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Decision trees– representing thestructure ofadecisionproblem 95
6.2.2 Branch probabilities andoutcome probabilities
Another key step in the analysis of a decision tree is calculating the probability for each of the possible outcomes of the
decision. These are called the outcome probabilities. Methods for determining an outcome probability depend on know-
ing the probabilities for each of the branches encountered on the path leading to that outcome. These are called the
branch probabilities.
Recall that a branch originating from a chance node corresponds to one of the possible values for the uncertainty
represented by the node. Examples of chance node values include the presence of a disease, a test result, or the effect of
a treatment. The branch probability for a chance node value is expressed as a conditional probability, which was first
discussed back in Chapter4. A conditional probability measures the likelihood of an event when another event is known
or assumed to be true. For example, the test result sensitivity for a disease is the conditional probability measuring the
likelihood of the result if the disease is present. In this case, the presence of the disease is called the conditioning event.
A branch probability measures the likelihood of the corresponding disease, test result, or treatment outcome,
conditioned on the collection of decision node alternatives and chance node values encountered on the path leading to
that branch. In other words, that collection of decision node alternatives and chance node values constitutes the
conditioning event for the branch probability.
Because a branch probability depends on what happens earlier on the path leading to that branch, it is not surprising
that the outcome probabilities share those same dependencies. Examples discussed later in this chapter demonstrate
how an outcome probability is calculated. However, the defining property of an outcome probability is that it meas-
ures the likelihood of the outcome conditioned on what happens on the path leading to that outcome.
6.2.3 Expected value calculations andlife expectancy
The final step in any decision analysis is to compare what are called the expected values for each of the decision alterna-
tives. The assumption behind this final step is that the best decision alternative will have the greatest expected value.
The expected value is calculated for a decision alternative by multiplying the outcome probabilities and outcome
values for each of the possible outcomes and summing the results.
Expected value calculations require a notation that identifies the numerous outcome values and probabilities.
Suppose that a decision involves two alternatives, which will be denoted by A and B. Also, suppose that n possible
outcomes can result from the choice between A or B. We will note these possible outcomes by the subscripted variables
x
1
, ..., x
n
.The following notion will represent the outcome probabilities conditioned on which alternative ischosen:
Probability outcome is if decision is
P
robability
xAPx A
ii
|
outcome is if decision is
xB
Px B
ii
|
In other words, the decision alternative becomes the conditioning event for the outcome probability.
The value associated with outcome x
i
will be denoted by V(x
i
) and the expected value calculated for the two alterna-
tives will be denoted by EV(A) and EV(B). The calculations used to determine the expected values can then be expressed
as follows:
Expected value for AEVA Px AV xPxAVx Px
n
1122
AAV x
BEVB Px BV xPxBVx
n
E
xpected value for
1122
Px BV x
nn
The final step for analyzing this decision is to compare EV(A) and EV(B). Decision alternative A is preferred to alter-
native B if and only if EV(A) is greater than EV(B).
Of course, the validity of this approach depends on how the values for the outcome values are determined. As noted
earlier, in this chapter and the next, outcome is measured by the length of the patient’s life. This is called life expectancy
analysis. In later chapters, we will see an alternative outcome value called utility. Analysis based on utility is called
expected utility analysis.
Definition: branch probabilities andoutcome probabilities
Branch probability: The likelihood of the corresponding random variable value conditioned on all of the
decision alternatives and chance node values encountered earlier in the decision tree.
Outcome probability: The likelihood of a final outcome conditioned on the decisions leading to that outcome.
https://t.me/medicina_free
96 Medical decision making
As already suggested, life expectancy analysis is problematic because length of life is not the only concern when
making patient care decisions. However, life expectancy is commonly used for the outcome value, and it makes sense
when the goal is to maximize the patient’s life expectancy. Moreover, much of the mathematics used to analyze deci-
sions does not care how outcomes are measured. Therefore, life expectancy will provide a suitable outcome value for
the discussion in these two chapters.
6.3 Constructing thedecision tree fora hypothetical decision problem
This section uses a simple hypothetical decision problem to illustrate how the concepts described in the previous
section are represented in a decision tree. This sample problem will be used again in the next chapter to demonstrate
the analysis of decision trees. Therefore, we will give this problem a name– Problem 1. As mentioned in the previous
section, the outcomes for this decision problem will be represented by the patient’s corresponding length of life.
The initial decision in Problem 1 is whether to treat a patient suspected of having a possible disease. If the patient
has that disease, treatment will increase the length of life from 4 to 12 years. Without the disease, the treatment will
decrease the length of life from 20 to 16 years. Currently, the physician believes the probability of the disease is 0.40.
Atest is available that has a true- positive rate of 0.90 and a false- positive rate of 0.20. However, the test also has a
risk,which reduces the patient’s life by 1 year. Assume this reduction in the length of life is faced regardless of the test
outcome and the presence or absence of disease.
In summary, Problem 1has two decisions: (1) whether to test and (2) whether to treat. This problem also has three
uncertainties: (1) the disease state, (2) the test result, and (3) the disease state after the test result is learned. This third
uncertainty is an example of the interaction between decisions and uncertainties mentioned in the introduction to
thischapter.
Figure6.1 shows a decision tree for Problem 1. This book adopts the convention of drawing decision nodes as
squares and chance nodes as circles. The nodes are organized so that the initial decision is placed on the left side of the
diagram and the final outcomes are placed on the right. Often the other nodes are organized from left to right accord-
ing to the temporal order in which the decisions are made, and the uncertainties are resolved. However, we will see
that there is some flexibility in the ordering of the nodes.
Notice in Figure6.1 that the nodes have been given labels, enclosed in parenthesis. For example, A1 is the label for
the initial decision node in the tree. These labels will simplify the following narrative.
Let us examine each of the nodes in the decision tree. Figure6.2 focuses on the initial test decision. Node A1 repre-
sents the decision of whether to use the test. This node can be thought of as the root for the decision tree since all other
Expected value calculations andlife expectancy
Suppose the possible outcomes for alternative A are x
1
, ..., x
n
. Let V(x
i
) denote the outcome value for outcome x
i
and denote the outcome probabilities as follows:
P[x
i
∣ A]=Probability outcome isx
i
if the decision isA.
The expected value for alternative A, denoted EV(A), is computed as follows:
EV APxAVx Px AV xPxAVx
nn
1122
When V(x
i
) is the length of the patient’s life with outcome x
i
, EV(A) is called the life expectancy.
Definition: Problem 1
Decide whether to test and treat a patient who may have a disease
• Probability of disease is 0.40.
• With disease, treatment increases length of life from 4 to 12 years.
• Without disease, treatment decreases length of life from 20 to 16 years.
• Test has a true- positive rate of 0.90 and a false- positive rate of 0.20.
• Test reduces length of life by 1 year.
https://t.me/medicina_free
Decision trees– representing thestructure ofadecisionproblem 97
P[D
+
|No test, No treatment]
P[D
–
|No test, No treatment]
P[D
+
|Test
+
, No treatment]
P[D
–
|Test
+
, No treatment]
P[D
+
|Test
–
, No treatment]
P[D
–
|Test
–
, No treatment]
P[D
+
|Test
–
, Treatment]
P[D
–
|Test
–
, Treatment]
P[D
+
|Test
+
, Treatment]
P[D
–
|Test
+
, Treatment]
P[D
+
|No test, Treatment]
P[D
–
|No test, Treatment]
P[Test
+
]
P[Test
–
]
No treatment
Treatment
No treatment
Treatment
No treatment
Treatment
D
+
and no treatment
D
–
and no treatment
D
+
and treatment
D
–
and treatment
D
+
and no treatment and test
D
–
and no treatment and test
D
+
and no treatment and test
D
–
and no treatment and test
D
+
and treatment and test
D
–
and treatment and test
D
+
and treatment and test
D
–
and treatment and test
Test
No test
(A1)
(A2)
(A3)
(A4)
(A6)
(A7)
(A8)
(A9)
(A10)
(A11)
(A5)
Figure6.1 Decision tree for Problem 1, which has one treatment, one disease, and a test with two possible outcomes.
P[D
+
|No test, No treatment]
D
+
and no treatment
D
–
and no treatment
D
+
and treatment
D
–
and no treatment and test
D
+
and no treatment and test
D
–
and no treatment and test
D
+
and no treatment and test
D
–
and treatment and test
D
+
and treatment and test
D
–
and treatment and test
D
+
and treatment and test
D
–
and treatment
P[D
–
|Test
+
, No treatment]
P[D
+
|Test
+
, No treatment]
P[D
–
|Test
–
, No treatment]
P[D
+
|Test
–
, No treatment]
P[D
–
|Test
+
, Treatment]
P[D
+
|Test
+
, Treatment]
P[D
–
|Test
–
, Treatment]
P[D
+
|Test
–
, Treatment]
P[D
+
|No test, Treatment]
P[D
–
|No test, Treatment]
P[D
–
|No test, No treatment]
P[Test
+
]
P[Test
–
]
No treatment
No treatment
Treatment
Treatment
No treatment
Treatment
Test
No test
(A1)
(A2)
(A3)
(A4)
(A6)
(A7)
(A8)
(A9)
(A10)
(A11)
(A5)
P[Test
–
]
P[Test
+
]
No treatment
Treatment
Test
No test
(A1)
(A2)
(A5)
No treatment
Treatment
(A6)
No treatment
(A9)
Figure6.2 Portion of the decision tree in Figure6.1 showing the initial test decision.
https://t.me/medicina_free
98 Medical decision making
decisions and uncertainties follow this initial decision. There are two possibilities: (1) Notest and (2) Test. These pos-
sibilities are represented by the two branches originating at node A1.
More complicated problems can have decision nodes with more than two possibilities, in which case additional
branches would be used. However, recall that the essential feature of a decision node is that each of the emanating
branches must represent exactly one of the possible alternatives. In other words, the alternatives represented by any
two branches cannot overlap and all possible alternatives must each be represented by exactly one of the branches.
Figure6.3 focuses on the portion of the decision tree immediately following the decision to forgo the test. In Figure6.2,
this is the portion of the decision tree starting with the branch from node A1 to node A2. Node A2 represents the treat-
ment decision following the decision to forgo the test. Once again, Problem 1has been designed so that there are only
two possibilities: (1) Notreatment and (2) Treatment. A more complicated treatment decision might require additional
treatment alternatives. The only requirement is that each of the possible treatments must be represented by one and
only one of the branches originating from the decision node.
Figure6.4 focuses on the branch corresponding to the decision not to treat. Referring back to Figure6.3, this branch
connects the decision node A2 with chance node A3. Node A3 represents the uncertainty about the disease state. More
precisely, node A3 represents the uncertainty about the disease state if there is no treatment. There are two possibilities:
(1) the disease can be present, denoted D
+
, and (2) the disease can be absent, denoted D
−
.
As before, this hypothetical problem has been designed with only two possible disease states, which are indicated
by the pair of branches originating from node A3. As with decision nodes, the branches originating from a chance node
No treatment
P[D
+
]
P[D
+
]
P[D
–
]
P[D
–
]
Treatment
No test
(A2)
(A3)
(A4)
P[D
+
|No test, No treatment]
D
+
and no treatment
D
–
and no treatment
D
+
and treatment
D
–
and no treatment and test
D
+
and no treatment and test
D
–
and no treatment and test
D
+
and no treatment and test
D
–
and treatment and test
D
+
and treatment and test
D
–
and treatment and test
D
+
and treatment and test
D
–
and treatment
P[D
–
|Test
+
, No treatment]
P[D
+
|Test
+
, No treatment]
P[D
–
|Test
–
, No treatment]
P[D
+
|Test
–
, No treatment]
P[D
–
|Test
+
, Treatment]
P[D
+
|Test
+
, Treatment]
P[D
–
|Test
–
, Treatment]
P[D
+
|Test
–
, Treatment]
P[D
+
|No test, Treatment]
P[D
–
|No test, Treatment]
P[D
–
|No test, No treatment]
P[Test
+
]
P[Test
–
]
No treatment
No treatment
Treatment
Treatment
No treatment
Treatment
Test
No test
(A1)
(A2)
(A3)
(A4)
(A6)
(A7)
(A8)
(A9)
(A10)
(A11)
(A5)
Figure6.3 Portion of the decision tree in Figure6.1 showing the treatment decision node following the decision to forgo testing.
D
–
and no treatment and test
D
–
and no treatment and test
D
+
and no treatment and test
D
–
and treatment and test
D
+
and treatment and test
P[D
+
|No test, No treatment]
P[D
–
|No test, No treatment]
D
+
and no treatment
D
–
and no treatment
(A3)
P[D
+
|No test, No treatment]
D
+
and no treatment
D
–
and no treatment
D
+
and treatment
D
+
and no treatment and test
D
–
and treatment and test
D
+
and treatment and test
D
–
and treatment
P[D
–
|Test
+
, No treatment]
P[D
+
|Test
+
, No treatment]
P[D
–
|Test
–
, No treatment]
P[D
+
|Test
–
, No treatment]
P[D
–
|Test
+
, Treatment]
P[D
+
|Test
+
, Treatment]
P[D
–
|Test
–
, Treatment]
P[D
+
|Test
–
, Treatment]
P[D
+
|No test, Treatment]
P[D
–
|No test, Treatment]
P[D
–
|No test, No treatment]
P[Test
+
]
P[Test
–
]
No treatment
No treatment
Treatment
Treatment
No treatment
Treatment
Test
No test
(A1)
(A2)
(A3)
(A4)
(A6)
(A7)
(A8)
(A9)
(A10)
(A11)
(A5)
Figure6.4 Portion of the decision tree in Figure6.1 showing the disease state chance node following the decisions to forgo testing and treatment.
https://t.me/medicina_free
Decision trees– representing thestructure ofadecisionproblem 99
must correspond to a collection of mutually distinct and collectively exhaustive possibilities. The probabilities placed
on the branches originating from a chance node– the branch probabilities– provide a test of this requirement. Each of
these probabilities quantifies the likelihood that the possibility represented by the corresponding branch is true.
Therefore, the probabilities for all branches originating from a chance node must sum to one. Not all possibilities have
been represented if the sum is less than one. Conversely, more than one of the possibilities can be true, at the same time,
if the sum is greater than one.
Determining the values for the branch probabilities often is the most challenging step in developing a decision tree.
For this simple problem, the probability of disease is 0.40. That is
PD
040.
Recall that the probabilities for the branches originating from a chance node are conditioned on the nodes and
branches encountered earlier in the decision tree. Therefore, the probability for the upper branch shown in Figure6.4
quantifies the possibility that disease is present if there is no treatment and the test has not been performed. This con-
ditional probability is denoted
P
DNotestNotreatment||
,
.
However, in Problem 1, treatment– or lack of treatment– does not affect the disease state that was present before
the treatment decision is made. Nor does the decision to forgo the test. Once again, using the concept of independence
introduced in Chapter3, the disease state is independent of the test decision and the treatment decision. This means
P DP
D
|NotestNotreatment
,.
040
Similarly
P DP
DPD
|NotestNotreatment
,.10
60
Finally, notice that the branches originating from node A3 each conclude with one of two final outcomes. Recall that
node A3 represents the uncertainty about the disease state if there is no treatment. Therefore, those two possible final
outcomes are (1) disease present and no treatment and (2) disease absent and no treatment. From the definition of
Problem 1, the corresponding lengths of life are 2 and 10 years, respectively.
Figure6.5 focuses on the portion of the decision tree representing what happens if treatment is selected after decid-
ing to forgo the test. Note that, the portion of the decision tree following chance node A4 is almost identical to the
structure after chance node A3, as shown in Figure6.4. The only difference is how the treatment decision affects the
lengths of life for the final outcomes. From the problem statement, the patient’s life will last 6 years when disease is
present, and the treatment is chosen. The patient’s life will last 8 years when disease is absent, and the treatment is
chosen.
D
–
and no treatment and test
D
–
and no treatment and test
D
+
and no treatment and test
D
–
and treatment and test
D
+
and treatment and test
P[D
+
|No test, Treatment]
P[D
–
|No test, Treatment]
D
+
and treatment
D
–
and treatment
(A4)
P[D
+
|No test, No treatment]
D
+
and no treatment
D
–
and no treatment
D
+
and treatment
D
+
and no treatment and test
D
–
and treatment and test
D
+
and treatment and test
D
–
and treatment
P[D
–
|Test
+
, No treatment]
P[D
+
|Test
+
, No treatment]
P[D
–
|Test
–
, No treatment]
P[D
+
|Test
–
, No treatment]
P[D
–
|Test
+
, Treatment]
P[D
+
|Test
+
, Treatment]
P[D
–
|Test
–
, Treatment]
P[D
+
|Test
–
, Treatment]
P[D
+
|No test, Treatment]
P[D
–
|No test, Treatment]
P[D
–
|No test, No treatment]
P[Test
+
]
P[Test
–
]
No treatment
No treatment
Treatment
Treatment
No treatment
Treatment
Test
No test
(A1)
(A2)
(A3)
(A4)
(A6)
(A7)
(A8)
(A9)
(A10)
(A11)
(A5)
Figure6.5 Portion of the decision tree in Figure6.1 showing the disease state chance node following the decisions to treat after forgoing the test.
https://t.me/medicina_free
100 Medical decision making
Figure6.6 focuses on the portion of the decision representing what happens if the initial decision is to perform the
test. This decision leads to node A5, which represents the uncertainty about the test result. In Problem 1, there are only
two possibilities: (1) a positive test result, denoted Test
+
, and (2) a negative test result, denoted Test
−
.
Determining the corresponding branch probabilities requires some thought. Consider the probability that the test
result is positive. This can happen in two ways. First, the test result can be positive when the disease is present– a true
positive. Second, the test result can be positive when the disease is absent– a false positive. Together these two possi-
bilities are the only way a test can be positive. Moreover, only one of these possibilities can occur at the same time.
Therefore, as was explained back in Chapter3,
P
PD
PD
Test Test andTestand
Recall the definition of conditional probability
P
D
PD
PD
Test
Test and
|
This expression can be rearranged as
P
DP DPDTest andTest
|
The term P[Test
+
∣D
+
] is the true- positive rate, or sensitivity, for the test, which is given as 0.90. As stated in the defini-
tion of the problem, the term P[D
+
] is 0.40. So
P
DTest and
090040 03
6
.. .
Similarly,
P
DP DPDTest andTest
|
The term P[Test
+
∣ D
−
] is the false- positive rate, or one minus the specificity, which is given as 0.20. By definition,
P[D
−
] is one minus P[D
−
]. So
P
DTest and
020060 01
2
.. .
No treatment
Treatment
Treatment
No treatment
(A6)
(A9)
(A5)
Test
(A1)
P[Test
+
]
P[Test
–
]
D
–
and no treatment and test
D
–
and no treatment and test
D
+
and no treatment and test
D
–
and treatment and test
D
+
and treatment and test
P[D
+
|No test, No treatment]
D
+
and no treatment
D
–
and no treatment
D
+
and treatment
D
+
and no treatment and test
D
–
and treatment and test
D
+
and treatment and test
D
–
and treatment
P[D
–
|Test
+
, No treatment]
P[D
+
|Test
+
, No treatment]
P[D
–
|Test
–
, No treatment]
P[D
+
|Test
–
, No treatment]
P[D
–
|Test
+
, Treatment]
P[D
+
|Test
+
, Treatment]
P[D
–
|Test
–
, Treatment]
P[D
+
|Test
–
, Treatment]
P[D
+
|No test, Treatment]
P[D
–
|No test, Treatment]
P[D
–
|No test, No treatment]
P[Test
+
]
P[Test
–
]
No treatment
No treatment
Treatment
Treatment
No treatment
Treatment
Test
No test
(A1)
(A2)
(A3)
(A4)
(A6)
(A7)
(A8)
(A9)
(A10)
(A11)
(A5)
(A1)
(A2)
(A3)
(A4)
(A6)
(A7)
(A8)
(A9)
(A10)
(A11)
(A5)
Figure6.6 Portion of the decision tree in Figure6.1 showing test result chance node following the decision to perform the test.
https://t.me/medicina_free
Decision trees– representing thestructure ofadecisionproblem 101
Combining these two results
P
Test
036012 04
8
...
Since the test result must be either positive or negative
P
PTest Test
1052
.
Figure6.7 focuses on the portion of the decision tree showing the treatment decision following a positive test result.
Notice that the structure for this portion of the tree is the same as for the treatment decision when the test is not
performed (see Figures6.3, 6.4, and6.5). The only differences are in the branch probabilities and the outcome values.
First, the probability of disease is changed because of the positive test result. Second, the lengths of life for the final
outcomes will be reduced by 1 year because of the risk associated with the test in this hypothetical problem.
As noted earlier, in Problem 1, the treatment does not affect the probability of disease. Therefore, the branch
probabilities can be simplified by noting that
P DP
D
||
Test andNotreatment Test
and
P DP
D
||
Test andNotreatment Test
Similarly,
P DP
D
||
Test andTreatment Test
and
P DP
D
Test andTreatment Test
(A9)
No treatment
Treatment
(A6)
(A7)
(A8)
D
+
and no treatment and test
D
–
and no treatment and test
D
+
and treatment and test
D
–
and treatment and test
P[Test
+
]
P[D
+
|Test
+
, No treatment]
P[D
–
|Test
+
, No treatment]
P[D
+
|Test
+
, Treatment]
P[D
–
|Test
+
, Treatment]
D
–
and no treatment and test
D
–
and no treatment and test
D
+
and no treatment and test
D
–
and treatment and test
D
+
and treatment and test
P[D
+
|No test, No treatment]
D
+
and no treatment
D
–
and no treatment
D
+
and treatment
D
+
and no treatment and test
D
–
and treatment
P[D
–
|Test
+
, No treatment]
P[D
+
|Test
+
, No treatment]
P[D
–
|Test
–
, No treatment]
P[D
+
|Test
–
, No treatment]
P[D
–
|Test
+
, Treatment]
P[D
+
|Test
+
, Treatment]
P[D
+
|No test, Treatment]
P[D
–
|No test, Treatment]
P[D
–
|No test, No treatment]
P[Test
+
]
P[Test
–
]
No treatment
No treatment
Treatment
Treatment
No treatment
Test
No test
(A1)
(A2)
(A3)
(A4)
(A6)
(A7)
(A8)
(A10)
(A5)
Figure6.7 Portion of the decision tree in Figure6.1 showing the treatment decision following a positive test result.
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102 Medical decision making
Bayes’ theorem, which was described in Chapter4, computes the new probability of disease after a positive test
result.
P
D
PDPD
PDPD P
|Test
Test
Test
|
|
TTest
|DPD
090040
090040 020060
075
..
....
.
Since the disease must be either present (D
+
) or absent (D
−
)
P
DPD
Test Test
11
0750
25..
Also, with Problem 1, the effect of the test on the length of life for the final outcomes is determined by subtracting
1year from the length of life faced by the patient if the test was not performed. Therefore, without treatment, after
testing the length of life with or without the disease is 1 year or 9 years, respectively. Similarly, with treatment,
after
testing the length of life with or without the disease is 5 or 7 years, respectively.
Almost the same analysis applies to the treatment decision following a negative test result– node A9 in Figure6.1.
The lengths of life for the final outcomes are the same since, in this simple problem, the risk to the patient is assumed
to be the same for both test results. The only difference is how the negative test result affects the probability of disease.
Once again, Bayes’ theorem can be used to determine the new probability of disease.
0.40
0.60
0.40
0.60
0.75
0.25
0.75
0.25
0.08
0.92
0.08
0.92
0.48
0.52
No treatment
Treatment
No treatment
Treatment
No treatment
Treatment
Test
No test
P [Test
+
]
P [Test
–
]
(A1)
(A2)
(A3)
P[D
+
]
P [D
+
|Test
+
]
P [D
+
|Test
+
]
P [D
+
|Test
–
]
P [D
+
|Test
–
]
P [D
–
|Test
–
]
P [D
–
|Test
–
]
P [D
–
|Test
+
]
P [D
–
|Test
+
]
P[D
+
]
P[D
–
]
P[D
–
]
(A4)
(A6)
(A7)
(A8)
(A9)
(A10)
(A11)
(A5)
D
+
and no treatment (4 years)
D
+
and treatment (20 years)
D
–
and treatment (16 years)
D
–
and no treatment (12 years)
D
+
and no treatment and test (3 years)
D
–
and no treatment and test (11 years)
D
+
and treatment and test (19 years)
D
–
and treatment and test (15 years)
D
+
and treatment and test (19 years)
D
–
and treatment and test (15 years)
D
+
and no treatment and test (3 years)
D
–
and no treatment and test (11 years)
Figure6.8 Decision tree for sample problem, showing branch probabilities and the corresponding lengths of life for the final outcomes. The path
to the outcome disease present, test performed, and treatment selected has been highlighted for the decisions to perform the test and select the
treatment if the test result is positive.
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Decision trees– representing thestructure ofadecisionproblem 103
P
D
PDPD
PDPD P
Test
Test
Test
TTest
DPD
1
11
PDPD
PDPD P
Test
Test Tes
tt
DPD
1090 040
1090 0401020060
00
8
..
.. ..
.
In summary, Figure6.8 shows the decision tree for the sample problem with the branch probabilities and the length
of life for the final outcomes. The decision tree in Figure6.8 fully describes this simple problem.
Notice that the decision tree in Figure6.8 can be used to determine the outcome probabilities for a given set of deci-
sions. For example, suppose that decisions are to (1) perform the test and (2) start treatment if the test result is positive.
Assuming these two decisions, what is the probability that a patient with the disease will be both tested and treated?
The pathway through the tree for this outcome has been highlighted in Figure6.8. Since the patient will be tested and
treatment always follows a positive result, the probability of both being tested and treated is just the probability of a
positive result. However, the outcome in question also includes having the disease. Therefore, with this set of deci-
sions, the probability of the outcome in question is the probability of both a positive test result and the presence of
disease. From the definition of conditional probability
P
DP PDTest andTes
tT
est
Note thatP[Test
+
] is the probability for the branch from chance node A5 to decision node A6. Similarly, P[D
+
∣ Test
+
]
is the probability for the branch from chance node A8 to the outcome in question. In short, the probability of the path
from the initial decision node to a final outcome is the product of the branch probabilities encountered along the path.
This demonstration of how an outcome probability can be computed illustrates the fundamental requirement for
assigning probabilities to the branches comprising a path. Each of the successive branch probabilities is conditioned
onthe uncertainties and decisions that precede the branch on a path from the initial decision to a final outcome.
Often an uncertainty is independent of some of those preceding uncertainties and decisions. These independencies
can simplify the branch probabilities. For example, the probability of disease is independent of treatment. So
P
D
Test andTreatment becomes
P
D
Test .
6.4 Constructing thedecision tree fora medical decision problem
By using hypothetical tests, treatments, and diseases, Problem 1has avoided much of the challenge involved in devel-
oping the decision tree for a real clinical decision. The probabilities and life expectancies were simply stated without
any attempt at empirical justification. The discussion now turns to a more realistic clinical problem involving the man-
agement of coronary artery disease.
Coronary artery disease was selected for this first medical example because of the extensive body of published data
that will simplify the determination of the decision tree parameters. The introduction of decision analysis to the medi-
cal community coincided with important advancements in the diagnosis and treatment of coronary artery disease in
the 1960s and 1970s. Because of this coincidence, a large amount of data about test performance and patient outcomes
for coronary artery disease appeared in the medical literature. The discussion that follows uses that empirical founda-
tion to describe an analysis of an important medical problem. Some of the data that will be used is out of date. However,
the important principle that complex medical decisions can be analyzed will be illustrated by these examples.
In addition, some simplifying assumptions will be used to reduce the complexity of the example that will be
discussed. This means that what follows will not be a complete exploration of current coronary artery disease man-
agement. However, the simplified decisions described for the treatment of this important disease will illustrate the
process of going from published results to decision tree parameters.
6.4.1 Management ofcoronary artery disease overview
Many of the model parameters used in this discussion will be taken from the Coronary Artery Surgery Study (CASS)
(Taylor etal.,1989). This large multicenter study conducted a randomized clinical trial of the effects of surgery- based
therapies and drug- based therapies on the survival of patients with coronary artery disease.
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104 Medical decision making
The ACC/AHA Coronary Arteriography Guidelines (Scalon etal.,1999) list the evaluation of stable angina as a reason
for using arteriography. Management of stable coronary artery disease typically starts with drug-
based therapies.
Arteriography can show the anatomical findings needed for more aggressive treatment alternatives.
1
These include
coronary artery bypass graft surgery (CABG). Other alternatives, known collectively as percutaneous coronary interven-
tion (PCI), involve the placement of stents in partially blocked arteries that have been opened using angioplasty. The
development of drug-
eluting stents has resulted in types of PCI that approach the effectiveness of CABG while avoid-
ing the risks of open-
heart surgery.
Studies show that the effectiveness of the treatment alternatives varies with the location and extent of the narrowing
in the coronary arteries. The patient’s survival resulting from the choice of treatment also depends on other factors,
such as the presence of arrhythmias, conduction defects, and other damage to the heart. However, the examples in this
chapter condense this complex array of diagnostic endpoints into a pair of health states, denoted CAD and NOCAD.
Similarly, the treatment alternatives will be grouped (1) as drug- based therapies (MED), (2) more aggressive
arteriography-
based treatments (PCI/CABG), and (3) no treatment (NONE).
The specific patient used in this example is a hypothetical 65-
year- old woman, who will be called Ms. Maple. Suppose
that both a beta-
blocker and a statin drug have been prescribed for Ms. Maple. However, she dislikes the side effects
of these drugs and has not adhered well to this typical medical regimen. Moreover, her older sister recently died from
a heart attack. This raises Ms. Maple’s concerns about her diagnosis and increases the likelihood that she may face a
similar outcome. Therefore, Ms. Maple’s physician is considering coronary arteriography as a first step toward a more
aggressive therapy.
The outcomes in Problem 1were specific lengths of life, such as the patient will live 3 years. The example discussed
in this section will use a slightly more sophisticated view of how long the patient will live with a given outcome. In
most cases a medical decision results in a prognosis that can only make a probabilistic statement about how long the
patient will live. Chapter11 provides a more detailed discussion of what a probabilistic statement about a patient’s
survival means. For now, we will use a life expectancy to represent the uncertainty about how long a patient will live.
The goal will still be to find the decision that maximizes Ms. Maple’s life expectancy.
For example, based on actuarial data published by the US Social Security Administration, a 65-
year- old woman has
a life expectancy of 24.2 years. What does this mean? It does not mean that the woman will live exactly 24.2 years. The
woman could step off the curb in front of a car and die tomorrow. She also could get lucky, live for another 35 years,
and become a centenarian. Each possible length of life has a probability. As we saw in the previous section, the wom-
an’s life expectancy is the sum of each possible life length weighted by the corresponding probability.
In general, denoting life expectancy by LE, someone’s life expectancy, in years, is computed by the following
summation:
LE
PP P
Dienow yearsDie in year yearsDie in years0112
23
3
44
yearsDie in yearsyears
Dieinyears years
P
P
P Dieinyears years55
In theory, the summation should be extended forever. However, since no one appears to live longer than about
120years, for a 65- year- old woman, like Ms. Maples, the sum only needs to consider the next 55 terms. The probabili-
ties for the remaining terms are essentially zero.
When computing someone’s actuarial life expectancy, the probabilities are determined using the life tables pub-
lished by the Social Security Administration. Chapter11 will describe how to make those calculations. The result is an
age- specific life expectancy for a member of the general population.
Of course, Ms. Maple is not necessarily a “typical” member of the general population. She has a chest pain that her
physician believes is caused by coronary artery disease. She certainly has a clinical presentation that would justify the
risk and expense of arteriography. The CASS study reported the survival rates for patients found to not have coronary
artery disease based on the results of arteriography (Meyer etal.,1985). These published survival rates can be adjusted
according to age and sex to estimate a life expectancy of 19.2 years for a 65- year- old woman with a clinical presentation
that justifies arteriography (Fisher etal.,1982).
Therefore, the Social Security Administration life tables and the CASS imply different life expectancies for
Ms.Maple. How to proceed with this difference partly depends on how the analysis will be used. Suppose the goal is
1
The examples in this chapter ignore important alternatives to arteriography that would be considered for the evaluation of an actual
patient. Computerized tomography coronary arteriography can visualize blockages with the use of peripherally injected radiopaque dye,
thereby avoiding the need for catheterization.
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