ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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The posterior mean need not be a small-ball MAP

Example

Let $θ$ have the uniform prior on $[0, 1]$ , regarded as a Borel probability on $R$ , and let an observation $Y \in {0, 1}$ satisfy $P (Y = 1 ∣ θ) = θ^{2} (1 - θ)$ . After observing $Y = 1$ , the posterior has density $12 θ^{2} (1 - θ)$ on $[0, 1]$ and zero outside. Its unique density mode, at $θ = 2 / 3$ , is a strong small-ball MAP, while its posterior mean is $3 / 5$ . Thus the posterior mean and a small-ball MAP differ.

Example references

the strong small-ball MAP is defined by small-ball probabilities relative to their global supremum

the fixed-data posterior normalizes likelihood-weighted prior mass

Facts & Assumptions

Given: The Borel probability prior that is uniform on $[0, 1]$ , the binary sampling rule with success probability $θ^{2} (1 - θ)$ , and observation $Y = 1$ .

[F1]

The posterior at a fixed datum is the likelihood-weighted prior divided by its evidence when the evidence is positive and finite. Misfit evidence and the potential posterior law

[F2]

A point is a strong small-ball MAP when its posterior ball probability divided by the supremum over all centers tends to $1$ . Small-ball MAP conventions and Cameron–Martin relative modes

[F3]

Let $π$ be the stipulated uniform Borel probability on $[0, 1]$ . Its Lebesgue density is $1_{[0, 1]}$ ; thus $π ([a, b]) = b - a$ for $0 \leq a \leq b \leq 1$ , every singleton has mass zero, and $π (J \cap [0, 1]) \leq length (J)$ for every interval $J$ . Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$

[F4]

The nonnegative Lebesgue integral is monotone and agrees with the simple integral on nonnegative simple functions. The nonnegative Lebesgue integral, Monotonicity and nonnegative homogeneity of the nonnegative integral, The nonnegative integral agrees with the simple integral on simple functions

[F5]

The Lebesgue integral is linear on integrable functions. The Lebesgue integral is linear on $L^{1} (μ)$

Verification

1.1F3F4givenalgebra

For $k = 1, 2, 3, 4$ , put $a_{k} = \int_{[0, 1]} θ^{k} d π$ . These bounded continuous functions are integrable because $0 \leq θ^{k} \leq 1$ on the support of the probability $π$ . For each $N \geq 1$ , partition $[0, 1]$ into $I_{j, N} = [j / N, (j + 1) / N)$ for $j < N - 1$ and $I_{N - 1, N} = [(N - 1) / N, 1]$ . The simple lower and upper step functions with values $(j / N)^{k}$ and $((j + 1) / N)^{k}$ on $I_{j, N}$ bound $θ^{k}$ pointwise; [F3] gives $π (I_{j, N}) = 1 / N$ , and [F4] therefore gives $L_{N, k} : = N^{- (k + 1)} \sum_{j = 0}^{N - 1} j^{k} \leq a_{k} \leq N^{- (k + 1)} \sum_{j = 1}^{N} j^{k} = : U_{N, k} .$ Their difference is $U_{N, k} - L_{N, k} = 1 / N$ . Also the binomial identity telescopes to $N^{k + 1} = (k + 1) \sum_{j = 0}^{N - 1} j^{k} + \sum_{i = 0}^{k - 1} (\binom{k + 1}{i}) \sum_{j = 0}^{N - 1} j^{i} .$ For each $i < k$ , $\sum_{j = 0}^{N - 1} j^{i} \leq N^{i + 1}$ , so after division by $N^{k + 1}$ all terms in the second sum tend to zero. Hence $L_{N, k} \to 1 / (k + 1)$ , and the squeeze gives $a_{k} = 1 / (k + 1)$ . Also $\int_{[0, 1]} 1 d π = 1$ .

1.2givenalgebra

On $[0, 1]$ , $p (θ) = θ^{2} (1 - θ) \geq 0$ . Its derivative is $p^{'} (θ) = θ (2 - 3 θ)$ , so its maximum is $p (2 / 3) = 4 / 27 < 1$ . Therefore $p$ and $1 - p$ define valid conditional probabilities for $Y = 1$ and $Y = 0$ .

2.1F1F3F5step 1.1givenalgebra

By [F1] and [F5], the evidence for $Y = 1$ is $Z = \int_{[0, 1]} p (θ) d π = \int θ^{2} d π - \int θ^{3} d π = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} > 0,$ using step 1.1. The posterior formula [F1] gives density $p / Z = 12 θ^{2} (1 - θ)$ relative to $π$ . Since $π$ has Lebesgue density $1_{[0, 1]}$ , the posterior Lebesgue density is $q (θ) = 12 θ^{2} (1 - θ)$ on $[0, 1]$ , extended by zero on $R ∖ [0, 1]$ .

3.1F2F3step 2.1givenalgebra

On $(0, 1)$ , $q^{'} (θ) = 12 θ (2 - 3 θ)$ , so $q$ increases up to $2 / 3$ and decreases after it; $q (0) = q (1) = 0$ . Thus $q_{\max} = q (2 / 3) > 0$ is the unique density maximum. For any center $x \in R$ and $r > 0$ , the posterior density relative to $π$ and [F3] give $m_{r} (x) \leq q_{\max} π ([x - r, x + r] \cap [0, 1]) \leq 2 r q_{\max}$ . For $0 < r < 1 / 3$ , $[2 / 3 - r, 2 / 3 + r] \subseteq [0, 1]$ and its $π$ -mass is $2 r$ . Continuity of $q$ then gives $∣ \frac{m_{r} (2 / 3)}{2 r} - q_{\max} ∣ \leq \sup_{∣ θ - 2 / 3 ∣ \leq r} ∣ q (θ) - q_{\max} ∣ ⟶ 0.$ Hence $1 \geq \frac{m_{r} (2 / 3)}{\sup_{x \in R} m_{r} (x)} \geq \frac{m_{r} (2 / 3)}{2 r q_{\max}} ⟶ 1.$ By [F2], $2 / 3$ is a strong small-ball MAP.

4.1F1F5step 1.1step 2.1step 3.1givenalgebra∎

By [F1], [F5], and steps 1.1 and 2.1, the posterior mean is $\int_{[0, 1]} θ q (θ) d π = 12 (\int θ^{3} d π - \int θ^{4} d π) = 12 (\frac{1}{4} - \frac{1}{5}) = \frac{3}{5} .$ Since $3 / 5 \neq 2 / 3$ , the posterior mean differs from the strong small-ball MAP. Continuity at $3 / 5$ gives $m_{r} (3 / 5) / (2 r) \to q (3 / 5)$ , while step 3.1 gives $\sup_{x} m_{r} (x) / (2 r) \to q_{\max}$ . Consequently $\frac{m_{r} (3 / 5)}{\sup_{x} m_{r} (x)} ⟶ \frac{q (3 / 5)}{q (2 / 3)} = \frac{216 / 125}{16 / 9} = \frac{243}{250} < 1,$ so the posterior mean is not a strong small-ball MAP.

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Sources

M. Dashti, K. J. H. Law, A. M. Stuart, and J. Voss, MAP Estimators and Their Consistency in Bayesian Nonparametric Inverse Problems (standard reference, not scraped)
H. Kekkonen and Y. Korolev, Inverse Problems, lecture notes (standard reference, not scraped)