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A finite-class M-estimator and its uniform law

Example

Let $0 < p \leq 1$ and let $X_{1}, X_{2}, \dots$ be IID Bernoulli $(p)$ . Write ${\overset{ˉ}{X}}_{n} = n^{- 1} \sum_{i = 1}^{n} X_{i}$ . Take $Θ = {0, 1}$ with distance $d (θ, η) = ∣ θ - η ∣$ , and set $m_{0} (x) = 0$ and $m_{1} (x) = x$ . Let ${\hat{θ}}_{n} = 1$ when ${\overset{ˉ}{X}}_{n} > 0$ and ${\hat{θ}}_{n} = 0$ when ${\overset{ˉ}{X}}_{n} = 0$ . Then ${\hat{θ}}_{n}$ is an exact M-estimator and is consistent for the unique population maximizer $θ_{0} = 1$ . This finite criterion class has a uniform law; the proof below gives the Bernoulli strong law directly.

Facts & Assumptions

Given: $0 < p \leq 1$ , the IID Bernoulli observations, the finite parameter set $Θ = {0, 1}$ , and the displayed criterion and estimator.

[F1]

M-estimators: A data-measurable estimator satisfying $P_{n} m_{{\hat{θ}}_{n}} = \sup_{θ \in Θ} P_{n} m_{θ}$ is an exact maximizing M-estimator.

[F2]

Identical distribution and IID families: IID means identically distributed with the whole family mutually independent.

[F3]

Pointwise consistency in probability: Consistency at a fixed target means the probability of every fixed positive metric error tends to zero.

[F4]

The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded: The usual metric on $R$ is $d (x, y) = ∣ x - y ∣$ ; on $Θ = {0, 1}$ the distance used here is its restriction, which remains a metric because the metric axioms persist on subsets.

[F5]

Estimator sequences in IID experiments: The IID sample-size- $n$ space is a finite product space, and an estimator is a measurable map on that space.

[A1]

The observations are Bernoulli $(p)$ , so $E X_{i} = p$ and $Y_{i} = X_{i} - p$ satisfies $E Y_{i} = 0$ and $∣ Y_{i} ∣ \leq 1$ .

[A2]

The criterion class consists only of the two functions $m_{0} (x) = 0$ and $m_{1} (x) = x$ .

Verification

1.1A1A2F1F2F5given

The population criterion is $M (0) = 0$ and $M (1) = E X_{1} = p > 0$ , so its unique maximum is $θ_{0} = 1$ . The empirical criteria are $M_{n} (0) = 0$ and $M_{n} (1) = {\overset{ˉ}{X}}_{n} \geq 0$ ; the displayed selection therefore attains their maximum for every sample. By [F5], the sample space is ${0, 1}^{n}$ with its product sigma-algebra. Each singleton vector is a measurable rectangle, and every subset is a finite union of singletons, so every data map is measurable. The estimator is exact by [F1].

1.2A1A2F2algebra

Independence and the Bernoulli finite sample law imply that expectations of products of functions of distinct coordinates factor: the joint mass at each $(x_{1}, \dots, x_{n}) \in {0, 1}^{n}$ is $\prod_{i} P (X_{i} = x_{i})$ , and summing this product over the finite sample space factors. Expanding $(\sum_{i} Y_{i})^{4}$ , every term with an index occurring once has expectation zero by $E Y_{i} = 0$ and this factorization; the remaining terms are $\sum_{i} E Y_{i}^{4} + 6 \sum_{i < j} E Y_{i}^{2} E Y_{j}^{2} \leq n + 6 (\binom{n}{2}) \leq 3 n^{2}$ . For $Z \geq 0$ and $t > 0$ , $t 1_{{Z > t}} \leq Z$ , so integrating gives $P (Z > t) \leq E Z / t$ . Applying this to $Z = (\sum_{i} Y_{i})^{4}$ gives $P (∣ {\overset{ˉ}{X}}_{n} - p ∣ > ε) \leq 3 / (n^{2} ε^{4})$ .

2.1step 1.2givenalgebra

For fixed $ε > 0$ , the bound in step 1.2 is summable in $n$ . By the union bound, $P (⋃_{n \geq N} {∣ {\overset{ˉ}{X}}_{n} - p ∣ > ε}) \leq \sum_{n \geq N} 3 / (n^{2} ε^{4}) \to 0$ ; hence only finitely many such deviations occur almost surely. Applying this to $ε = 1 / j$ for every positive integer $j$ gives ${\overset{ˉ}{X}}_{n} \to p$ almost surely.

3.1A1A2F3F4step 1.2step 2.1given∎

For $0 < ε < 1$ , the metric-error event is ${{\hat{θ}}_{n} = 0} = {{\overset{ˉ}{X}}_{n} = 0}$ , which is contained in ${∣ {\overset{ˉ}{X}}_{n} - p ∣ > p / 2}$ . Step 1.2 bounds its probability by $3 / (n^{2} (p / 2)^{4}) \to 0$ ; for $ε \geq 1$ the error event is empty. Hence ${\hat{θ}}_{n}$ is consistent by [F3] in the metric of [F4]. Also, since $p > 0$ , step 2.1 implies that ${\hat{θ}}_{n} = 1$ eventually almost surely, and $\sup_{θ \in Θ} ∣ M_{n} (θ) - M (θ) ∣ = ∣ {\overset{ˉ}{X}}_{n} - p ∣ \to 0$ almost surely, displaying the finite-class uniform law.

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Sources

Peter Bartlett, Theoretical Statistics, Stat 210B Lecture 15, consistency of M-estimators, PDF pp. 6–10 (standard reference, not scraped)