EMGT 269 - Elements of Problem Solving and Decision Making
10. USING DATA
1. Constructing Probability Distributions with Data
A. Discrete Case
The Empirical Probability Mass function is constructed using
relative frequencies of events
Relative Frequency 
# of occurences of particular outcome
Total number of occurrence s
Estimation of Discrete Empirical Probability Mass Function
# Accidents
0
1
2
3
4
# Occurrences
N0
N1
N2
N3
N4
Pr(#Accidents)
N0/M
N1/M
N2/M
N3/M
N4/M
4
Total
M=
N
i 0
i
1.0
Maintenance Example
You own a manufacturing plant and are trying to design a
maintenance policy. You would like to design your
maintenance interval such that you balance the failure time
of a machine to the maintenance interval.
 Maintenance interval is short  costly due to frequent
maintenance and machines never fail.
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 126
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
 Maintenance interval is too long the machines fail,
interrupting production resulting into high cost.
Suppose you suggest an interval of 260 days. You want to
calculate the probability of #machine failures per day in this
period and use this in your interval selection.
No Failures
One Failure
Two Failures
Total # Days
Pr(# Failures in a Day)
217/260 = 0.835
32/260 = 0.123
11/260 = 0.042
1
Pr(# Failures in a Day)
# Days
217
32
11
260
Pr(# Failures in a Day)
Also important in selecting a maintenance1.000
interval is whether
these "two failure days" happen towards the end of the 260
day period.
Graph of Probability Mass Function
1.000
0.800
0.600
0.400
0.200
0.000
Pr(# Failures in a
Day)
0.500
0.000
No Failures
0.835
Pr(# Failures in a
No Failures
Day)One Failure Two Failures
0.835
0.123
0.042
No Failure (0.835)
One Failure (0.123)
Two Failures (0.042)
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 127
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
One
EMGT 269 - Elements of Problem Solving and Decision Making
Notes:
 Make sure you have enough data for accuracy (at least 5
observations in each category)
 Always ask: Does past data represent future uncertainty?
B. Continuous Case
Estimation of Emperical Continuous Distribution Function
Y = Failure Time of Machine
1. Given data: yi , i = 1,…,n.
2. Order data such that
y(1)  y( 2 )    y( n 1)  yn
3. Estimate y Min . In this case, we may set: yMin  0
4. Set:
F ( yMin )  Pr(Y  yMin )  0
1
F ( y(1) )  Pr(Y  y(1) ) 
n
2
F ( y( 2 ) )  Pr(Y  y( 2 ) ) 
n

n 1
n
1
F ( y( n ) )  Pr(Y  y( n ) ) 
n
5. Plot the points ( y min , F ( y min )), ( y(1) , F ( y(1) )), , ( y( n ) , F ( y( n ) )) in
a graph.
6. Connect these points by a straight line.
F ( y( n 1) )  Pr(Y  y( n1) ) 
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 128
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
Above procedure may be referred to as
STRAIGHT LINE APPROXIMATION
EXAMPLE: HALFWAY HOUSE
Y(i)
0
52
76
100
136
137
186
196
205
250
257
264
280
282
283
303
313
317
325
345
373
384
400
402
408
417
422
472
480
643
693
732
749
750
791
891
# Observations <=
Y(i)
Estimated Minimum
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
Pr(Yearly Bed Cost <=
y(i))
0.00
0.03
0.06
0.09
0.11
0.14
0.17
0.20
0.23
0.26
0.29
0.31
0.34
0.37
0.40
0.43
0.46
0.49
0.51
0.54
0.57
0.60
0.63
0.66
0.69
0.71
0.74
0.77
0.80
0.83
0.86
0.89
0.91
0.94
0.97
1.00
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 129
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
Pr(Yearly Bed Cost <= X)
EMGT 269 - Elements of Problem Solving and Decision Making
1.00
0.80
0.60
0.40
0.20
0.00
0
100
200
300
400
500
600
700
800
900
X
 What if we observe ties in the data?
Observation
Y(i)
# Observations <= Y(i)
Pr(Y<=Y(i))
0
0
Estimated Minimum
0
1
1
1
1/7
2
3
2
2/7
3
7
skip
skip
4
7
skip
skip
5
7
5
5/7
6
9
6
6/7
7
11
7
7/7
y
0
1
3
7
9
11
Pr(Y<=y)
0.00
0.14
0.29
0.71
0.86
1.00
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 130
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
1.00
Pr(Y<=y)
0.80
0.60
0.40
0.20
0.00
0
1
2
3
4
5
6
7
8
9 10 11 12
y
How do we use Empirical CDF in Decision Trees?
Pr(Yearly Bed Cost <= X)
As before, use discrete approximation e.g.
 Extended Pearson Tukey Method
 The Bracket Median Method
95%
1.00
0.80
0.60
0.40
0.20
0.00
50%
5%
0
100 200 300 400 500 600 700 800 900
75
320
X
750
Cost= 75 (0.185)
Cost = 320 (0.630)
Cost = 70 (0.185)
Discrete Approximation - Extended Pearson Tukey Method
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 131
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
C. Using Data to fit Theoretical Probability Models
Method of Moments
Let Y be a random variable e.g. failure time of a machine
1. Given data: yi , I=1,…,n.
2. Calculate the Sample Mean (=First Moment):
1 n
y   yi
n i 1
3. Calculate the Sample Variance (=Second Moment)
1 n
2
s 
(
y

y
)
 i
n  1 i 1
2
4. Select a Theoretical Probability Model with CDF
F ( y | 1 , 2 ) , where (1 , 2 ) are the parameters
5. Calculate the theoretical expressions for
E[Y ]  g(1 , 2 ),Var(Y )  h(1 , 2 )
6. Solve for the parameters (1 , 2 ) by setting
 g ( 1 , 2 )  y
 E [Y ]  y


2
2
h
(

,

)

s
Var
(
Y
)

s

 1 2
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 132
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
HALFWAY HOUSE EXAMPLE CONTINUED
Observation
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
Total
Sample Mean
Divide by n=35
X
52
76
100
136
137
186
196
205
250
257
264
280
282
283
303
313
317
325
345
373
384
400
402
408
417
422
472
480
643
693
732
749
750
791
891
13314
X - Sample Mean
-328.4
-304.4
-280.4
-244.4
-243.4
-194.4
-184.4
-175.4
-130.4
-123.4
-116.4
-100.4
-98.4
-97.4
-77.4
-67.4
-63.4
-55.4
-35.4
-7.4
3.6
19.6
21.6
27.6
36.6
41.6
91.6
99.6
262.6
312.6
351.6
368.6
369.6
410.6
510.6
Total
(X - Sample Mean)^2
107846.56
92659.36
78624.16
59731.36
59243.56
37791.36
34003.36
30765.16
17004.16
15227.56
13548.96
10080.16
9682.56
9486.76
5990.76
4542.76
4019.56
3069.16
1253.16
54.76
12.96
384.16
466.56
761.76
1339.56
1730.56
8390.56
9920.16
68958.76
97718.76
123622.56
135865.96
136604.16
168592.36
260712.36
1609706.40
380.4
Sample Variance
47344.31
Sample St. Dev.
217.59
Divide by (n-1)=34
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 133
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
Random Variable Y =
Yearly Bed Cost in Half Way House
4. Propose Normal Probability Model: i.e Y N (  , )
2
5. E[Y] = g (  , )   , Var(Y) = h(  , )  
6.
 E[Y ]  380.4
   380.4
 2

Var(Y )  47344.31   47344.31
   380.4

  217.59
1.00
Pr(Yearly Bed Cost <= x)
0.80
0.60
0.40
0.20
0.00
0
100
200
300
400
500
600
700
800
900
x
Empirical Pr(Yearly Bed Cost <= X)
Theoretical Normal Approximation
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 134
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
EXAMPLES "METHOD OF MOMENTS"
FOR CONTINUOUS DISTRIBUTIONS
Theoretical
Distribution
Normal:
N (  , )
Gamma:
G ( ,  )
Exponential:
Exp(  )
Beta:
Beta ( r, n )
Theoretical
Expressions
Parameter
Solutions
 E[Y ]  

2
Var(Y )  
 y

2
  s


 E[Y ]  

Var(Y )  
2

  y 2 / s 2

   y / s 2
1

E[Y ] 



1
Var(Y ) 

( ) 2
r

E[Y ] 

n

r  (n  r )
Var(Y )  2

n ( n  1)

1
y


n  r  y 3 ( y  1)  s 2 ( y  1) 2 / y 3

 y * (n  r )

r

y 1

EXAMPLES "METHOD OF MOMENTS"
FOR DISCRETE DISTRIBUTIONS
Theoretical
Distribution
Theoretical
Expressions
Binomial:
Bin ( n, p )
E[Y ]  n  p


Var(Y )  n  p  (1  p )
Poisson:
Poisson (m)
 E[Y ]  m

Var (Y )  m
Geometric:
Geo( p )
1 p

 E[Y ]  p

1 p
Var(Y ) 
( p) 2

Parameter
Solutions
p
y
n
m y
p
1
y 1
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 135
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
Fitting Theoretical Distributions using quantile estimates
Y= Yearly bed cost in Half Way House
1. Given data: yi , I=1,…,n.
2. Order the data such that
y(1)  y( 2 )    y( n 1)  yn
3. Set:
1
n
2
p2  Pr(Y  y ( 2 ) ) 
n
p1  Pr(Y  y (1) ) 

n 1
n
1
pn  Pr(Y  y ( n ) ) 
n
pn 1  Pr(Y  y ( n 1) ) 
7. Fit a Theoretical Probability Model with CDF F ( y | 1 , 2 )
by selecting parameters (1 , 2 ) such that
 F ( y
n
i 1
|  1 , 2 )  pi 
2
(i )
is minimized.
Note:
 Above procedure requires the use of numerical algorithms
to calculate the parameters (1 , 2 ) .
 Software BESTFIT not only determines optimal
parameters but also test multiple theoretical distributions.
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 136
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
HALFWAY HOUSE EXAMPLE USING BESTFIT
Comparison of Input Distribution and
Normal(3.80e+2,2.18e+2)
1.2
1
0.8
0.6
0.4
0.2
0
1.0444 2.0931 3.1419 4.1906 5.2394 6.2881 7.3369 8.3856
User Input
Normal(3.80e+2,2.18e+2)
Comparison of Input Distribution and
Gamma(2.89,1.31e+2)
1.2
1
0.8
0.6
0.4
0.2
0
1.0444 2.0931 3.1419 4.1906 5.2394 6.2881 7.3369 8.3856
User Input
Gamma(2.89,1.31e+2)
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 137
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
Uncertainty about Parameters and Bayesian Updating
A. Discrete Case
B = {Killer in a Murder Case}
B  {B1, B2, B3}, where; B1 = Hunter, B2 = Near Sighted Man,
B3 = Sharp Shooter
After interrogations, interviews with witnesses, we are able
to establish the following prior distribution.
Pr(B= B1)=0.2, Pr(B= B2)=0.7, Pr(B= B3)=0.1.
Evidence A becomes available, being that the victim was
shot from 2000 ft. We establish the following probability
model.
Pr(A|B1)=0.7, Pr(A|B2)=0.1, Pr(A|B3)=0.9.
We update our prior distribution using the evidence into a
posterior distribution using Bayes Theorem.
Pr(A) = Pr(A|B1)Pr(B1)+ Pr(A|B2)Pr(B2)+ Pr(A|B3)Pr(B3)
= 0.70.2+0.10.7+0.90.1=0.30
Pr( A | B1 ) Pr( B1 ) 0.7  0.2

 0.47
Pr( A)
0.3
Pr( A | B2 ) Pr( B2 ) 0.1  0.7
Pr( B2 | A) 

 0.23
Pr( A)
0.3
Pr( A | B3 ) Pr( B3 ) 0.9  0.1
Pr( B3 | A) 

 0.30
Pr( A)
0.3
Pr( B1 | A) 
Conclusion:
Refocus investigation on Hunter and Sharp shooter.
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 138
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
Choose a theoretical
probability model,
P( X  k |  )
for the physical process
interest
Assess uncertainty about
parameter  by
specifying a prior
distribution f ( )
Uncertainty about X has two parts:
•Uncertainty due to the
process itself P ( X  k |  )
•Uncertainty about the
parameter  , though f ( )
Uncertainty about X can be
collapsed into one source by
applying Law of Total Probability
to calculate the prior predictive
distribution P ( X  k )
MODELING + PAST DATA + EXPERT JUDGEMENT
B. Continuous Case
Observe Data D1
Reassess uncertainty
parameter  by
using Bayes Theorem to
calculate posterior
distribution f ( | D1 )

•Uncertainty due to the
process itself P ( X  k |  )
•Uncertainty about the
parameter  , though f ( | D1 )
Uncertainty about X can be
collapsed into one source by
applying Law of Total Probability
to calculate the posterior predictive
distribution P ( X  k | D1 )
FUTURE DATA + ANALYSIS
Uncertainty about X has two parts:
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 139
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
Two calculations in above diagram have not been specified:
1. Calculating the Predictive Distribution
Probability Model: Pr( X  x |  ) , e.g. X Bin(N,p).
Prior distribution on  : f ( ) e.g. f(p) = Beta(n0,r0)
To calculate predictive distribution apply Law of Total
Probability for the continuous case:
Pr( X  x ) 
Pr( X  x |  ) f ( )d



SOFTE PRETZLE EXAMPLE CONTINUED
Y = # Customers out of N that buy your pretzle, Y Bin(N,p),
Where p is your market percentage. You are uncertain about
p and you decide to model your uncertainty using a Beta
distribution. p  Beta(n0,r0).
N 
( n0 )
 p r0 1 (1  p ) n0  r0 1 dp
Pr( Y  k | N )     p k  (1  p ) N  k
 ( r0 )   ( n 0  r0 )
k
0 
1
N
(n0 )
   
 p r0 k 1 (1  p) n0  N r0 k 1 dp
k  ( r0 )  (n0  r0 )
0
1
N
(n0 )
  
p r0 k 1 (1  p) n0  N r0 k 1 dp

 k  ( r0 )  (n0  r0 ) 0
1
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 140
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
But:
p r0 k 1 (1  p)n0  N r0 k 1
looks like a Beta(n0+N ,r0+k) distribution without the term
 ( n0  N )
( r0  k )  ( n0  N  r0  k )
Thus:
1
r k 1
n  N r k 1
 p 0 (1  p) 0 0 dp 
0
( r0  k )  (n0  N  r0  k )
(n0  N )
Finally:
N
(n0 )
( r0  k )  (n0  N  r0  k )
Pr(Y  k | N )   
(n0  N )
 k  ( r0 )  (n0  r0 )
Note:
 ( n )  ( n  1)! , n  1,2,3,...
In Soft Pretzel Example you decide to set f(p) = Beta(4,1) or
in other words n0,=4, r0=1. Thus,
 N  (4) (1  k )  (4  N  1  k )
Pr(Y  k | N )   
( 4  N )
 k  (1)  (3)
 N  3! k!( N  k  2)!
 N  k!( N  k  2)!
Pr(Y  k | N )   
 3    
( N  3)!
 k  1  2! ( N  3)!
k
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 141
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
Use Excel Spreadsheet to perform the Calculations
n0
r0
4
1
N
20
k
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Pr(Y=k|N)
0.1304
0.1186
0.1073
0.0966
0.0864
0.0768
0.0678
0.0593
0.0514
0.0440
0.0373
0.0311
0.0254
0.0203
0.0158
0.0119
0.0085
0.0056
0.0034
0.0017
0.0006
Pr(Y<=k|N)
0.1304
0.2490
0.3563
0.4529
0.5392
0.6160
0.6838
0.7431
0.7945
0.8385
0.8758
0.9068
0.9322
0.9526
0.9684
0.9802
0.9887
0.9944
0.9977
0.9994
1.0000
Conclusion:
E.g. Pr(Y>10|N) = 1- 0.8758=0.1242, you believe there is
approximately a 12.5% chance that you will sell more than
10 pretzels.
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 142
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
2. Calculating the Posterior Distribution
Probability Model: Pr( X  x |  ) , e.g. X Bin(N,p).
Prior distribution on  : f ( ) e.g. f(p) = Beta(n0,r0)
Observed data: D
To calculate the posterior distribution apply Bayes Theorem
for the continuous case:
f ( | D ) 
Pr( D |  ) f ( )
Pr( D |  ) f ( )d



Pr( D |  ) f ( )
Pr( D )

SOFTE PRETZLE EXAMPLE CONTINUED
Y = # Customers out of N that buy your pretzle, Y Bin(N,p),
Where p is your market percentage. You are uncertain about
p and you decide to model your uncertainty using a Beta
distribution. p  Beta(n0,r0).
f ( p) 
( n0 )
 p r0 1 (1  p ) n0 r0 1
( r0 )  ( n0  r0 )
Suppose you observe the following data: D = (N,k), i.e. k out
of N customers bought your pretzle. Then
f ( p | D) 
Pr( D | p ) f ( p )
1
 Pr( D | p) f ( p)dp

Pr( D | p ) f ( p )
Pr( D)
0
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 143
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making

f ( p) 
( n0 )
 p r0 1 (1  p ) n0 r0 1
( r0 )  ( n0  r0 )
N k
Pr(
D
|
p
)

Pr(
Y

k
|
N
,
p
)

  p (1  p ) N  k

k
 Pr( D )  Pr(Y  k | N ) , the predictive distribution that we
just calculated.
N k
( n0 )
 p r0 1 (1  p ) n0 r0 1
  p (1  p ) N k
k
( r0 )  ( n0  r0 )
f ( p | D)   
N
( n0 )
( r0  k )  ( n0  N  r0  k )
 
( n0  N )
 k  ( r0 )  ( n0  r0 )
N k
( n0 )
 p r0 1 (1  p ) n0  r0 1
  p (1  p ) N k
k
( r0 )  ( n0  r0 )
f ( p | D)   
N
( n0 )
( r0  k )  (n0  N  r0  k )
 
( n0  N )
 k  ( r0 )  (n0  r0 )
f ( p | D) 
( n 0  N )
p r0  k 1 (1  p ) n0  N  r0  k 1
( r0  k )  (n0  N  r0  k )
Conclusion:
The posterior distribution is ALSO a beta distribution but with
parameters (n0+N,r0+k).
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 144
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
Definition:
The prior distribution and the theoretical probability model
are such that the prior distribution and the posterior
distribution belong to the same family of distributions



Prior Distribution and Theoretical Probability Model are
Conjugate distributions.
SOFT PRETZLE EXAMPLE CONTINUED:
In Soft Pretzel Example you decide to set f(p) = Beta(4,1) as
your prior distribution on the market percentage p or in other
words n0,=4, r0=1. You observed that out of 20 potential
customers 7 bought your pretzle. Thus D=(20,7). Thus the
posterior distribution of the market percentage p is
f(p|D)= Beta(4+20,1+7)= Beta(24,8)
4.50
4.00
3.50
3.00
2.50
2.00
1.50
1.00
0.50
0.00
0.00
0.20
0.40
Prior: Beta(4,1)
0.60
0.80
1.00
Posterior: Beta(24,8)
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 145
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
Conclusion:
The good news is that with the observed data D=(20,7) I am
becoming more certain about my market percentage p.
However, what does this mean with respect to my
investment in a soft pretzle stand. Is there some bad news
too?
What about the posterior predictive distribution?
Recall:
In Soft Pretzel Example you decide to set f(p) = Beta(4,1) or
in other words n0,=4, r0=1. Thus,
N
( 4 )
(1  k )  (4  N  1  k )
Pr(Y  k | N )   
( 4  N )
 k  (1)  (4  1)
After observing data D:
f(p|D) = Beta(24,8) or in other words
N
(24)
(8  k )  (24  N  8  k )
Pr(Y  k | N , D )   
(24  N )
 k  (8)  (24  8)
 N  23! (k  7)!( N  k  15)!
Pr(Y  k | N )   
( N  23)!
 k  7!15!
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 146
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
n0+20
r0+7
24
8
N
20
k
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Pr(Y=k|N)
0.0034
0.0155
0.0389
0.0707
0.1032
0.1279
0.1386
0.1338
0.1165
0.0920
0.0662
0.0433
0.0257
0.0138
0.0066
0.0028
0.0010
0.0003
0.0001
0.0000
0.0000
Pr(Y<=k|N)
0.0034
0.0188
0.0577
0.1284
0.2316
0.3595
0.4981
0.6319
0.7483
0.8403
0.9065
0.9498
0.9756
0.9893
0.9959
0.9986
0.9996
0.9999
1.0000
1.0000
1.0000
Conclusion:
After observing the data Pr(Y>10|N, D) = 1- 0.9065=0.0935.
Thus your updated belief says there is approximately a 9.3%
chance that you will sell more than 10 pretzels. In addition,
you observe that your posterior has less uncertainty than the
prior. Hence you are becoming more certain that selling soft
pretzles may not be a good investment.
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 147
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
Conjugate Analysis for Normal Distributions
Predictive Analysis Table
Probability Model
(Y |  , ) 
Prior
 
Predictive
(Y |  ) 
Normal (  , )
Normal(m0 ,  0 )
Normal(m0 ,  2   0 )
2
Posterior Analysis Table
Probability Model
Prior
(Y |  , ) 
 
Posterior given
D = (n1 , y )
(  | D,  ) 
Normal (  , )
Normal(m0 ,  0 )
Normal(m* ,  * )
Where:
 2 / n1
 02
m  2
m0  2
y
2
2
 / n1   0
 / n1   0
*
 0 2 2 / n1
 
 2 / n1   0 2
*
Assignment:
Study conjugate analysis for Halfway House Example on
pages 392-396.
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 148
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
Question 10.19
A comptroller was preparing to analyze the distribution of
balances in the various accounts receivable for her firm. She
knew from studies in previous years that the distribution
would be normal with a standard deviation of $1500, but she
was unsure of the mean value  She thought carefully
about her uncertainty about this parameter and assessed a
normal distribution for  with mean m0 = 10,000, 0 = 800.
Over lunch, she discussed this problem with her friend, who
also worked in the accounting division. Her friend
commented that she also was unsure of  but would have
place it somewhat higher. The friend said that “better”
estimates for m0 and 0 would have been $12000 and $750,
respectively.
Define:
Y:= Balance in Accounts Receivable
Then:
(Y |  , )  Normal (  , ) ,  = $1500
a. Find P(  $11000) for both prior distributions.
For the Comptroller:
PN( > 11,000 | m0 = 10,000, 0 = 800)
= P(Z > (11,000-10,000)/800) = P(Z > 1.25) = 0.1056.
For her friend,
PN( > 11,000 | m0 = 12,000, 0 = 750)
= P(Z > (11,000-12,000)/750) = P(Z > -1.33) = 0.9082.
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 149
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
b. That afternoon, the comptroller randomly chose nine
accounts and calculated x =$11,003. Find the posterior
distributions of  Find P(  $11000) for both posterior
distributions.
For the Comptroller:
 15002 
10,000 
 11,003 8002 

 9 
m* 
 10,721
15002
 8002
9
15002
(8002 )
9
* 
 424
15002
2
 800
9
Thus,
PN( > 11,000 | m* = 10,721, * = 424)
= P(Z > (11,000-10,721)/424) = P(Z > 0.66) = 0.2546.
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 150
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
For the friend:
 15002 
12,000 
 11,003  7502 

 9 
m* 
 11,310
15002
 7502
9
15002
(7502 )
9
* 
 416
15002
 7502
9
Thus,
PN( > 11,000 | m* = 11,310, * = 416)
= P(Z > (11,000-11,310)/416) = P(Z > -0.75) = 0.7734.
c. A week later the analysis had been completed. Of a
total of 144 accounts (including the nine reported in part
b), the average was x =$11,254. Find the posterior
distributions of  Find P(  $11000) for both posterior
distributions.
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 151
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
For the Comptroller:
 15002 
10,000 
 11, 254 8002 

 144 
m** 
 11, 224
15002
 8002
144
15002
(8002 )
 *  1442
 123.5
1500
 8002
144
Thus,
PN( > 11,000 | m** = 11,224, ** = 123.5)
= P(Z > (11,000-11,224)/123.5) = P(Z > -1.81) = 0.9649.
For the friend:
 15002 
12,000 
 11, 254  7502 

 144 
m** 
 11, 274
15002
 7502
144
15002
(7502 )
 **  1442
 123.3
1500
 7502
144
Thus, PN( > 11,000 | m** = 11,274, ** = 123.3)
= P(Z > (11,000-11,274)/123.3) = P(Z > -2.22) = 0.9868
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 152
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen
EMGT 269 - Elements of Problem Solving and Decision Making
d. Discuss your answers to parts a,b, and c. What can you
conclude?
Eventually the data overwhelm any prior information. In the
limit, as more data are collected, the comptroller and her
friend will end up with the same posterior distribution.
Lecture notes by: Dr. J. Rene van Dorp
Session 9 - Page 153
Source: Making Hard Decisions, An Introduction to Decision Analysis by R.T. Clemen