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Bayes’ Theorem Special Type of Conditional Probability Recall- Conditional Probability P(Y T C|S) will be used to calculate P(S|Y T C) P(Y T C|F) will be used to calculate P(F|Y T C) 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 partition of 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 (N R ) c P(N)=1-P(R)=0.8 Our goal is to compute P(A). Using Venn diagram –Method1 S=RN Event A is the disjoint union of A event R A & R N event N A 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 Let us examine P(A|R) S=RN Consider P(A|R) The conditional probability that sale is A successful given that it rains Using conditional R N probability formula P( R A ) P( A | R ) P( R ) Bayes’, Partitions Tree Diagram-Method 2 Conditional Probability Event Probability Probability 0.3 A RA 0.20.3 = 0.06 R P(R ) P(A|R) 0.2 0.7 U RU 0.20.7 = 0.14 Saturday 0.6 A NA 0.80.6 = 0.48 0.8 N P(N ) P(A|N) 0.4 U NU 0.80.4 = 0.32 *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 P( R A ) P( A | R ) P( R ) P( R | A ) P( A ) P( A | R ) P( R ) P( A | N ) P( N ) 0.3 0.2 = 0.3 0.2 0.6 0.8 *show slide 7 = 0.1111 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 T are complementary events (C T ) c P(C)=1-P(T)=0.6 Our goal is to compute the conditional probability of a Light truck accident given that it is fatal P(T|F). Goal P(T|F) S=CT Consider P(T|F) Conditional probability F of a Light truck accident given that it is fatal C T Using conditional probability formula P( T F ) P( T | F ) P( F ) 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 P( T F ) P( F | T ) P( T ) P( T | F ) P( F ) P( F | T ) P( T ) P( F | C ) P( C ) ( 0.00025 )( 0.4 ) = ( 0.00025 )( 0.4 ) ( 0.0002 )( 0.6 ) = 0.4545 Tree Diagram- Method2 Conditional Probability Event Probability Probability 0.0002 F CF 0.6 0.0002 = .00012 C 0.6 0.9998 N C N 0.6 0.9998 = 0.59988 Vehicle 0.00025 F TF 0.40.00025= 0.0001 0.4 T 0.99975 N T N 0.40.99975= .3999 Tree Diagram- Method2 P( T F ) P( T F ) P( T | F ) P( F ) P( T F ) P( C F ) ( 0.00025 )( 0.4 ) = ( 0.00025 )( 0.4 ) ( 0.0002 )( 0.6 ) = 0.4545 Partition S A B1 B2 B3 P( A) P( A | B1 ) P( B1 ) P( A | B2 ) P( B2 ) P( A | B3 ) P( B3 ) 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. P( A) P( A S ) P( A ( B1 B2 Bn )) P(( A B1 ) ( A B2 ) ( A Bn )) Law of Total Probability P(( A B1 ) ( A B2 ) ( A Bn )) P( A B1 ) P( A B2 ) P( A Bn ) P( A | B1 ) P( B1 ) P( A | B2 ) P( B2 ) P( A | Bn ) P( Bn ) n P( A | Bi ) P( Bi ) i 1 n P ( A) P ( A | Bi ) P ( Bi ). i 1 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 1 k n we have P A | Bk PBk PBk | A PA | B PB n j j j 1 Focus on the Project Recall P(Y T C|S) will be used to calculate P(S|Y T C) P(Y T C|F) will be used to calculate P(F|Y T C) How can Bayes’ Theorem help us with the decision on whether or not to attempt a loan work out? Partitions 1. Event S 2. Event F Given P(Y T C|S) P(Y T C|F) Need P(S|Y T C) P(F|Y T C) Using Bayes Theorem P(Y T C | S ) P( S ) P S | Y T C P(Y T C | S ) P( S ) P(Y T C | F ) P( F ) (0.022) (0.464) (0.022) (0.464) (0.021) (0.536) P(S|Y T C) 0.477 LOAN FOCUS EXCEL-BAYES P(Y T C | F ) P( F ) P F | Y T C P(Y T C | S ) P( S ) P(Y T C | F ) P( F ) (0.021) (0.536) . (0.022) (0.464) (0.021) (0.536) P(F|Y T C) 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 ) $4,000,000 P ( Z $4,000,000) $250,000 P ( Z $250,000) $4,000,000 P ( S | Y T C ) $250,000 P ( F | Y T C ) $4,000,000 (0.477) $250,000 (0.523) 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 Years In Years In Education State Of Loan Paid Former Bank Business Business Level Economy Back BR >=6 <=8 yes 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(Y T C|S) 0.073 and P(Y T C|F) 0.050. Calculations P(Y T C|S) 0.073 P(Y T C|F) 0.050 P(S|Y T C) 0.558 P(F|Y T C) 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.

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posted: | 11/13/2010 |

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