First, a quick retrieval check — different concept, new setting.
Two servers, in different data centers, each independently have a 3% chance of failing on any given day (Lesson 4’s independence: learning about one tells you nothing about the other). What’s the probability at least one of the two fails today? (Hint: it’s easier to find P(neither fails) first.)
(Worked out in the solution.)
Lesson 15 gave you the beta-binomial update rule: start with Beta(α, β), observe k successes out
of n trials, and the posterior is Beta(α+k, β+(n−k)) — addition of counts. Every worked example so
far updated on one clean batch of data at a time. Real evidence rarely arrives that tidily — it
trickles in across days, sources, and reports.
Setup. You start with prior Beta(2, 3) (a mild pull toward lower rates — 2 virtual successes, 3
virtual failures). Two batches of real evidence exist, and you don’t get to choose which one shows up
first:
- Batch A: 5 successes, 3 failures (8 trials)
- Batch B: 2 successes, 4 failures (6 trials)
Part 1 — Process A, then B. Starting from Beta(2,3), update on Batch A to get an intermediate
posterior. Treat that as your new prior, and update on Batch B. Write down the final Beta(α, β)
and its mean.
Part 2 — Process B, then A. Start over from Beta(2,3) again, but update on Batch B first, then
treat that result as your prior and update on Batch A. Write down this final Beta(α, β) and its
mean.
Part 3 — Compare (the numeric answer above). Are the two final posteriors identical? Give the shared posterior mean as a decimal. In one sentence, say why — what property of “just add counts” guarantees the order genuinely can’t matter here?
Part 4 — Break it. Order-independence quietly assumes something: that both batches are evidence
about the same, fixed underlying rate p. Describe a realistic scenario where Batch A and Batch B
come from a system whose true rate actually changed between the two batches — and explain why
“just add up all the counts, order doesn’t matter” would then be the wrong thing to do, even though
the arithmetic would run exactly the same way and produce a number that looks equally confident.