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Frequentist Properties of Bayesian Multiplicity Control for Multiple Testing of Normal Means

Abstract

We consider the standard problem of multiple testing of normal means, obtaining Bayesian multiplicity control by assuming that the prior inclusion probability (the assumed equal prior probability that each mean is nonzero) is unknown and assigned a prior distribution. The asymptotic frequentist behavior of the Bayesian procedure is studied, as the number of tests grows. Studied quantities include the false positive probability, which is shown to go to zero asymptotically. The asymptotics of a Bayesian decision-theoretic approach are also presented.

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Acknowledgments

Supported in part by NSF grants DMS-1007773 and DMS-1407775

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Appendix

Appendix

Fact 6.1 (Normal tail probability).

Let Φ(t) be the cumulative distribution function of a standard normal random variable. Then as \(t \rightarrow \infty \),

$$ \frac{t\frac{1}{\sqrt{2\pi}} e^{-t^{2}/2} }{t^{2}+1}\leq\ 1-{\Phi}(t)\leq \frac{\frac{1}{\sqrt{2\pi}} e^{-t^{2}/2} }{t} , $$
$$ 1-{\Phi}(t) =\frac{\frac{1}{\sqrt{2\pi}} e^{-t^{2}/2}}{t}(1+o_{p}(1)) . $$

The proof can be found in Durrett (2010).

Fact 6.2 (Weak law for triangular arrays).

For each m, let Xm,k, 1 ≤ km be independent random variables. Let βm > 0 with \(\beta _{m} \rightarrow \infty \) and let \(\bar {X}_{m,k}=X_{m,k}1_{\{|X_{m,k}| \leq \beta _{m} \}}\). Suppose that as \(m\rightarrow \infty \)

$$\sum \limits^{m}_{k=1} P(|X_{m,k}|>\beta_{m}) \rightarrow 0 \quad \text{and} \quad \frac{1}{{\beta_{m}^{2}}}\sum \limits^{m}_{k=1} E\bar{X}_{m,k}^{2} \rightarrow 0 .$$

Defining Sm = Xm,1 + ... + Xm,m and \(\alpha _{m}=\sum \limits ^{m}_{k=1} E\bar {X}_{m,k}\), it follows that

$$ \frac{(S_{m}-\alpha_{m})}{\beta_{m}} \rightarrow 0 \quad \text{in probability} . $$

See Durrett (2010) for a proof.

Lemma 6.3.

If \(Y_{j}=\frac {1}{\sqrt {1+\tau ^{2}}}\exp \{\frac {\tau ^{2}}{2(1+\tau ^{2})}{X_{j}^{2}}\}\) and τ2 > 1/3, then

$$ m^{-2\tau^{2} /(1+\tau^{2})}\sum\limits_{j=1}^{m} {Y_{j}^{2}} \rightarrow 0 \quad \text{ when } m\rightarrow \infty . $$

Proof.

First we need the following order estimate before applying Fact 6.2: If \(\lim _{m\rightarrow \infty } a_{m} \in (0,\infty ]\), \(\lim _{m\rightarrow \infty }b_{m}= \infty \), then

$$ {\int}^{b_{m}}_{0} e^{\frac{a_{m} x^{2}}{2}}dx = O\bigg(\frac{e^{a_{m}{b_{m}^{2}}/2}}{a_{m}b_{m}}\bigg) . $$
(6.1)

To show (6.1) is true, notice that \({\int \limits }^{b_{m}}_{0} e^{\frac {a_{m}x^{2}}{2}}dx = \frac {1}{\sqrt {a_{m}}}{\int \limits }_{0}^{\sqrt {a_{m}}b_{m}} e^{\frac {x^{2}}{2}}dx\) and by L’Hôpital’s rule:

$$\begin{array}{llll} &\lim\limits_{m\rightarrow \infty} \frac{\frac{1}{\sqrt{a_{m}}}{\int}^{\sqrt{a_{m}}b_{m}}_{0} e^{\frac{x^{2}}{2}}dx} {e^{\frac{a_{m}{b_{m}^{2}}}{2}}/(a_{m} b_{m})} = \lim\limits_{m\rightarrow \infty} \frac{{\int}^{\sqrt{a_{m}}b_{m}}_{0} e^{\frac{x^{2}}{2}}dx} {e^{\frac{a_{m}{b_{m}^{2}}}{2}}/(\sqrt{a_{m}} b_{m})}\\ &= \lim\limits_{v\rightarrow \infty}\frac{{{\int}^{v}_{0}} e^{\frac{x^{2}}{2}}dx}{e^{\frac{v^{2}}{2}}/v} \quad \text{ where } v = \sqrt{a_{m}}b_{m}\\ &= \lim\limits_{v\rightarrow \infty} \frac{\frac{d}{dv}{{\int}^{v}_{0}} e^{x^{2}/2}dx}{\frac{d}{dv}\frac{e^{v^{2}/2}}{v}} = \lim\limits_{v \rightarrow \infty} \frac{1}{1-\frac{1}{v^{2}}} =1 . \end{array}$$

Take \(X_{m,j}= \exp \{\frac {\tau ^{2}}{1+\tau ^{2}} {x_{j}^{2}}\}\) and \(\beta _{m} = m^{2\tau ^{2}/(1+\tau ^{2})}\) in Fact 6.2. Both assumptions in Fact 6.2 hold since

$$\sum \limits_{j=1}^{m} P(|X_{m,j}|>\beta_{m}) = O\bigg(\frac{1}{\sqrt{log m}}\bigg) \rightarrow 0 \qquad \text{ by Fact } 6.1 ,$$

and

$$\begin{array}{llll} \frac{1}{{\beta_{m}^{2}}} \sum \limits^{m}_{j=1} E\bar{X}_{m,j}^{2} &=\frac{m}{m^{4\tau^{2}/(1+\tau^{2})}} \int\limits_{|x|\leq \sqrt{ 2 \log m} }\frac{1}{\sqrt{2\pi}}e^{(\frac{2\tau^{2}}{1+\tau^{2}} -1) \frac{x^{2}}{2}}dx\\ &= \left\{\begin{array}{llll} O\big(m^{-\frac{2\tau^{2}}{1+\tau^{2}}} \frac{1}{\sqrt{log m }}\big) &\text{ if } \tau^{2}>1, \text{ by } (6.1) \\ O\big(m^{\frac{1-3\tau^{2}}{1+\tau^{2}}}\big) &\text{ else,} \end{array}\right. \\&= o(1) \qquad \text{ if } \tau^{2}>1/3 . \end{array} $$

The limit is

$$ \begin{array}{llll} \frac{1}{\beta_{m}} \sum \limits^{m}_{j=1} E\bar{X}_{m,j} &= m^{-2\tau^{2}/(1+\tau^{2})} \frac{m}{\sqrt{2\pi}}\int\limits_{|x|<\sqrt{2 \log m}} \exp\left\{ \frac{x^{2}}{2}(\frac{2\tau^{2}}{1+\tau^{2}}-1) \right\} \\ &= m^{(1-\tau^{2})/(1+\tau^{2})} O\bigg(m^{(\tau^{2}-1)/(1+\tau^{2})} \frac{1}{\sqrt{log m}} \bigg) \text{ by } (6.1)\\ &= O\bigg(\frac{1}{\sqrt{log m}}\bigg) \rightarrow 0 . \end{array} $$

Fact 6.4 (Marcinkiewicz and Zygmund).

Let X1,X2,... be i.i.d. with EX1 = 0 and \(E|X_{1}|^{p} < \infty \) where 1 < p < 2. If Sm = X1 + ... + Xm then

$$\frac{S_{m}}{m^{1/p}} \rightarrow 0 \text{ a.s. } $$

The proof can be found in Durrett (2010).

Lemma 6.5.

If \(Y_{j}=\frac {1}{\sqrt {1+\tau ^{2}}}\exp \{\frac {\tau ^{2}}{2(1+\tau ^{2})}{X_{j}^{2}}\}\) and k = O(mr), with \( r\in [0, \frac {1}{1+\max \limits \{1,\tau ^{2}\}})\), then

$$ \bigg(\frac{1}{m} \sum\limits_{j=1}^{m} Y_{j} \bigg)^{k} = 1+o(1) . $$

Proof.

If τ2 < 1, then E[Yj] = 1 and Yj has finite variance, so

$$ \bigg(\frac{1}{m} \sum\limits_{j=1}^{m} Y_{j} \bigg)^{k} = (1+O(m^{-1/2}))^{k} = 1+ O(m^{(r-1/2)}) $$

and the result holds.

If τ2 ≥ 1, set Xi = Yi − 1 and p = 1 + τ− 2𝜖. Fact 6.4 then implies that

$$\frac{{{\sum}_{1}^{m}} Y_{j} - m} {m^{1/p}} =\bigg(\frac{{{\sum}_{1}^{m}} Y_{j}} {m^{1/p}} - m^{1-1/p} \bigg)=o(1) ,$$

so that

$$ \frac{{{\sum}^{m}_{1}} Y_{j}}{m} -1 = \frac{1}{m^{1-1/p}}\bigg(\frac{\sum Y_{j}}{m^{1/p}} -m^{1-1/p} \bigg) = o\bigg(\frac{1}{m^{1-1/p}}\bigg) . $$
(6.2)

Thus

$$ \left( \frac{{{\sum}^{m}_{1}} Y_{j}}{m} \right)^{k} = (1+o(m^{-(1-1/p)}))^{k} = (1+o(m^{(r-1+1/p)})) . $$
(6.3)

Computation shows that r − 1 + 1/p < 0 for small enough 𝜖, proving the result. □

Proof Proof of Lemma 3.1.

Let \(Y_{j} = \frac {1}{\sqrt {1+\tau ^{2}}} e^{{X_{j}^{2}}\tau ^{2}/2(1+\tau ^{2})}\) and expand \((\frac {1}{m}{{\sum }_{1}^{m}} Y_{j})^{k}\):

$$ \begin{array}{llll} \bigg[\frac{{{\sum}_{1}^{m}} Y_{j}}{m} \bigg]^{k} &= \sum \limits_{k_{1}+...+k_{m} = k} \binom{k}{k_{1},k_{2},...,k_{m}}\prod\limits_{1\leq j \leq m} \frac{Y_{j}^{k_{j}}}{m^{k_{j}}} \\ &= \underbrace{\sum \limits_{\begin{array}{llll}k_{1}+...+k_{m} = k \\ max|k_.|=1 \end{array}} \frac{k!}{m^{k}} \prod\limits_{1\leq j \leq m} Y_{j}^{k_{j}}}_{I} + \underbrace{\sum \limits_{\begin{array}{llll}k_{1}+...+k_{m} = k \\ max|k_.|\geq 2 \end{array}} \frac{k!}{k_{1}!...k_{m}!} \prod\limits_{1\leq j \leq m} \frac{Y_{j}^{k_{j}}}{m^{k_{j}}}}_{II} . \end{array} $$
(6.4)

Notice that

$$ \begin{array}{llll} I &= \frac{(m-1)...(m-k+1)}{m^{k-1}}\frac{1}{\binom{m}{k}} \sum \limits_{\begin{array}{llll}k_{1}+...+k_{m} = k \\ max|k_.|=1 \end{array}} \prod\limits_{1\leq j \leq m} \exp \left\{ \frac{\tau^{2} k_{j}}{2(1+\tau^{2})} {X_{j}^{2}}\right\}\\ &= \frac{(m-1)...(m-k+1)}{m^{k-1}} \frac{1}{\binom{m}{k}} \sum\limits_{|\boldsymbol{\gamma}| =k} B_{\boldsymbol{\gamma}0}\\ &= \bigg[\frac{1}{\binom{m}{k}}\sum\limits_{|\boldsymbol{\gamma}| =k} B_{\boldsymbol{\gamma}0}(\boldsymbol{X})\bigg](1+O(k^{2}/m))= \bigg[\frac{1}{\binom{m}{k}}\sum\limits_{|\boldsymbol{\gamma}| =k} B_{\boldsymbol{\gamma}0}(\boldsymbol{X})\bigg](1+o(1)) , \end{array} $$

where the order of the O term can be bounded by a constant independent of k, as established by a detailed Stirling’s approximation argument.

$$ \begin{array}{llll} II &= \frac{1}{m^{k}}\sum \limits_{\begin{array}{llll}k_{1}+...+k_{m} = k \\ max|k_.|\geq 2 \end{array}} \frac{k!}{k_{1}!...k_{m}!} \prod\limits_{1\leq j \leq m} Y_{j}^{k_{j}} \\ &\leq \frac{k(k-1)}{m^{k}}\bigg({\sum\limits_{1}^{m}} {Y_{i}^{2}}\bigg) \sum\limits_{\tilde{k}_{1}+...+\tilde{k}_{m}=k-2} \frac{(k-2)!}{\tilde{k}_{1}!...\tilde{k}_{m}!} Y_{1}^{\tilde{k}_{1}}...Y_{m}^{\tilde{k}_{m}} \\ &= \bigg(\frac{k(k-1)}{m^{2/(1+\tau^{2})}}\bigg) \bigg(\frac{{{\sum}_{1}^{m}} {Y_{i}^{2}}}{m^{2\tau^{2}/(1+\tau^{2})}}\bigg) \bigg(\frac{{{\sum}_{1}^{m}} Y_{i}}{m}\bigg)^{k-2} . \end{array} $$

To conclude that \(II\rightarrow 0\) when \(m\rightarrow \infty \), note that \(\frac {k(k-1)}{m^{2/(1+\tau ^{2})}}\) converges to 0 because of the k = O(mr) assumption;

By Lemma 6.3, when τ2 > 1/3, \(\frac {{{\sum }_{1}^{m}} {Y_{i}^{2}}}{m^{2\tau ^{2}/(1+\tau ^{2})}}=o_{p}(1)\);

By Lemma 6.5, \(\big (\frac {{{\sum }_{1}^{m}} Y_{i}}{m}\big )^{k-2}= 1+o_{p}(1)\).

Hence, the right hand side of Eq. 6.4 converges to \(1/\binom {m}{k}{\sum }_{|\boldsymbol {\gamma }| =k}B_{\boldsymbol {\gamma }0}(1+o(1))\). By Lemma 6.5, the left hand side goes to 1. Hence,

$$ \lim \limits_{m\rightarrow \infty} \frac{1}{\binom{m}{k} }\sum\limits_{|\boldsymbol{\gamma}| =k}B_{\boldsymbol{\gamma} 0} = 1 . $$

Lemma 6.6.

$$\frac{1}{\binom{m}{k}} \sum \limits_{|\boldsymbol{\gamma}| =k} B_{\boldsymbol{\gamma}0}(x) \text{ is a U-statistic} . $$

Proof.

By definition,

$$\begin{array}{llll} B_{\boldsymbol{\gamma}0}(x_{1},x_{2},...,x_{m}) = \frac{1}{(1+\tau^{2})^{\frac{|\boldsymbol{\gamma}| }{2}}} \exp \left\{ \frac{\tau^{2}}{2(1+\tau^{2})} \sum \limits_{i:\gamma_{i}\neq 0 } {x_{i}^{2}} \right\} \end{array}$$
$$h^{k}(x_{1},x_{2},...,x_{k})= \frac{1}{(1+\tau^{2})^{\frac{k}{2}}} \exp \left\{ \frac{\tau^{2}}{2(1+\tau^{2})} \sum \limits_{i=1}^{k} {x_{i}^{2}} \right\} . $$

Notice that an iteration over models is the same as an iteration through inputs:

$$\sum \limits_{\gamma:|\boldsymbol{\gamma}| =k} B_{\boldsymbol{\gamma}0}(x) = \sum \limits_{{C^{m}_{k}}} h^{k}(x_{i_{1}},x_{i_{2}},...,x_{i_{k}}) .$$

Therefore,

$$\frac{1}{\binom{m}{k}} \sum \limits_{|\boldsymbol{\gamma}| =k} B_{\boldsymbol{\gamma}0}(x) \text{ is a U-statistics with kernel } h^{k} .$$

Definition 6.7.

For c ∈{0,1,...,k},

$$\begin{array}{llll} {h^{k}_{c}}(x_{1},...,x_{c})&=\mathbbm{E}(h^{k}(x_{1},x_{2},...,x_{c},X_{c+1},...,X_{k}))\\ ({\sigma_{c}^{k}})^{2} &= Var({h^{k}_{c}}(X_{1},...,X_{c})) . \end{array}$$

Note that \({h^{k}_{k}} = h^{k}\), \(({\sigma ^{k}_{k}})^{2} =Var(h^{k})={\sigma _{k}^{2}}\).

Fact 6.8.

Let U be the U-statistics with kernel hk. If \({\sigma _{k}^{2}}<\infty \), then

$$\sqrt{m}\big[U-\mathbbm{E}(h^{k}(X_{1},...,X_{k}))\big] \rightarrow N(0,k^{2}({\sigma_{1}^{k}})^{2}) .$$

The asymptotic theory of U-statistics can be found in Ferguson (2003).

Theorem 6.9.

If τ2 < 1, and kmax = o(m), then under the null model,

$$\sqrt{m} \bigg[ \frac{1}{\binom{m}{k}}\sum\limits_{|\boldsymbol{\gamma}| =k} B_{\boldsymbol{\gamma}0}(x)- 1\bigg] \rightarrow N\bigg(0,k^{2}\bigg[\frac{1}{\sqrt{1-\tau^{4}}}-1\bigg]\bigg) .$$

Proof.

By Lemma 6.6, \(1/\binom {m}{k}{\sum }_{|\boldsymbol {\gamma }| =k}B_{\boldsymbol {\gamma }0}\) is a U-statistics, then evaluate the expectation and variance in Definition 6.7, one can get under the null model:

$$ \begin{array}{llll} \mathbbm{E}(B_{\boldsymbol{\gamma}0}(X))&= \mathbbm{E}({h^{k}_{1}}(X)) = 1 ,\\ \mathbbm{E}(B_{\boldsymbol{\gamma}0}(X)^{2}) &= \left\{\begin{array}{llll} \frac{1}{[(1+\tau^{2})(1-\tau^{2})]^{\frac{|\boldsymbol{\gamma}| }{2}}} &\text{ if } \tau^{2}<1 , \\ \infty &\text{ else, } \end{array}\right.\\ ({\sigma^{k}_{1}})^{2} &= \left\{\begin{array}{llll} \frac{1}{\sqrt{1-\tau^{4}}}-1 &\text{ if } \tau^{2}<1 ,\\ \infty &\text{ else, } \end{array}\right.\\ ({\sigma^{k}_{k}})^{2} &= \left\{\begin{array}{llll} \frac{1}{[(1+\tau^{2})(1-\tau^{2})]^{\frac{k}{2}}}-1 &\text{ if } \tau^{2}<1 , \\ \infty &\text{ else. } \end{array}\right. \end{array} $$

then apply Fact 6.8. □

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Chang, S., Berger, J.O. Frequentist Properties of Bayesian Multiplicity Control for Multiple Testing of Normal Means. Sankhya A 82, 310–329 (2020). https://doi.org/10.1007/s13171-019-00192-1

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Keywords

  • Multiple hypothesis testing
  • spike-and-slab prior
  • multiplicity control
  • false positive probability.

AMS (2000) subject classification.

  • Primary 62C10; Secondary 62J15