Decision Science II
ORL 30306 — Wageningen University
Worked Solutions
Fully worked answers with step-by-step explanations for all exercises in
the companion exercise collection.
Part Topic Solutions
1 Bayesian updating (DA2/DA3) 1.1 – 1.5
2 Expected utility & certainty equivalents (Lecture 7) 2.1 – 2.3
3 Utility application: fertilizer decision (DA3) 3.1
4 Eliciting your own utility function (self-test) 4.1
5 Bayesian decision analysis: cattle purchase (DA12) 5.1
6 Portfolio theory (DA10) 6.1 – 6.2
7 Stochastic dominance (DA10) 7.1
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, Decision Science II · ORL 30306 Worked Solutions
Part 1 — Bayesian Updating: Solutions
Every exercise in this part uses the same three-step recipe: (1) write down the priors P(H) for
each hypothesis; (2) write down the likelihoods P(E | H) of the observed evidence under each
hypothesis; (3) multiply to get the joints P(E and H) = P(E | H)·P(H), and divide each joint by
their sum (the marginal P(E)) to get the posteriors.
Solution 1.1 — Medical test
Notation: D+/D− = diseased/healthy; T+/T− = positive/negative test.
Step 1 — Priors
P(D+) = 1/1000 = 0.001 P(D-) = 999/1000 = 0.999
Step 2 — Likelihoods
P(T+ | D+) = 0.99 (sensitivity) P(T- | D+) = 0.01
P(T+ | D-) = 0.02 (false positives) P(T- | D-) = 0.98
Step 3 — Joints and posterior
P(D+ and T+) = 0.001 x 0.99 = 0.00099
P(D- and T+) = 0.999 x 0.02 = 0.01998
P(T+) = 0.00099 + 0.01998 = 0.02097
P(D+ | T+) = 0..02097 = 0.0472
Answer: P(D+ | T+) ≈ 0.047, i.e. only about a 4.7% chance of actually having the
disease.
Intuition: the disease is so rare that among 100,000 people we expect about 99 true positives (99%
of 100 diseased) but roughly 1,998 false positives (2% of 99,900 healthy). Most positives are false
— the low prior dominates the accurate test.
Solution 1.2 — Iterated testing
The posterior of the first test becomes the prior of the second test — this is exactly what
'updating your degree of belief on the basis of new information' means.
(a) Second positive test — prior now 0.047
P(D+ and T+) = 0.047 x 0.99 = 0.04653
P(D- and T+) = 0.953 x 0.02 = 0.01906
P(D+ | T+) = 0.04653 / (0.04653 + 0.01906) = 0..06559 = 0.71
Answer: After two positive tests P(D+ | T+) ≈ 0.71.
(b) Third positive test — prior now 0.71
P(D+ and T+) = 0.71 x 0.99 = 0.7029
P(D- and T+) = 0.29 x 0.02 = 0.0058
P(D+ | T+) = 0.7029 / (0.7029 + 0.0058) = 0.99
Answer: After three positive tests P(D+ | T+) ≈ 0.99 — the evidence has
accumulated enough to make the disease almost certain.
Solution 1.3 — Spam filter
Step 1 — Prior
P(Spam) = 0.2 P(Not spam) = 0.8
Step 2 — Likelihoods
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