1. One-shot decision (no test)

Patient: 95 years old, lung tumor, 90% chance it’s malignant.

Given cancer vs no cancer, they give expected (quality-adjusted) survival under each treatment:

  • If no cancer → 34.8 months.
  • If cancer:
    • Radiotherapy → 16.7
    • Surgery → 20.3 if he survives surgery, but 35% chance of immediate death
    • No treatment → 5.6

Also: any treatment (radio or surgery) costs him 1 month of quality due to discomfort.

So compute expected quality-adjusted life for each option, using prior Pr(cancer) = 0.9.

Radiotherapy

$0.9 \times 16.7 + 0.1 \times 34.8 – 1 = 17.5 \text{ months}$

Surgery

  • 35%: dies immediately → 0
  • 65%: then apply the cancer/no-cancer mix, minus 1

$0.65 \times (0.9 \times 20.3 + 0.1 \times 34.8 – 1) = 13.5 \text{ months}$

No treatment

$0.9 \times 5.6 + 0.1 \times 34.8 = 8.5 \text{ months}$

So radiotherapy wins: 17.5 > 13.5 > 8.5.

If there were only this decision, we’d be done.

2. Add a test (second decision point)

Now introduce bronchoscopy:

  • If cancer: test is positive 70% of the time (sensitivity 0.70).
  • If no cancer: test is (falsely) positive 2% of the time (false positive 0.02).
  • The test itself can kill him: 5% fatal complication.

Question: is it worth doing the test first? To answer that, we must ask: after we see the test result, what treatment would we pick?

So we do Bayes to update Pr(cancer | test result).

2a. If test is positive:

Posterior

$\Pr(\text{cancer} \mid T{+}) = \frac{0.9 \times 0.7}{0.9 \times 0.7 + 0.1 \times 0.02} = 0.997 \text{ (≈ certain)}$

Then recompute the three options, but now with 0.997 instead of 0.9:

  • Radio: $0.997 \times 16.7 + 0.003 \times 34.8 – 1 = 15.8$
  • Surgery: $0.35 \times 0 + 0.65 \times (0.997 \times 20.3 + 0.003 \times 34.8 – 1) = 12.6$
  • No treatment: $0.997 \times 5.6 + 0.003 \times 34.8 = 5.7$

So if test is positive → still choose radiotherapy (15.8 is largest).

2b. If test is negative:

Posterior

$\Pr(\text{cancer} \mid T{-}) = \frac{0.9 \times 0.3}{0.9 \times 0.3 + 0.1 \times 0.98} = 0.734$

Recompute:

  • Radio: $0.734 \times 16.7 + 0.266 \times 34.8 – 1 = 20.5$
  • Surgery: $0.35 \times 0 + 0.65 \times (0.734 \times 20.3 + 0.266 \times 34.8 – 1) = 15.1$
  • No treatment: $0.734 \times 5.6 + 0.266 \times 34.8 = 13.4$

So even if test is negative → still choose radiotherapy (20.5 is largest).

This is the key structural observation: the test never changes the optimal treatment. Before seeing the test, radiotherapy is best; after a positive test, radiotherapy is best; after a negative test, radiotherapy is best.

When a test can’t change the downstream action, the test has no value (except it can hurt you).

3. Now, what is the expected value of “test first”?

We still finish the math to show it clearly.

First, probability of each test outcome:

  • Positive test: $0.9 \times 0.7 + 0.1 \times 0.02 = 0.632$
  • Negative test: $0.9 \times 0.3 + 0.1 \times 0.98 = 0.368$

(check: they add to 1)

Each outcome then gets multiplied by 0.95 because of the 5% chance the test kills him.

  • If positive: quality = $0.95 \times 15.8 = 15.0$ months
  • If negative: quality = $0.95 \times 20.5 = 19.5$ months

So total expected quality with “do test, then optimal treatment” is

$0.632 \times 15.0 + 0.368 \times 19.5 = 16.6 \text{ months}$

But “just do radiotherapy now” was 17.5 months.

So: testing makes him worse off.

4. What this example is teaching

  1. Multistage decisions = decision + uncertainty + decision → draw as a tree.
  2. In multistage decisions, Bayesian updating is the natural way to get “what we will believe later” (posterior after test).
  3. Value of information: a test is only valuable if it can actually change what you do. Here, both posteriors still lead to radiotherapy, so the test has zero (actually negative) value.
  4. You must also account for risks/costs of the test itself (here: 5% mortality).

So the “surprising” result is not an arithmetic trick; it’s a general decision-analytic principle: don’t buy information that can’t change your decision.