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Conjugate families and Gaussian conditioning — Examples

1 · Prerequisites

2 · Summary

The examples compute one Beta–Binomial evidence and its endpoint posteriors, show a zero-exposure Gamma–Poisson update and its impossible-data boundary, pool two Normal observations by precision, evaluate a three-category Dirichlet–Multinomial evidence exactly, and apply the Gaussian conditioner to a singular state covariance with a positive-definite observed block.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Gaussian conditioning with singular state covariance

Statement

Assume AC. Let $Z = (Z_{1}, Z_{2}) \sim N_{2} (0, I_{2})$ , where $I_{2}$ is the $2 \times 2$ identity matrix, and set $X = (Z_{1}, Z_{1})$ and $Y = Z_{1} + Z_{2}$ . The state covariance is singular, $Var (Y) = 2 > 0$ , and the conditional law at every $y \in R$ is $X ∣ Y = y \sim N_{2} ((y / 2, y / 2), (\begin{array}{cc} 1 / 2 & 1 / 2 \\ 1 / 2 & 1 / 2 \end{array})) .$

Facts & Assumptions

Given: AC, $Z \sim N_{2} (0, I_{2})$ , and the definitions of $X$ and $Y$ in the Statement.

[A1]

AC is assumed by the multivariate-normal construction and conditional-law theorem. The Axiom of Choice.

[F1]

Under AC, $N_{d} (m, Σ)$ exists for every positive-semidefinite covariance, has the stated mean and covariance, and is characterized by its Gaussian linear projections. Multivariate normal law, including singular covariance.

[F2]

If a jointly Gaussian pair has block covariance $(\begin{array}{cc} A & C \\ C^{T} & B \end{array})$ with $B$ positive definite, its conditional kernel has mean $a + C B^{- 1} (y - b)$ and covariance $A - C B^{- 1} C^{T}$ , which may be singular. Conditional law of a jointly Gaussian block with positive-definite observed covariance.

Verification

Given: AC and the Gaussian vector and affine definitions in the Statement, with facts [A1], [F1], [F2].

1.1A1F1givenalgebra

For any $u_{1}, u_{2}, v \in R$ , $u_{1} X_{1} + u_{2} X_{2} + v Y = (u_{1} + u_{2} + v) Z_{1} + v Z_{2}$ is Gaussian by [F1]. Its mean is zero. Computing covariances from $Cov (Z) = I_{2}$ gives $A = (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ , $C = (\begin{array}{c} 1 \\ 1 \end{array})$ , and $B = (2)$ . Since $\det A = 0$ , the state covariance is singular.

2.1step 1.1algebra

The observed block $B = (2)$ is positive definite. Its inverse is $(1 / 2)$ , so $A - C B^{- 1} C^{T} = (\begin{array}{cc} 1 / 2 & 1 / 2 \\ 1 / 2 & 1 / 2 \end{array})$ and $C B^{- 1} (y - 0) = (y / 2, y / 2)$ . The conditional covariance is also singular, with determinant zero.

3.1A1F2step 2.1given∎

Apply [F2] with these blocks. It gives the displayed conditional Gaussian kernel for every $y$ , despite singularity of $A$ and the conditional covariance; no inverse or pseudoinverse of either is used.

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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.

ExampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

Gamma-Poisson updating with a zero exposure

Statement

Assume Countable Choice. Take $λ \sim Gamma (2, 1)$ and two independent counts with exposures $(e_{1}, e_{2}) = (2, 0)$ . For observed counts $(y_{1}, y_{2}) = (3, 0)$ , the evidence is $32 / 243$ and the posterior is $Gamma (5, 3)$ . If the zero-exposure count were instead $y_{2} = 1$ , its evidence would be zero and no posterior quotient would be defined.

Facts & Assumptions

Given: Countable Choice, shape $a = 2$ , rate $b = 1$ , exposures $(2, 0)$ , and counts $(3, 0)$ .

[A1]

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

[F1]

For compatible zero-exposure data, the evidence is $c b^{a} Γ (a + S) / [Γ (a) (b + E)^{a + S}]$ and the posterior is Gamma $(a + S, b + E)$ ; an impossible zero-exposure count has zero evidence. Exposure-adjusted Gamma-Poisson conjugate update.

[F2]

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

Proof

Given: Countable Choice, the stated prior, exposures, and counts.

1.1A1F1F2algebra

Here $S = 3$ , $E = 2$ , and $c = 2^{3} / 3! = 4 / 3$ because the zero-exposure observation contributes factor one. By [F2], $Γ (5) = 24$ and $Γ (2) = 1$ . Substitution in [F1] gives $m (3, 0) = \frac{4}{3} \frac{Γ (5)}{Γ (2) 3^{5}} = \frac{4}{3} \frac{24}{243} = \frac{32}{243} .$

1.2A1F1givenalgebra

The updated shape is $2 + 3 = 5$ and the updated rate is $1 + 2 + 0 = 3$ , so [F1] gives $Gamma (5, 3)$ .

2.1A1F1cases∎

If $y_{2} = 1$ at exposure $e_{2} = 0$ , that observation is impossible under every $λ$ by [F1]. The likelihood is identically zero, its evidence is zero, and the finite-positive condition for a quotient fails.

ExampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

Normal precision pooling with two observations

Statement

Assume AC. Let the prior be $μ \sim N (0, 4)$ , and, conditionally on $μ$ , observe independent $Y_{1}, Y_{2} \sim N (μ, 1)$ . For data $(y_{1}, y_{2}) = (1, 3)$ , the evidence is finite and positive and the posterior is $N (16 / 9, 4 / 9)$ .

Facts & Assumptions

Given: AC, the stated Gaussian model, and observed data $(1, 3)$ .

[A1]

The Axiom of Choice is required by the Gaussian density normalization used in the update theorem. The Axiom of Choice.

[F1]

For a proper prior $N (μ_{0}, τ_{0}^{2})$ and $n \geq 1$ conditionally IID $N (μ, σ^{2})$ data, the evidence is finite positive and the posterior variance and mean are $V_{n} = (τ_{0}^{- 2} + n σ^{- 2})^{- 1}$ and $V_{n} (μ_{0} / τ_{0}^{2} + n \overset{ˉ}{y} / σ^{2})$ . Known-variance Normal-Normal conjugate update.

Verification

Given: The hypotheses and data in the Statement, with [A1] and [F1].

1.1givenalgebra

The sample mean is $\overset{ˉ}{y} = (1 + 3) / 2 = 2$ . The prior precision is $1 / 4$ and the two observations contribute total precision $2 / 1 = 2$ , so the posterior precision is $9 / 4$ .

2.1A1F1step 1.1algebra∎

Thus $V_{2} = (9 / 4)^{- 1} = 4 / 9$ , and the posterior mean is $V_{2} (0 / 4 + 2 \cdot 2 / 1) = (4 / 9) 4 = 16 / 9$ . Applying [F1] gives posterior $N (16 / 9, 4 / 9)$ and finite positive evidence.

ExampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-6-luna)Open item page →

A three-category Dirichlet-Multinomial update

Statement

Assume Countable Choice. With prior $p \sim Dirichlet (1, 2, 1)$ , sample size $n = 3$ , and observed counts $y = (2, 1, 0)$ , the evidence is $1 / 10$ and the posterior is $Dirichlet (3, 3, 1)$ .

Facts & Assumptions

Given: Countable Choice, the prior parameters $(1, 2, 1)$ , sample size $3$ , and counts $(2, 1, 0)$ .

[A1]

Countable Choice is used by the Dirichlet simplex normalization theorem. The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F1]

For positive Dirichlet parameters and admissible counts, the posterior adds the counts coordinatewise and the evidence is the Multinomial coefficient times the ratio of Dirichlet normalizers. Dirichlet-Multinomial conjugate update.

[F2]

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

Verification

Given: The hypotheses in the Statement and facts [A1], [F1], [F2].

1.1A1F1F2givenalgebra

Here $α_{+} = 4$ , $n = 3$ , and $3! / (2! 1! 0!) = 3$ . By [F2], $Γ (4) = 6$ , $Γ (1) = Γ (2) = 1$ , $Γ (3) = 2$ , and $Γ (7) = 720$ . Substitution into [F1] gives $m (y) = 3 \cdot 6 \cdot 4 / 720 = 1 / 10$ .

2.1A1F1givenalgebra∎

The updated coordinates are $(1 + 2, 2 + 1, 1 + 0) = (3, 3, 1)$ , all strictly positive, so [F1] gives the proper posterior $Dirichlet (3, 3, 1)$ . The zero count in the third category leaves its positive prior coordinate unchanged.

Conjugate families and Gaussian conditioning — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Gaussian conditioning with singular state covariance

Statement

Facts & Assumptions

Verification

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

Statement

Facts & Assumptions

Proof

Gamma-Poisson updating with a zero exposure

Statement

Facts & Assumptions

Proof

Normal precision pooling with two observations

Statement

Facts & Assumptions

Verification

A three-category Dirichlet-Multinomial update

Statement

Facts & Assumptions

Verification

Sources