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Bayes Theorem - For Friday_ Oct 4

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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=RN
                               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=RN               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   RA         0.20.3 = 0.06
                            R                                    P(R ) P(A|R)
                 0.2
                                           0.7       U   RU         0.20.7 = 0.14
 Saturday
                                           0.6       A   NA         0.80.6 = 0.48
                 0.8
                            N                                    P(N ) P(A|N)

                                           0.4       U   NU         0.80.4 = 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=CT                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 CF       0.6 0.0002 = .00012
                        C
               0.6
                                0.9998    N C  N 0.6 0.9998 = 0.59988
Vehicle
                               0.00025    F TF       0.40.00025= 0.0001
               0.4
                        T
                               0.99975    N T N      0.40.99975= .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 PBk 
         PBk | A 
                        PA | B PB 
                       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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