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Topic I. Hypothesis testings

General idea

Assuming:

  • Null hypothesis is fixed (not data dependent)
  • Can compute valid p-values for testing each null, and the function mapping to p-value is fixed in advance.

Formulation of Multiple testing problem:

  • There are a list of null hypothesis \(H_{o1},H_{o2},\dots,H_{on}\), and the p-values for each hypothesis are \(p_1,p_2,\dots,p_n\).

  • A Testing procedure is a function mapping \(\{p_1,\dots,p_n\}\) to a subset of \(\{1,2,\dots,n\}\) (discoveries/rejections).

Possible goals to control false positives:

  1. FamilyWise Error Rate (FWER): Bound Pr(Any false discoveries)

  2. Bound Pr(More than …% false discoveries)

  3. Bound Pr(Number of false discoveries)

  4. Bound Pr(Proportion of false discoveries(FDP) )

  5. Bound Pr (False discoveries rate (FDR) )

1, 4, 5 are more common (they have abbreviation!)

Question 1. Global null hypothesis

  • Goal: test \(H_{o,global}=\) all n nulls are true \((=\bigcap_{i=1,2,\dots,n}H_{oi})\).

Notice: if \(H_{o,global}\) is true, then every \(P_i\sim unif(0,1)\) (but may not be independent)

Method 1. Bonferroni correction

Statistic \(= \min\{p_1,p_2,\dots,p_n\}.\)

Reject if \(\min_i \{p_i\}\leq c\), where \(c = \frac \alpha n\), \(\alpha\) is the level of test.

Proof (upperbound of type_I error): \(\mathbb P(\min_i p_i\leq c)\leq \sum_i\mathbb P(p_i\leq c)=nc\), using the property \(\mathbb P(A\cup B)\leq \mathbb P(A)+\mathbb P(B)\).

Very conservative: the bound in the proof is hard to reach.

Method 2. Fisher’s test

Statistic \[F = -2\sum_i \log(p_i).\]

Under \(H_o,global\), \(p_i\sim unif (0,1)\), if additionally \(p_i\)’s are independent, then \(F\sim \chi_{2n}^2\).

Reject if \(F\geq (1-\alpha)\) -quantile of \(\chi^2_{2n}\).

3. Simes test:

  • Reject \(H_{o,global}\) if:

    • At least 1 \(p_i\leq\frac\alpha n\)
    • and/or at least 2 \(p_i\leq\frac{2\alpha} n\)
    • \(\vdots\)
    • and/or at least n \(p_i\leq\frac{n\alpha} n\)

  • Theorem: Under \(H_{o,global}\), if \(p_i\)’s are independent. Then \(\mathbb P\)(Simes rejects at level \(\alpha\))\(=\alpha\).