DECISION ANALYSIS WITH SAMPLE INFORMATION
In the previous section, we saw how probability information about the states of nature affects the
expected value calculations and therefore the decision recommendation. To make the best possible
decision, the decision maker sometimes cannot simply rely on preliminary or prior probabilities. He may
need to seek additional information about the states of nature. This new information may be used to revise
or update the prior probabilities so that more accurate probability estimates are obtained in order to make
the final decision.
Additional information is usually obtained by performing designed experiments. Examples of
methods that are used to update the state-of-nature probabilities are raw material sampling, product
testing and market research. Revised probabilities are known as posterior probabilities (as opposed to
prior). Information obtained through research or experimentation is referred to as an indicator. The
experiment consists of taking a statistical sample so that the new information is known as sample
information.
Let I 1 = favourable market research report and
I 2 = unfavourable market research report
We denote the probability of a state of nature s j , given an indicator I k , as P ( s j | I k ) . This is the
conditional probability that s j will occur given that the outcome of the market research study is indicator
Ik .
Since test marketing and expert opinions are never 100% accurate, we cannot say, for example,
that, if the market is favourable, then the revised estimate of the probability of buying an expensive
system is 1. Test marketing is prone to sampling error and experts are not always right. This inability of
indicators to predict the correct states of nature can be quantified by using the conditional probabilities
P ( I k | s j ) . For example, given that the state of nature turns out to be favourable, what is the probability
that the market research will result in a favourable report? To carry out the analysis, we need conditional
probabilities for all indicators, given all the states of nature, that is, P ( I 1 | s1 ) , P ( I 2 | s1 ) , P ( I 1 | s 2 ) and
P( I 2 | s 2 ) .
Let us assume that the following estimates are available for the conditional probabilities:
Market research
State of nature
Favourable market ( s1 )
Favourable ( I 1 )
P ( I 1 | s1 ) = 0.90
Unfavourable ( I 2 )
P ( I 2 | s1 ) = 0.10
Unfavourable market ( s 2 )
P ( I 1 | s 2 ) = 0.25
P ( I 2 | s 2 ) = 0.75
Note that these estimates provide a reasonable degree of confidence in the market research study.
The reason for a high figure of 0.25 is that, when potential customers hear about using an expensive and
sophisticated system, their enthusiasm may lead them to overstate their real interest in it. However, when
they hear about the cost of using such a system, their responses may change very quickly to a ‘No, thank
you!’.
Once the indicator reliabilities are determined, the posterior probabilities can be found by using
Bayes’ Theorem:
P( A | B) =
P( A ∩ B)
P( B)
NOTE
There is a direct relationship between P ( I j | s k ) and P ( I k | s j ) . From definition,
P( I k | s k ) =
P{I k ∩ s j )
P( s j )
and P ( s j | I k ) =
P{s j ∩ I k )
P( I k )
.
The numerators of the right-hand-side expressions being the same, we can write
P( I k | s j ) P( s j ) = P( s j | I k ) P( I k ) .
We can now write the prior and posterior probabilities concerning I 1 for Carl and Betty’s
problem in the following table:
Probabilities
State of nature ( s j )
Prior P ( s j )
Conditional P ( I 1 | s j )
Favourable market ( s1 )
0.4
0.90
Unfavourable market ( s 2 )
0.6
0.25
From these, we can find the probabilities based on indicator I 1 :
P ( s1 ∩ I 1 ) = P( I 1 | s1 ) P ( s1 ) = (0.90)(0.4) = 0.36
P ( s 2 ∩ I 1 ) = P( I 1 | s 2 ) P ( s 2 ) = (0.25)(0.6) = 0.15
and also the probability of indicator I 1 :
P ( I 1 ) = P ( I 1 ∩ s1 ) + P ( s 2 ∩ I 1 ) = 0.36 + 0.15 = 0.51 .
Now, we can also find the posterior probabilities P ( I 1 | s1 ) and P ( I 1 | s 2 ) :
P{s1 ∩ I 1 ) 0.36
=
= 0.7059 and
P( I 1 )
0.51
P{s 2 ∩ I 1 ) 0.15
P(s 2 | I 1 ) =
=
= 0.2941 .
P( I 1 )
0.51
P ( s1 | I 1 ) =
All these probabilities can now be summarise in the following table:
Probabilities
Conditional
Joint
States of nature
Prior
Posterior
sj
P(s j )
P( I 1 | s j )
P( I 1 ∩ s j )
P(s j | I 1 )
s1
s2
0.4
0.90
0.36
0.7059
0.6
0.25
0.15
0.2941
P ( I 1 ) = 0.51
The same can be done for I 2 :
Probabilities
Conditional
Joint
States of nature
Prior
Posterior
sj
P(s j )
P( I 2 | s j )
P( I 2 ∩ s j )
P(s j | I 2 )
s1
s2
0.4
0.10
0.04
0.0816
0.6
0.75
0.45
0.9184
P ( I 2 ) = 0.49
Example
Developing a small driving range for golfers of all abilities has long been a desire of John
Jenkins. John, however, believes that the chance of a successful driving range is only about 40%. A friend
of John has suggested that he conduct a survey in the community to get a better feeling of demand for
such a facility. There is a probability of 0.9 that the research will be favourable if the driving range
facility will be successful. Furthermore, it is estimated that there is a probability of 0.8 that the market
will be unfavourable if indeed the facility will be unsuccessful. John would like to determine the chances
of a successful driving range given a favourable result, from the marketing survey.
Solution
Market research
State of nature
Favourable ( I 1 )
Unfavourable ( I 2 )
Driving range successful ( s1 )
Driving range unsuccessful ( s 2 )
P ( I 1 | s1 ) = 0.9
P ( I 1 | s 2 ) = 0.2
P ( I 2 | s1 ) = 0.1
P ( I 2 | s 2 ) = 0.8
Probabilities
Conditional
Joint
States of nature
Prior
sj
P(s j )
P( I 1 | s j )
P( I 1 ∩ s j )
P(s j | I 1 )
s1
s2
0.4
0.9
0.36
0.75
0.6
0.2
0.12
0.25
P( I 1 ) = 0.48
Posterior
DEVELOPING A DECISION STRATEGY
A decision strategy is a rule to be followed by the decision maker. This rule recommends a
particular decision based on whether the market research report is favourable or unfavourable. We may
now use a decision tree to find the optimal decision strategy for Carl and Betty’s problem.
From left to right, the tree shows the natural or logical order of the decision-making process.
First, they must decide whether or not to have a market research study done and, after having obtained the
market research indicators, they must decide on which alternative to select and, finally, the state of nature
will occur.
The payoffs on the decision tree when no market research is done are the same as those that we
had originally in Table 2.1.2. However, on the branches stemming from the decision to do the market
research, the cost of doing the research must be taken into account (here, we are going to assume that its
cost is R 1 000). This means that all the payoffs related to market research will be decreased by a
thousand rands.
Nodes 1, 3, 4 and 5 are decision nodes whereas nodes 6 to 14 are state-of-nature nodes. The
probabilities for these branches must be found before a decision tree analysis can be carried out. These are
posterior probabilities. Node 2 is known as an indicator node. Since the indicator branches are not under
the control of the decision maker, but are determined by chance, the node is represented by a circle just
like the state-of-nature node.
Once the branch probabilities have been found, the expected value approach can be used to
determine the optimal decision strategy. We then work backward through the decision tree and calculate
the expected value at each state-of-nature node. The expected values are given in the tree below.
We conclude that, if the market research report is favourable, Carl and Betty should buy an
expensive system (expected value of R 3 706.20) and that, if the market research report is unfavourable,
they should not buy any system at all (expected value of –R 1 000).
Procedure for determining the optimal decision strategy
1.
Draw the decision tree consisting of decision, indicator and state-of-nature nodes and
branches that describe the sequential nature of the problem.
2.
Calculate posterior probabilities to establish indicator and state-of-nature branch
probabilities.
3.
Work backwards through the decision tree by computing expected values at state-ofnature and indicator nodes. Hence determine an optimal decision strategy and its
associated expected value.
Market favourable (s1)
Expensive system (d1)
[3 706.2]
6
P ( s1 | I 1 ) = 0.7059
Market unfavourable (s2)
P ( s 2 | I 1 ) = 0.2941
Market favourable (s1)
Favourable report (I1)
P ( I 1 ) = 0.51
3
Less expensive system (d2)
[3 470.8]
7
[3 706.2]
P ( s1 | I 1 ) = 0.7059
Market unfavourable (s2)
P ( s 2 | I 1 ) = 0.2941
Market favourable (s1)
No system (d3)
[– 1 000]
[1 400]
8
P ( s1 | I 1 ) = 0.7059
Market unfavourable (s2)
P ( s 2 | I 1 ) = 0.2941
2
Market favourable (s1)
Do market research
Expensive system (d1)
[1 400]
[–7 531.2]
9
P ( s1 | I 2 ) = 0.0816
Market unfavourable (s2)
P ( s 2 | I 2 ) = 0.9184
Market favourable (s1)
Unfavourable report (I2)
P ( I 2 ) = 0.49
[– 1000]
4
Less expensive system (d2)
[– 4 020.8]
10
P ( s1 | I 2 ) = 0.0816
Market unfavourable (s2)
P ( s 2 | I 2 ) = 0.9184
Market favourable (s1)
1
No system (d3)
[– 1 000]
11
P ( s1 | I 2 ) = 0.0816
Market unfavourable (s2)
P ( s 2 | I 2 ) = 0.9184
Market favourable (s1)
Expensive system (d1)
[– 800]
12
P ( s1 ) = 0.4
Market unfavourable (s2)
P ( s 2 ) = 0.6
Market favourable (s1)
Do no market research
5
[800]
Less expensive system (d2)
[800]
13
P ( s1 ) = 0.4
Market unfavourable (s2)
P ( s 2 ) = 0.6
Market favourable (s1)
No system (d3)
[0]
14
P ( s1 ) = 0.4
Market unfavourable (s2)
P ( s 2 ) = 0.6
9 000
– 9 000
7 000
– 5 000
– 1 000
– 1 000
9 000
– 9 000
7 000
– 5 000
– 1 000
– 1 000
10 000
– 8 000
8 000
– 4 000
0
0
Expected value of sample information
From the above example, the expected value of the optimal decision strategy is R 1 400. This is
also known as the expected value with sample information (EVwSI). Initially, when no market research
was carried out, the expected value was R 800 – that value is the expected value without sample
information (EVwoSI).
The expected value of sample information (EVSI) is defined as the absolute value (modulus) of
the difference between EVwSI and EVwoSI, that is
EVSI = | EVwSI – EVwoSI |
Note
1.
2.
3.
The expected value without sample information (EVwoSI) is equal to the expected value
without perfect information (EVwoPI). This is simply the expected value of the best decision
based on prior probabilities.
It is very useful to determine the EVSI and EVPI in order to decide whether it is necessary to
but additional information. It should be clear that the EVPI is the upper bound for the amount
that one would be willing to pay for information.
The same equation handles both maximisation and minimisation problems since the absolute
value function is used.
Efficiency of sample information
Efficiency of sample information is measured as a percentage of perfect information (which is
obviously 100% efficient!). The formula is
E=
EVSI
× 100
EVPI
When the efficiency of sample information is high, we have almost perfect information so that
additional information would not yield significantly better results. However, low efficiency ratings would
lead the decision-maker to look for other types of information.