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Finite-Sample Point Estimation — Examples

1 · Prerequisites

2 · Summary

These worked calculations show the binomial sample proportion and its exact finite-sample variance, a parameter-specific MSE improvement from a biased constant rule, and the Rao–Blackwellization of one Bernoulli observation by the sample sum. Endpoint laws are included in the discrete conditional calculation.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

The binomial sample proportion

Example

Let $X_{1}, \dots, X_{n}$ be independent Bernoulli $(p)$ observations, where $n \geq 1$ and $p \in [0, 1]$ . For $\overline{X} = n^{- 1} \sum_{i = 1}^{n} X_{i}$ , $E_{p} [\overline{X}] = p, {Var}_{p} (\overline{X}) = {MSE}_{p} (\overline{X}) = \frac{p (1 - p)}{n} .$ The equalities include the degenerate endpoints $p = 0$ and $p = 1$ .

Facts & Assumptions

Given: The finite product Bernoulli experiment, $n \geq 1$ , and $p \in [0, 1]$ .

[F1]

An estimator is unbiased when its expectation equals the estimand at every parameter; MSE is the squared-error expectation. (Unbiased estimators and UMVU estimators, Bias and mean squared error)

[F2]

Variance of a finite sum is the sum of individual variances and twice the sum of pairwise covariances. (Variance of a finite sum as the sum of all variances and covariances)

[F3]

Independent square-integrable variables have zero covariance. (Independence forces covariance to vanish)

[F4]

On a finite probability space, variance is the expectation of the centered square. (Variance, standard deviation, and covariance on a finite probability space)

[F5]

Expectation is linear on finite linear combinations of integrable random variables. (The Lebesgue integral is linear on $L^{1} (μ)$ )

[F6]

For an unbiased estimator, MSE equals variance. (Bias–variance decomposition)

[F7]

A measurable real-valued data map to the named target is an estimator. (Statistics and estimators)

[F8]

A Bernoulli $(p)$ variable takes values $1$ and $0$ with probabilities $p$ and $1 - p$ , and mutual independence of finite-valued variables means their joint cell probabilities factor. (Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials, Pairwise and mutual independence of finite-valued random variables)

Verification

technique · direct

1.1givenF1F4F5F7F8

Each $X_{i}$ and $\overline{X}$ is measurable on the finite product space, hence an estimator by [F7]. The Bernoulli law [F8] gives $P_{p} (X_{i} = 1) = p$ and $P_{p} (X_{i} = 0) = 1 - p$ ; summing these two values gives $E_{p} X_{i} = p$ and ${Var}_{p} (X_{i}) = (1 - p) p^{2} + p (1 - p)^{2} = p (1 - p)$ , including $p = 0, 1$ . Thus [F5] gives $E_{p} \overline{X} = p$ , so $\overline{X}$ is unbiased by [F1].

2.1givenF2F3F6step 1.1∎

The independent $X_{i}$ are square-integrable, so [F3] makes every off-diagonal covariance zero. Applying [F2] gives ${Var}_{p} (\sum_{i} X_{i}) = n p (1 - p)$ and hence ${Var}_{p} (\overline{X}) = p (1 - p) / n$ . Since the bias is zero, [F6] gives the same value for MSE.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A biased estimator can have lower MSE

Example

Observe one $X \sim Bernoulli (p)$ and estimate $p$ . The unbiased estimator $T_{1} = X$ has ${MSE}_{p} (T_{1}) = p (1 - p)$ . The constant estimator $T_{0} = 0$ is biased when $p > 0$ , but at $p = 1 / 5$ , ${MSE}_{1 / 5} (T_{0}) = \frac{1}{25} < \frac{4}{25} = {MSE}_{1 / 5} (T_{1}) .$ This is a pointwise comparison at $p = 1 / 5$ , not a uniform improvement: at $p = 1$ the constant rule has MSE $1$ and $T_{1}$ has MSE $0$ .

Facts & Assumptions

Given: One Bernoulli observation $X$ with $p \in [0, 1]$ .

[F1]

The bias and MSE are expectation minus target and expected squared error, respectively. (Bias and mean squared error)

[F2]

Unbiasedness means expectation equals the target at every parameter. (Unbiased estimators and UMVU estimators)

[F3]

A Bernoulli $(p)$ variable has mean $p$ and variance $p (1 - p)$ , and takes values $1$ and $0$ with probabilities $p$ and $1 - p$ . (Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials, A Bernoulli $(p)$ variable has mean $p$ and variance $p (1 - p)$ ; a binomial $(n, p)$ variable has mean $n p$ and variance $n p (1 - p)$ )

[F4]

A measurable real-valued function of the data is an estimator of its named target. (Statistics and estimators)

Verification

technique · direct

1.1givenF1F2F3F4

Both $T_{1} = X$ and the constant $T_{0} = 0$ are measurable estimators by [F4]. The two-point law gives $E_{p} X = p$ and ${Var}_{p} (X) = p (1 - p)$ , so [F2] makes $T_{1} = X$ unbiased and [F1] gives ${MSE}_{p} (T_{1}) = p (1 - p)$ . For $T_{0} = 0$ , [F1] gives bias $- p$ and MSE $p^{2}$ .

2.1step 1.1choosealgebra∎

At $p = 1 / 5$ , the MSEs are $1 / 25$ and $(1 / 5) (4 / 5) = 4 / 25$ , respectively, so the biased rule has strictly smaller MSE there. At $p = 1$ , they are $1$ and $0$ , so the inequality is not uniform in $p$ .

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Rao–Blackwellizing one Bernoulli observation by the sample sum

Example

Work on the finite product sample space ${0, 1}^{n}$ with its discrete sigma-algebra. Let its coordinates $X_{1}, \dots, X_{n}$ be iid Bernoulli $(p)$ with $p \in [0, 1]$ , $n \geq 1$ , and $S = \sum_{i = 1}^{n} X_{i}$ . The unbiased estimator $U = X_{1}$ is Rao–Blackwellized by $S$ to $E_{p} [X_{1} ∣ S] = \frac{S}{n} .$ The statistic $S$ is sufficient: conditional on $S = s$ , the data vector is uniform over the vectors with $s$ ones when $0 < p < 1$ . At $p = 0$ and $p = 1$ , only the fibers $s = 0$ and $s = n$ are supported, respectively, and their conditional laws are point masses at the all-zero and all-one vectors. These conditional laws do not depend on $p$ .

Facts & Assumptions

Given: The finite iid Bernoulli experiment with $p \in [0, 1]$ and $n \geq 1$ .

[F1]

Rao–Blackwellization of an unbiased finite-variance estimator is its conditional expectation given a sufficient statistic. (Rao–Blackwell theorem for scalar finite-variance estimators)

[L1]

Sufficiency is defined by one probability kernel $K$ from the standard-Borel statistic space to the standard-Borel data space satisfying $P_{θ} (A \cap S^{- 1} (B)) = \int_{B} K (s, A) P_{θ}^{S} (d s)$ for every parameter $θ$ and measurable data and statistic events $A, B$ ; the kernel is independent of $θ$ . (Sufficient statistics)

[F2]

$X_{1}$ is a measurable estimator of $p$ , and it is unbiased when its expectation is $p$ . (Statistics and estimators, Unbiased estimators and UMVU estimators)

[F3]

The conditional-expectation notation is an almost-sure class; AC is inherited from its existence convention. (Conditional expectation as an ae class, The Axiom of Choice)

[F4]

[F5]

The finite discrete data and statistic spaces are standard Borel: their discrete metrics are complete, and each finite space is separable, so the identity map witnesses the definition. (Standard Borel spaces)

Verification

technique · direct

1.1givenF4L1construct

For $s \in {0, \dots, n}$ let $F_{s} = {x \in {0, 1}^{n} : \sum_{i} x_{i} = s}$ ; it has $(\binom{n}{s}) \geq 1$ elements. Define $K (s, A) = ∣ A \cap F_{s} ∣ / (\binom{n}{s})$ for every data event $A$ . Each section is the uniform probability measure on $F_{s}$ , and every event evaluation is measurable because the statistic space is finite and discrete; thus $K$ is a probability kernel independent of $p$ . By [F4], independence makes the joint law the product of the Bernoulli coordinate masses. For $0 < p < 1$ , each vector in $F_{s}$ has mass $p^{s} (1 - p)^{n - s}$ and $P_{p} (S = s) > 0$ , so $K (s, \cdot)$ is the conditional law given $S = s$ . At $p = 0$ or $p = 1$ , only $s = 0$ or $s = n$ is supported, respectively, and that fiber is a singleton on which the same kernel is the conditional law. Hence $K$ is a common regular conditional law and proves sufficiency by [L1].

2.1F1F2F3F4F5step 1.1algebra∎

For $0 < p < 1$ , the uniform fiber law is invariant under coordinate permutations, so the $n$ conditional means $E_{p} [X_{i} ∣ S = s]$ agree. Their sum is $s$ , hence each is $s / n$ . At $p = 0, 1$ , $X_{1} = S / n$ almost surely, so the same formula holds on the support. Since $0 \leq X_{1} \leq 1$ , $U = X_{1}$ is square-integrable; its Bernoulli law gives $E_{p} X_{1} = p \cdot 1 + (1 - p) \cdot 0 = p$ by [F4]. Both spaces are standard Borel by [F5] and $S$ is sufficient by step 1.1, so [F2] and [F1] identify $S / n$ as the Rao–Blackwellized unbiased estimator; [F3] records its conditional-expectation class convention.

Finite-Sample Point Estimation — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The binomial sample proportion

Example

Facts & Assumptions

Verification

A biased estimator can have lower MSE

Example

Facts & Assumptions

Verification

Rao–Blackwellizing one Bernoulli observation by the sample sum

Example

Facts & Assumptions

Verification

Sources