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Bayes_Theorem.ppt

• 1. Bayes’Theorem Special Type of Conditional Probability
• 2. 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
• 3. 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.
• 5. 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?
• 6. 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). ) R N ( c 
• 7. Using Venn diagram –Method1 Event A is the disjoint union of event R  A & event N  A S=RN R N A P(A) = P(R  A) + P(N  A)
• 8. 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
• 9. 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 ) R ( P ) A R ( P ) R | A ( P   S=RN R N A
• 10. Tree Diagram-Method 2 Bayes’, Partitions Saturday R N A R  A 0.20.3 = 0.06 A N  A 0.80.6 = 0.48 U R  U 0.20.7 = 0.14 U N  U 0.80.4 = 0.32 0.2 0.8 0.7 0.3 0.6 0.4 Probability Conditional Probability Probability Event *Each Branch of the tree represents the intersection of two events *The four branches represent Mutually Exclusive events P(R ) P(A|R) P(N ) P(A|N)
• 11. Method 2-Tree Diagram Using P(A) = P(R  A) + P(N  A) =0.06+0.48=0.54
• 12. 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 ) N ( P ) N | A ( P ) R ( P ) R | A ( P ) R ( P ) R | A ( P ) A ( P ) A R ( P ) A | R ( P        8 0 6 0 2 0 3 0 2 0 3 0 . . . . . .     = = 0.1111 *show slide 7
• 13. 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?
• 14. 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 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). ) T C ( c 
• 15. Goal P(T|F) Consider P(T|F) Conditional probability of a Light truck accident given that it is fatal Using conditional probability formula ) F ( P ) F T ( P ) F | T ( P   S=CT C T F
• 16. 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 ) C ( P ) C | F ( P ) T ( P ) T | F ( P ) T ( P ) T | F ( P ) F ( P ) F T ( P ) F | T ( P        ) . )( . ( ) . )( . ( ) . )( . ( 6 0 0002 0 4 0 00025 0 4 0 00025 0  = = 0.4545
• 17. Tree Diagram- Method2 Vehicle C T F C  F 0.6 0.0002 = .00012 F T  F 0.40.00025= 0.0001 N C  N 0.6 0.9998 = 0.59988 N T N 0.40.99975= .3999 0.6 0.4 0.9998 0.0002 0.00025 0.99975 Probability Conditional Probability Probability Event
• 19. Partition S B1 B2 B3 A ) ( ) | ( ) ( ) | ( ) ( ) | ( ) ( 3 3 2 2 1 1 B P B A P B P B A P B P B A P A P      
• 20. Law of Total Probability )) ( ) ( ) (( )) ( ( ) ( ) ( 2 1 2 1 n n B A B A B A P B B B A P S A P A P                 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.
• 21. Law of Total Probability                         n i i i n n n n B P B A P B P B A P B P B A P B P B A P B A P B A P B A P B A B A B A P 1 2 2 1 1 2 1 2 1 ) ( ) | ( ) ( ) | ( ) ( ) | ( ) ( ) | ( ) ( ) ( ) ( )) ( ) ( ) ((    . ) ( ) | ( ) ( 1     n i i i B P B A P A P
• 22. 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 n k   1              n j j j k k k B P B A P B P B A P A B P 1 | | |
• 23. 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)
• 24. 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)
• 25. Using Bayes Theorem P(S|Y  T  C)  0.477   ) 536 . 0 ( ) 021 . 0 ( ) 464 . 0 ( ) 022 . 0 ( ) 464 . 0 ( ) 022 . 0 ( ) ( ) | ( ) ( ) | ( ) ( ) | ( |                   F P F C T Y P S P S C T Y P S P S C T Y P C T Y S P   . ) 536 . 0 ( ) 021 . 0 ( ) 464 . 0 ( ) 022 . 0 ( ) 536 . 0 ( ) 021 . 0 ( ) ( ) | ( ) ( ) | ( ) ( ) | ( |                   F P F C T Y P S P S C T Y P F P F C T Y P C T Y F P LOAN FOCUS EXCEL-BAYES P(F|Y  T  C)  0.523
• 26. 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. ) 523 . 0 ( 000 , 250 \$ ) 477 . 0 ( 000 , 000 , 4 \$ ) | ( 000 , 250 \$ ) | ( 000 , 000 , 4 \$ ) 000 , 250 \$ ( 000 , 250 \$ ) 000 , 000 , 4 \$ ( 000 , 000 , 4 \$ ) (                   C T Y F P C T Y S P Z P Z P Z E E(Z)  \$2,040,000.
• 27. 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
• 28. 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(Y   T  C|S)  0.073 and P(Y   T  C|F)  0.050. Former Bank Years In Business Years In Business Education Level State Of Economy Loan Paid Back BR >=6 <=8 yes
• 29. 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.
• 30. 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.
• 31. Similarly can calculate E(Z  )  Make at a decision- Foreclose vs. Workout  Data indicates Loan work out
• 32. 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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