LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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MLR orders the power of randomized thresholds

Statement

Assume AC and let a dominated one-parameter experiment have MLR in the real statistic $T$ . For $c \in R$ and $γ \in [0, 1]$ , put $ϕ_{c, γ}^{+} = 1_{{T > c}} + γ 1_{{T = c}}, ϕ_{c, γ}^{-} = 1_{{T < c}} + γ 1_{{T = c}} .$ Then for every $θ_{1} < θ_{2}$ , $E_{θ_{1}} ϕ_{c, γ}^{+} \leq E_{θ_{2}} ϕ_{c, γ}^{+}, E_{θ_{1}} ϕ_{c, γ}^{-} \geq E_{θ_{2}} ϕ_{c, γ}^{-} .$

Facts & Assumptions

Given: AC, a dominated experiment with MLR in $T$ , real $c$ , $γ \in [0, 1]$ , and $θ_{1} < θ_{2}$ .

[F1]

For $θ_{1} < θ_{2}$ , there is a nondecreasing Borel function $r$ such that $r (T) = p_{θ_{2}} / p_{θ_{1}}$ $μ$ -almost everywhere where $p_{θ_{1}} > 0$ , and $r (T) = + \infty$ $μ$ -almost everywhere where $p_{θ_{1}} = 0 < p_{θ_{2}}$ . (Families with a monotone likelihood ratio)

[F2]

If $p$ is a nonnegative measurable density, then integrating against $p d μ$ equals integrating the product against $μ$ : $\int g d (p d μ) = \int g p d μ$ for nonnegative measurable $g$ . (Integrating against a density agrees with integrating the product)

[F3]

$P_{θ_{1}}$ and $P_{θ_{2}}$ are probability measures, so each has mass one. (Probability measures and probability spaces)

[F4]

A Borel function composed with the measurable statistic $T$ is measurable; hence the MLR ratio function $r (T)$ can be integrated. (Composition with a Borel measurable outer map preserves measurability)

[A1]

AC supplies the simultaneous likelihood-density versions used in [F1]. (The Axiom of Choice)

Proof

1.1givenA1F1F4choose

Choose simultaneous densities $p_{1} = p_{θ_{1}}$ and $p_{2} = p_{θ_{2}}$ using AC [A1], let $r$ be their nondecreasing Borel MLR function, and put $A = {p_{1} > 0}$ and $S = {p_{1} = 0 < p_{2}}$ with $s = P_{θ_{2}} (S)$ . Let $N$ be the union of the $μ$ -null exceptional sets in [F1]. For $x \in S ∖ N$ and $y \in A ∖ N$ , $r (T (x)) = + \infty$ while $r (T (y)) = p_{2} (y) / p_{1} (y) < + \infty$ ; monotonicity therefore gives $T (y) < T (x)$ . The two laws give $N$ measure zero, $P_{θ_{1}} (A) = 1$ , and $P_{θ_{2}} (S) = s$ . Since $ϕ_{c, γ}^{+}$ is nondecreasing in $T$ and $ϕ_{c, γ}^{-}$ is nonincreasing in $T$ , this support order gives $E_{θ_{2}} [ϕ_{c, γ}^{+} 1_{S}] \geq s E_{θ_{1}} ϕ_{c, γ}^{+}, E_{θ_{2}} [ϕ_{c, γ}^{-} 1_{S}] \leq s E_{θ_{1}} ϕ_{c, γ}^{-} .$ By [F4], $r (T)$ is measurable.

2.1givenF1F2F3step 1.1

Partition $X$ into $B_{-} = {T < c}$ , $B_{0} = {T = c}$ , and $B_{+} = {T > c}$ . Put $a_{j} = P_{θ_{1}} (B_{j})$ and $b_{j} = P_{θ_{2}} (B_{j} \cap A)$ . On $A$ , [F1] gives $p_{2} = r (T) p_{1}$ $μ$ -almost everywhere, so [F2] yields $b_{j} = \int_{B_{j} \cap A} p_{2} d μ = \int_{B_{j} \cap A} r (T) p_{1} d μ = \int_{B_{j} \cap A} r (T) d P_{θ_{1}}$ . The measurability of $r (T)$ was established in step 1.1. Also [F3] and step 1.1 give $\sum_{j} b_{j} = 1 - s$ and $P_{θ_{1}} (A) = 1$ , hence $P_{θ_{1}} (B_{j} \cap A) = a_{j}$ . For each block with $a_{j} > 0$ , set $r_{j} = b_{j} / a_{j}$ . Monotonicity [F1] orders $r (T)$ values across every pair of occupied blocks, so their $P_{θ_{1}}$ -averages satisfy $r_{i} \leq r_{j}$ whenever $i < j$ .

3.1step 1.1step 2.1algebra

For ordered block values $g_{-} \leq g_{0} \leq g_{+}$ , omitting zero-mass blocks, expansion gives $\sum_{j} a_{j} r_{j} g_{j} - (\sum_{j} a_{j} r_{j}) (\sum_{j} a_{j} g_{j}) = \sum_{i < j} a_{i} a_{j} (r_{j} - r_{i}) (g_{j} - g_{i}) \geq 0$ . Step 2.1 gives $\sum_{j} a_{j} = 1$ and $\sum_{j} a_{j} r_{j} = 1 - s$ . For the upper threshold $g = (0, γ, 1)$ , this identity makes $E_{θ_{2}} [ϕ_{c, γ}^{+} 1_{A}] \geq (1 - s) E_{θ_{1}} ϕ_{c, γ}^{+}$ . Step 1.1 gives $E_{θ_{2}} [ϕ_{c, γ}^{+} 1_{S}] \geq s E_{θ_{1}} ϕ_{c, γ}^{+}$ , and $P_{θ_{2}} (X ∖ (A \cup S)) = 0$ . Adding these bounds proves $E_{θ_{2}} ϕ_{c, γ}^{+} \geq E_{θ_{1}} ϕ_{c, γ}^{+}$ .

4.1step 1.1step 2.1step 3.1algebra∎

The lower threshold $g = (1, γ, 0)$ is nonincreasing, so the covariance identity in step 3.1 is nonpositive and gives $E_{θ_{2}} [ϕ_{c, γ}^{-} 1_{A}] \leq (1 - s) E_{θ_{1}} ϕ_{c, γ}^{-}$ . Step 1.1 gives $E_{θ_{2}} [ϕ_{c, γ}^{-} 1_{S}] \leq s E_{θ_{1}} ϕ_{c, γ}^{-}$ ; the remaining set has $P_{θ_{2}}$ measure zero. Adding these bounds proves $E_{θ_{2}} ϕ_{c, γ}^{-} \leq E_{θ_{1}} ϕ_{c, γ}^{-}$ .

Depends on

Used by

The Karlin–Rubin theorem for one-sided tests Theorem

Dependency tree · two levels

24 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jun Shao, Stat 610 Mathematical Statistics, Lecture 11, Lemma B and Theorem 8.3.17, PDF pp. 8–10 (standard reference, not scraped)
George Deligiannidis, Foundations of Statistical Inference, Theorem 13.5 and proof, PDF pp. 77–78 (standard reference, not scraped)