The document discusses using expected value in decision making and provides an example using a decision tree. It shows a scenario where an advertising budget can be small or large. If small, the expected profit is 46. If large, the expected profit is 68. Therefore, without sample information, the large budget should be chosen. The document then discusses taking a sample to get more information before making the decision.

1. 7/24/2012
If (A,B,C,D) has (multivariate)
Use of Expected Value in Decision Making Normal distribution with a
DECISION TREE variance covariance matrix
A B C D
A 2.000 1.061 0.490 0.022
B 1.061 1.000 0.043 0.158
C 0.490 0.043 0.750 0.068
D 0.022 0.158 0.068 0.100
Session IX
A Softwhite story
Calculate the chance of reaching under 16 hours in either route.
10
Case: A softwhite story
How to choose ‘right’ option
100
from a set of alternatives
Q. What should be the advertising budget?
Expected time = 10hr
Small or Large?
A
Exp. Time 5 hr
C B
Expected time = 12hr D Decision node
Expected time = 3.5hr
Chance node
10
.6
How to choose ‘right’ option So, if the advertising budget is taken to be small,
100 the expected profit would be
from a set of alternatives .4
0.6 × 10 + 0.4 × 100 = 46
N(10,2)
46 And, if the advertising budget is taken to be large,
N(5,1)
the expected profit would be
• Example: lottery N(12,0.75) 0.6 × (−20) + 0.4 × 200 = 68
68
N(3.5,0.1)
So, in absence of sample information,
• Utility function .6
the advertising budget should be large
.4
1

2. 7/24/2012
10 10
100 P[‘LOW’ market condition given
100
-20 that sample response is BAD] ?
200
Large budget
P[market condition is low] = 0.6
Take
P[BAD sample response given ‘LOW’ market condition]
sample
10 10
Binomial Distribution
100
100 low high
0.2 0.4 values
0.0115 0.0000 0 0.8042 0.1256 If the market condition is high
0.0576 0.0005 1
0.1369 0.0031 2
(40% of the users will buy SW),
L
0.2054 0.0123 3 what is the chance of 8 out of
0.2182 0.0350 4
0.1746 0.0746 5 the 20 randomly chosen HH
Take
0.1091 0.1244 6 0.1932 0.6297 purchasing SW?
0.0545 0.1659 7
0.0222 0.1797 8
0.0074 0.1597 9
20
sample 0.0020 0.1171 10 0.0026 0.2447 C 8 ×0.408 × 0.6012 = 0.1797
0.0005 0.0710 11
0.0001 0.0355 12
0.0000 0.0146 13
0.0000 0.0049 14
0.0000 0.0013 15
0.0000 0.0003 16
0.0000 0.0000 17
0.0000 0.0000 18
0.0000 0.0000 19
0.0000 0.0000 20
10
So now should compute expected profit .6? Probability through Venn Diagram
if sample information is collected and used
100
0.4 ×1256=0.05
Small budget
L
46 0.60×.8042=0.48
BAD 0.25
Take
AVERAGE
sample
0.10
68 0.60 ×0.1932=0.12
GOOD
Large budget
Low High
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3. 7/24/2012
EVPI, EVSI, efficiency
(posterior) probability distribution of
Market condition given the sample response • If one has perfect information, large advertising
budget will be used when the market condition
Sample Market condition is high and small budget when the market
response Low high condition is low.
Bad .48/.53 .05/.53 • Perfect information does not guarantee control
(about MC)
=.91 =.09
Average .12/.37 .25/.37 Expected profit with perfect information regarding market condition
=.32 =.68 0.6 × 10 + 0.4 × 200 = 86
good .02 .98 The expected value of perfect information (EVPI) = 86 – 68 = 18 lakh..
Sampling Efficiency = EVSI/EVPI = 9.44/18
=52.44%
10
So now should compute expected profit . 9057
if sample information is collected and used
100 Principles for solving
18.487 .0943
Small budget
Decision trees
• At any chance node compute expected gain
.3152
L
46
.5327
71.632 • At any decision node select the node which
.3152
Take .6848 gives the maximum expected gain among
77.44 .3678 available choices
sample 130.656
.6848 • Develop the solution from right to left
0.0994
68
196.546
Large budget
Summary of decisions
Inference problems: Estimation
• Should use the sample data
• Go for a large budget if the sample response is
average or good (at least 6 households out of 20 Insurance example
surveyed use SW)
• Expected value of sample information Bangalore Flat price
EVSI = 77.44 – 68 = 9.44 lakh
• Check that trimming the branches was ok Clear tone radios
Is Medworld cheating
3

4. 7/24/2012
A simple example: Overview
Parameter & Statistic
Population = all projects undertaken by a company
An unknown proportion π of them took longer than scheduled
• A characteristic of the • A characteristic of the
* population which is of
* Delay * sample (to estimate
* * * * interest in the study
** * the parameter)
* * • Fixed or Non-random
* * * • Random (because the
**
* * * * * sample is random)
• Unknown (because
typically you don’t have • Computable or
A random sample of n projects are selected. Y of them are found to be delayed.
information about all units known once you draw
(sample outcome) Y is random, but the randomness depends on π. of the population) the sample
Given the value of Y, one can make an objective inference about π.
Estimator/Estimate and
Some of the questions to be
its Bias, Standard Error and
answered
Sampling Distribution
• To what degree, the randomness in Y can be
• Value of the Estimator (statistic) for a given
attributed to sampling fluctuations? sample is your estimate
• How close is Y = p to π ? • Bias = Mean (expected value) of the Estimator
n
• If we want p to be within ±0.05 of π , how minus the parameter
many projects do we need to look at? • Standard Error = Standard deviation of the
Estimator
• Sampling Distribution is the probability
distribution of the Estimator
24
Simple Random Sampling
º º º × (SRS)
× º • Each unit in the population has equal chance of being
× ×× × included in the sample (even position-wise in the sample)
º ××
º º • SRS with replacement (SRSWR): unit already selected are
×× º returned before drawing subsequent ones. (Same unit may
× ×××
× × appear more than once). Not too realistic but most useful
º for theoretical treatment
N = 50 ×× º
º • SRS without replacement (SRSWOR):
n=5
º – same unit may not be included more than once
× º
– selections are not independent
º – if the population size is very large compared to sample size,
×
SRSWOR can be considered/approximated by SRSWR
Parameter (θ) = proportion of ‘×’ in the population
4

5. 7/24/2012
Unbiasedness and Standard error
of Sample Mean/Proportion
E( X ) = µ E ( p) = π
σ π (1 − π )
S .E.( X ) = S .E.( p ) =
n n
Estimated standard errors:
S p (1 − p )
S .E.( X ) = S .E.( p ) =
n n26
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