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✓ 4 results · all verified · 4 also independently AI-judged
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Loss, Risk and Decision Theory — Examples

1 · Prerequisites

2 · Summary

These worked cases compute squared-error risk, show how randomization can improve worst-case risk under zero-one loss, and verify pointwise domination of a constant rule. The counterexample uses an open action interval to show why posterior-risk minimization requires an attained measurable minimizer.

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 →

Squared-error risk for a Bernoulli observation

Example

Observe one Bernoulli variable X∈{0,1} with parameter θ∈[0,1], so Pθ({1})=θ and Pθ({0})=1−θ. Estimate θ by the rule d(X)=X under squared-error loss L(θ,a)=(a−θ)2. Its risk is R(θ,d)=θ(1−θ), which is zero at both endpoints and has maximum 1/4 at θ=1/2.

Facts & Assumptions

Given: The finite Bernoulli model, estimator, squared-error loss, and risk definition in the Example.

[F1]

A probability measure is a measure with total mass one (Probability measures and probability spaces).

[F2]

A statistic or estimator is a measurable data map to its target (Statistics and estimators).

[F3]

Risk is the integral of loss under the sampling law (Loss functions and risk).

[F4]

On a finite probability space, expectation is the probability-weighted sum (Expectation agrees with the published finite weighted sum).

Verification

technique · direct finite verification
1.1givenF1F2

For each θ∈[0,1], define Pθ(A)=θ1{1∈A}+(1−θ)1{0∈A} for A⊆{0,1}. Its singleton masses are nonnegative and sum to one; any disjoint sequence of subsets has at most two nonempty terms, and their masses add to the union mass. Thus Pθ is a probability measure by [F1]. The identity rule d(x)=x is measurable on the finite full-power-set spaces and estimates the target ψ(θ)=θ by [F2].

1.2F3F4algebra

The finite expectation formula [F4] and risk definition [F3] give R(θ,d)=(1−θ)(0−θ)2+θ(1−θ)2=θ(1−θ).

2.1step 1.2algebra∎

Since θ(1−θ)=14−(θ−12)2, the risk is at most 1/4, with equality at θ=1/2; substitution gives risk zero at θ=0 and θ=1.

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

Randomization lowers worst-case risk under zero-one loss

Example

Let Θ={0,1}, let the data space contain one point with the same sampling law at both parameters, and let the action space be {0,1}. Under zero-one loss L(θ,a)=1{a≠θ}, the deterministic rule that always chooses 0 has risks 0 and 1, hence worst-case risk 1. The randomized rule that chooses each action with probability 1/2 has risk 1/2 at both parameters and is minimax with value 1/2.

Facts & Assumptions

Given: The finite two-parameter decision problem and loss in the Example.

[F1]

A probability measure has total mass one (Probability measures and probability spaces).

[F2]

A randomized decision rule is a probability kernel from data to actions (Randomized procedures as probability kernels).

[F3]

Risk is expected loss; worst-case risk and minimax value use a supremum and infimum over rules (Loss functions and risk, Bayes risk and minimax risk).

[F4]

Finite expectations are probability-weighted sums (Expectation agrees with the published finite weighted sum).

Verification

technique · direct finite verification
1.1givenF1F2

The one-point data law assigns mass one to its only point under either parameter, so it is a probability measure by [F1]. For each q∈[0,1], define δq(∗,B)=q1{1∈B}+(1−q)1{0∈B} for B⊆{0,1}; its two singleton masses are nonnegative and sum to one, hence it is a probability measure. Evaluations are measurable on the one-point data space, so δq is a probability kernel by [F2]. Deterministic rules are the cases q=0 and q=1.

1.2F3F4algebra

By the finite expectation formula [F4] and the loss-risk definition [F3], the rule with probability q of action 1 has risks R(0)=q and R(1)=1−q. Its worst-case risk is therefore max⁡(q,1−q)≥(q+1−q)/2=1/2, with equality at q=1/2.

2.1step 1.2algebra∎

The two deterministic rules have q=0 or q=1, so each has worst-case risk 1. The fair randomized rule has worst-case risk 1/2, and step 1.2 shows no rule can do better. Thus the minimax value is 1/2 and randomization strictly improves worst-case risk over every deterministic rule.

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

A data-dependent rule dominates a constant rule

Example

Let Θ={0,1} and observe X∈{0,1} with Pθ({x})=1 when x=θ and zero otherwise. Under squared-error loss, the constant rule d0(x)=0 has risk 0 at θ=0 and 1 at θ=1. The data rule d1(x)=x has risk zero at both parameters, so it dominates d0.

Facts & Assumptions

Given: The finite model, two decision rules, squared-error loss, and domination definition in the Example.

[F1]

A probability measure has total mass one (Probability measures and probability spaces).

[F2]

A statistic or estimator is a measurable data map (Statistics and estimators).

[F3]

Risk is the expected loss under the sampling law (Loss functions and risk).

[F4]

Domination requires weakly lower risk at every parameter and strict improvement at one parameter (Risk domination and admissibility).

[F5]

Finite expectations are probability-weighted sums (Expectation agrees with the published finite weighted sum).

Verification

technique · direct finite verification
1.1givenF1F2

For each θ∈{0,1}, let Pθ be the point mass at θ. In any disjoint sequence of events at most one event contains θ, so this law is countably additive and has total mass one; it is a probability measure by [F1]. Both maps d0(x)=0 and d1(x)=x are measurable on the finite full-power-set spaces by [F2].

1.2F3F5algebra

Under squared-error loss, the only data value with positive Pθ-mass is x=θ. The finite-sum formula [F5] and risk definition [F3] therefore give R(θ,d0)=θ2=θ and R(θ,d1)=(θ−θ)2=0.

2.1F3F4step 1.2∎

The risks satisfy R(θ,d1)≤R(θ,d0) at both parameters, and the inequality is strict at θ=1. Thus d1 dominates the constant rule d0 by [F4].

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

An open action set need not contain a posterior-risk minimizer

Statement refuted

Every posterior expected-loss function on a nonempty open action space attains its infimum.

Facts & Assumptions

Given: The singleton prior, singleton observation, open action interval, and loss in the Counterexample.

[F1]

A probability measure on a singleton assigns mass one to its point (Probability measures and probability spaces).

[F2]

Risk is the nonnegative integral of loss (Loss functions and risk).

[F3]

A posterior is a regular conditional distribution (Regular conditional distribution).

[F4]

The Bayes-rule theorem assumes a measurable posterior minimizer is attained (A posterior-risk minimizer minimizes prior risk).

Counterexample

Given: A one-point parameter space Θ={0} with prior mass one, a one-point data space with sampling mass one, the open action interval D=(0,1), and loss L(0,a)=a.

1.1givenF1F2F3

The prior and data law are point masses on singleton standard-Borel spaces by [F1]; their joint law is the point mass on the singleton pair. Define the posterior to be the point mass at 0. The only conditioning events are empty or the whole singleton, and the conditional-law identity holds for both, so this is a regular conditional law by [F3]. Integrating L(0,a)=a against it gives posterior risk q(a)=a by [F2].

2.1F4step 1.1algebra∎

For every a∈(0,1), q(a)=a>0, while the sequence an=1/(n+1) lies in (0,1) and has q(an)→0. Thus inf⁡a∈(0,1)q(a)=0 and the infimum is not attained. The claimed universal minimizer does not exist, which refutes the statement and illustrates the explicit existence premise in [F4].