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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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 $p$ .

Depends on

Used by

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Sources

George Deligiannidis, SB2.1 Foundations of Statistical Inference, Chapter 5, PDF pp. 29–31 (standard reference, not scraped)
Karl B. Gregory, STAT 713 Lecture 6: Best Unbiased Estimators, Rao-Blackwell Theorem, PDF pp. 2–8 (standard reference, not scraped)