Statistical Evidence Evaluation
Suggested solutions to Exercises 3
1.
Build an Influence Diagram according to the following
The utility is 10000 if the merchandise is “old age” and sold. The loss is 5000 (the fine) if the
merchandise is a reproduction and sold. If the merchandise is “younger age” and sold the
customer will get his money back and so there is zero utility and zero loss. This is also the
case for all states when the merchandise is not sold.
Instantiate the test outcome to “old age”:
Hence, the decision maximising the expected posterior utility is to sell the merchandise.
(Note that the information about reasonable selling prices if the merchandise is “younger age”
and reproduction resp. is irrelevant for this problem since the selling price would always be
10000 if sold.)
2.
There are some typos in the text, but most important the information is not sufficient. The
probability of 3% is of course for the screening device to indicate to high blood alcohol
content when this is not the case (false positive rate).
Taking this into account the following Influence Diagram may be constructed:
This is the node in which we can elaborate
with the probabilities. We can for example try
with
There is no loss if the driver is sent for blood sampling and the outcome of this shows a too
high level (due to the depreciation of the cost).There is also no loss if the driver is not sent for
blood sampling and they do not cause any accident.
Note that there is no direct link from the node “Blood alcohol content > 0.2 per mille” to the
node “Loss”. This is so since the loss is given by a combination of the decision made and the
potential consequences of having a blood alcohol content above 0.2 per mille and the time lag
from screening to blood sampling.
Now we should instantiate the node “Outcome from screening device” to state “Too_high”:
The decision minimising the expected posterior loss is then to send for blood sampling.
Now, we can try to decrease the probability that a driver with too high blood alcohol content
will cause an accident:
Again instantiating the node “Outcome from screening device” to state “Too_high” gives
Still the optimal decision with respect to minimised posterior expected loss is to send for
blood sampling.
Further trials will eventually give that not until the probability of an accident due to drunk
driving is close to 0.001 the optimal decision will change to “not send”. Such a low
probability is however not reasonable since otherwise we would not be that concerned with
drunk-driving.
3.
Let
H0: “Patient has a flu”
H1: “Patient has a bacterial infection
E1 = Observed symptoms (sore throat and fever)
E2 = Lack of signs of a bacterial infection in patient’s throat
And the possible decisions (actions) are
d1 = treatment for bacterial infection
d2 = no treatment
We are informed that
P(Ho ) = 80P(H1 )
P(E1 | H0 ) = 0.99
P(E1, E2| H1 ) = 0.5
A reasonable assumption is that P(E1, E2 | H0 )  0.99 since we can write
P(E1, E2 | H0 ) = P(E1| E2, H0 )P(E2 | H0 )
and lack of signs of a bacterial infection is to be considered almost certain when H0 is true. The
relevance of E2 for E1 may be explained away given that H0 is true, which thengives
P(E1| E2, H0 ) = P(E1| H0 )
Then Bayes’ theorem gives
PH 0 E1 , E2 
PH 1 E1 , E2 

PE1 , E2 H 0  PH 0  0.99


 80  158.4
PE1 , E2 H 1  PH 1  0.5
Loss function:
H0 true H1 true
treatment for bacterial infection
c
0
no treatment
0
2c
The expected posterior losses are
L treatment   c  PH 0 E1 , E 2   0  PH 1 E1 , E2   c  PH 0 E1 , E 2 
L no treatment   0  PH 0 E1 , E2   2c  PH 1 E1 , E 2   2c  PH 0 E1 , E2 
and
L treatment 
c PH 0 E1 , E2  1


  158.4  79.2
L no treatment  2c PH 1 E1 , E2  2
Hence, the expected posterior loss is 79.2 times higher with action “treatment” than with action ”no
treatment” so the action with lowest expected posterior loss is “no treatment”.