# 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

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,
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
Using the range
Years In    Years In   Education   State Of   Loan Paid
BR          >=6         <=8                                yes

Let Z be the random variable giving the amount of
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
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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 views: 100 posted: 11/13/2010 language: English pages: 32