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

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Leave-one-out identity for the BH step-up rule

Statement

Let $m \geq 1$ , $q \in (0, 1)$ , and let $P_{1}, \dots, P_{m}$ be measurable $[0, 1]$ -valued functions on one measurable space. Apply the algorithm in Benjamini–Hochberg step-up procedure. For each $i \in {1, \dots, m}$ and vector $x_{- i} \in [0, 1]^{m - 1}$ , order the remaining values as $x_{- i, (1)} \leq \dots \leq x_{- i, (m - 1)}$ and define

$s_{i} (x_{- i}) : = \max ({0} \cup {k \in {1, \dots, m - 1} : x_{- i, (k)} \leq (k + 1) q / m}) .$

When $m = 1$ , set $s_{i} = 0$ . Let $S_{i} = s_{i} (P_{- i})$ , let $R$ be the number of rejections, and let $I_{i}$ be the indicator that hypothesis $i$ is rejected. Then $S_{i}$ is measurable with respect to the sigma-algebra generated by the other coordinates and, pointwise,

$\frac{I_{i}}{R \lor 1} = \frac{1 {P_{i} \leq (S_{i} + 1) q / m}}{S_{i} + 1} .$

Facts & Assumptions

Given: $m \geq 1$ , $q \in (0, 1)$ , measurable functions $P_{i} : Ω \to [0, 1]$ , and the step-up algorithm specified in the Statement.

[F1]

Benjamini–Hochberg step-up procedure: order the p-values and let $R$ be the largest rank $k$ satisfying $p_{(k)} \leq k q / m$ ; reject ranks $1, \dots, R$ , taking $R = 0$ when no rank qualifies.

[F2]

A measurable function between measurable spaces: a function is measurable when the preimage of every measurable codomain set is measurable.

[F3]

The Borel sigma-algebra of a topological space: the Borel sigma-algebra is generated by the open sets; in particular, closed half-lines and their finite intersections and unions are Borel.

Proof

1.1givenF2F3construct

For every $k \in {1, \dots, m}$ and real $a$ , ${P_{(k)} \leq a} = ⋃_{J \subseteq {1, \dots, m}, ∣ J ∣ = k} ⋂_{j \in J} {P_{j} \leq a}$ , since the $k$ th smallest value is at most $a$ exactly when at least $k$ coordinates are at most $a$ ; the half-line is Borel by [F3], so [F2] makes this finite union of measurable intersections measurable, and the same formula works after deleting any coordinate.

2.1F2step 1.1

Each $A_{k} = {P_{(k)} \leq k q / m}$ is measurable by step 1.1; its indicator and the simple function $k 1_{A_{k}}$ are measurable by [F2], and the finite maximum of these functions and $0$ is measurable because its sublevel sets are finite intersections. The same construction with $P_{- i, (k)}$ and thresholds $(k + 1) q / m$ shows each level set of $S_{i}$ lies in the sigma-algebra generated by the remaining coordinates.

3.1F1step 2.1algebra

If $R = 0$ , no $P_{i}$ is zero because then $P_{(1)} = 0 \leq q / m$ , so no value is at most $R q / m = 0$ . If $0 < R < m$ , maximality gives $P_{(R)} \leq R q / m$ and $P_{(R + 1)} > (R + 1) q / m > R q / m$ , so exactly the first $R$ ranks have values at most $R q / m$ ; if $R = m$ , all values are at most $P_{(m)} \leq q = m q / m$ . Thus the rejected indices are exactly ${i : P_{i} \leq R q / m}$ .

4.1givenF1step 3.1choose

Fix $i$ and write $s = S_{i}$ . If $s \geq 1$ , then $P_{- i, (s)} \leq (s + 1) q / m$ , and if $s = 0$ there are no other values to check. When $P_{i} \leq (s + 1) q / m$ , at least $s + 1$ full-family values meet that threshold, so $R \geq s + 1$ . If $R \geq s + 2$ , the qualifying rank $R$ has $P_{(R)} \leq R q / m$ ; deleting $P_{i}$ leaves at least $R - 1$ other values at most $R q / m$ , making $R - 1 \geq s + 1$ qualify in the definition of $S_{i}$ , a contradiction. Hence $R = s + 1$ , and step 3.1 shows that $i$ is rejected.

4.2F1step 3.1algebra

If $P_{i} > (s + 1) q / m$ but $i$ is rejected, step 3.1 gives $P_{i} \leq R q / m$ ; write $r = R \geq 1$ . At least $r - 1$ other values are at most $r q / m$ , so $r - 1$ qualifies in the definition of $S_{i}$ (with $0$ automatically qualifying when $r = 1$ ). Thus $r - 1 \leq s$ , whence $P_{i} \leq r q / m \leq (s + 1) q / m$ , a contradiction; therefore $i$ is not rejected.

5.1step 4.1step 4.2algebra∎

Step 4.1 shows that $P_{i} \leq (s + 1) q / m$ implies $I_{i} = 1$ and $R = s + 1$ , while step 4.2 shows that otherwise $I_{i} = 0$ ; substituting these two cases gives the stated pointwise identity.

Depends on

Used by

Benjamini–Hochberg FDR under independence Theorem

Dependency tree · two levels

7 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

Yoav Benjamini and Yosef Hochberg, Controlling the False Discovery Rate: A Practical and Powerful Approach to Multiple Testing, JRSS B 57(1) (1995), Theorem 1 and Appendix A (standard reference, not scraped)
Yoav Benjamini and Daniel Yekutieli, The Control of the False Discovery Rate in Multiple Testing under Dependency, The Annals of Statistics 29(4) (2001), §5, PDF pp. 16–17 (standard reference, not scraped)
STATS 200: Introduction to Statistical Inference, Lecture 11: Testing Multiple Hypotheses, Stanford University, PDF pp. 32–36 (standard reference, not scraped)