Lesson 16 left you holding a posterior, Beta(9, 5), mean ≈ 0.643 — your current belief about the
true success rate p for some process (say, a coin, or a conversion rate). A natural next question:
if you had to predict actual future data, not just describe your belief about p, what would you
say?
Posterior predictive distribution: the probability of an observable future outcome, averaged
over everything the posterior currently believes about p. It’s the honest answer to “what do you
expect to see next?” — honest because it doesn’t pretend p is known; it accounts for your
remaining uncertainty about p itself.
The trap: the tempting shortcut is to plug in a single number — the posterior mean, 0.643 — and
treat it as if it were p, then compute ordinary probabilities from there. Under that shortcut,
“both of the next two trials succeed” would be estimated as 0.643 × 0.643 ≈ 0.413.
That shortcut is wrong, because it throws away the fact that p itself is still uncertain. The
honest calculation integrates over the whole posterior distribution for p, not just its mean.
For a beta posterior Beta(a, b), the exact probability that the next two independent trials are
both successes is:
P(next two both succeed) = [ a / (a+b) ] × [ (a+1) / (a+b+1) ]
(This comes from asking the question one trial at a time: the first trial succeeds with probability
equal to the posterior mean a/(a+b); given that success, the posterior itself updates to
Beta(a+1, b) — exactly like Lesson 15’s update rule — before you ask about the second trial. You’re
chaining two posterior means, not squaring one.)
Your task: using a = 9, b = 5, compute the exact probability that the next two trials are
both successes. Give your answer to three decimal places, and compare it to the naive
“square-the-mean” shortcut above.