|YTC) 0.477 Using Bayes Theorem LOAN FOCUS EXCEL-BAYES P(F|YTC) 0.523
RECALL • Z is the random variable giving the amount of money, in dollars, that Acadia Bank receives from a future loan work out attempt to borrowers with the same characteristics as Mr. Sanders, in normal times. E(Z) $2,040,000.
Decision EXPECTED VALUE OF A WORKOUT=E(Z) $2,040,000 FORECLOSURE VALUE- $2,100,000 RECALL FORECLOSURE VALUE> EXPECTED VALUE OF A WORKOUT DECISION FORECLOSURE
Further Investigation I • let Y be the event that a borrower has 6, 7, or 8 years of experience in the business. Using the range Let Z be the random variable giving the amount of money, in dollars, that Acadia Bank receives from a future loan work out attempt to borrowers with Y and a Bachelor’s Degree, in normal times. When all of the calculations are redone, with Y replacing Y, we find that P(YTC|S) 0.073 and P(YTC|F) 0.050.
Calculations P(Y TC|S) 0.073 P(Y TC|F) 0.050 P(S|Y TC) 0.558 P(F|Y TC) 0.442 The expected value of Z is E(Z ) $2,341,000. Since this is above the foreclosure value of $2,100,000, a loan work out attempt is indicated.
Further Investigation II • Let Y" be the event that a borrower has 5, 6, 7, 8, or 9 years of experience in the business • Let Z" be the random variable giving the amount of money, in dollars, that Acadia Bank receives from a future loan work out attempt to borrowers with 5, 6, 7, 8, or 9 years experience and a Bachelor's Degree, in normal times. Redoing our work yields the follow results.
Similarly can calculate E(Z ) • Make at a decision- Foreclose vs. Workout • Data indicates Loan work out
Close call for Acadia Bank loan officers Based upon all of our calculations, we recommend that Acadia Bank enter into a work out arrangement with Mr. Sanders.