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A Beta(2,3) prior updated by five Bernoulli trials

Statement

Assume Countable Choice. With prior $P \sim Beta (2, 3)$ and $n = 5$ Binomial trials, an observed success count $y = 3$ has evidence $4 / 21$ and posterior $Beta (5, 5)$ . The boundary counts $y = 0$ and $y = 5$ give proper posteriors $Beta (2, 8)$ and $Beta (7, 3)$ .

Facts & Assumptions

Given: Countable Choice, the prior parameters $(2, 3)$ , trial count $5$ , and success count $3$ .

[A1]

Countable Choice is used by the Beta density normalization in the update theorem. The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F1]

For positive $α, β$ , the evidence is $(\binom{n}{y}) B (α + y, β + n - y) / B (α, β)$ and the posterior is Beta $(α + y, β + n - y)$ ; endpoint counts are allowed. Beta-Binomial conjugate update.

[F2]

For every natural number $r$ , $Γ (r + 1) = r!$ . $Γ (n + 1) = n!$ for every natural number $n$ .

[F3]

For $u, v > 0$ , $B (u, v) = Γ (u) Γ (v) / Γ (u + v)$ . The real Beta--Gamma identity.

Proof

Given: Countable Choice, $P \sim Beta (2, 3)$ , $n = 5$ , and $y = 3$ .

1.1A1F1F2F3algebra

By [F3] and [F2], $B (5, 5) = 4! 4! / 9! = 1 / 630$ and $B (2, 3) = 1! 2! / 4! = 1 / 12$ . Thus [F1] gives $m (3) = (\binom{5}{3}) \frac{B (5, 5)}{B (2, 3)} = 10 \frac{1 / 630}{1 / 12} = \frac{4}{21} .$ This number is finite and positive.

1.2A1F1givenalgebra

The posterior parameters in [F1] are $2 + 3 = 5$ and $3 + (5 - 3) = 5$ , so the posterior is $Beta (5, 5)$ .

2.1A1F1casesalgebra∎

At $y = 0$ the updated shapes are $(2, 8)$ ; at $y = 5$ they are $(7, 3)$ . Both pairs are positive, so [F1] gives finite-positive evidence and the stated proper endpoint posteriors.

Depends on

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Sources

Johnson, Ott & Dogucu, Bayes Rules!, Chapter 3, §3.1 (standard reference, not scraped)
Gelman et al., Bayesian Data Analysis, 3rd ed., §2.1 (standard reference, not scraped)