Assessing mean and median filters in multiple testing for large-scale imaging data

Abstract

A new multiple testing procedure, called the FDR _L procedure, was proposed by Zhang et al. (Ann Stat 39:613–642, 2011) for detecting the presence of spatial signals for large-scale 2D and 3D imaging data. In contrast to the conventional multiple testing procedure, the FDR _L procedure substitutes each p-value by a locally aggregated median filter of p-values. This paper examines the performance of another commonly used filter, mean filter, in the FDR _L procedure. It is demonstrated that when the p-values are independent and uniformly distributed under the true null hypotheses, (i) in view of estimating the resulting false discovery rate, the mean filter better alleviates the “lack of identification phenomenon” of the FDR _L procedure than the median filter; (ii) in view of signal detection, the median filter enjoys the “edge-preserving property” and lends support to its better performance in detecting sparse signals than the mean filter.

Appendix

Throughout the proof, since $\alpha_{\infty}^{\mathrm{FDR}}$ has been treated in the Appendix of Zhang et al. (2011), derivations will be confined to $\alpha_{\infty}^{\mathrm{FDR}_{L}}$ of the mean filter.

Condition A

• A0 The neighborhood size k is an integer not depending on n.

• A1 lim _n→∞ n ₀/n=π ₀ exists and π ₀<1.

Proof of Theorem 1

Note that for the original p-values,

$$ G_0(t)=t, \qquad G_1(t)=1-F_1 \bigl(F_0^{-1}(1-t)\bigr). $$

(A.1)

Also, for the “mean filtered” p ^∗(υ)-values, for any $\upsilon \in\mathcal{V}_{j}$, where j=0,1, we obtain

$$\begin{aligned} & \mathrm{P}\bigl\{p^*(\upsilon )\le t\bigr\} \\ &\quad = \mathrm{P}\bigl\{ \mathrm{mean}\bigl(p_1(\upsilon ),\ldots,p_k(\upsilon ) \bigr)\le t\bigr\} \\ &\quad = \mathrm{P}\bigl\{\mathrm{mean}\bigl(G_j \bigl(p_1(\upsilon )\bigr),\ldots ,G_j \bigl(p_k(\upsilon )\bigr)\bigr)\le G_j(t)\bigr\} \\ &\quad = \mathrm{P}\bigl\{\mathrm{mean}(\mathrm{U}_1,\ldots, \mathrm{U}_k)\le G_j(t)\bigr\} \\ &\quad = G_0^* \bigl(G_j(t)\bigr) = \begin{cases} G_0^*(t), & \mbox{if}\ \upsilon \in\mathcal{V}_0, \\ G_0^*(G_1(t)) = G_1^*(t), & \mbox{if}\ \upsilon \in\mathcal{V}_1. \end{cases} \end{aligned}$$

(A.2)

Thus, for p ^∗(υ)-values under true H ₀(υ), $G_{0}^{*}(t)$ is the C.D.F. corresponding to the p.d.f. in (3.7); for p ^∗(υ)-values under true H ₁(υ), $G_{1}^{*}(t)$ is the C.D.F. corresponding to the p.d.f. in (3.7) with t replaced by G ₁(t).

Part I. So for the FDR _L procedure, by (3.2),

$$\begin{aligned} \frac{dG_0^*(t)}{dt} =& \frac{k}{(k-1)!}\sum_{j=0}^{k} (-1)^{j} \binom{k}{j} \bigl\{(k t-j)_+\bigr\}^{k-1} \, \mathrm{I}(k t\ge j), \\ \frac{dG_1^*(t)}{dt} =& \frac{dG_0^*(G_1(t))}{dG_1(t)} \frac{dG_1(t)}{dt}, \\ =& \frac{k}{(k-1)!}\sum_{j=0}^{k} (-1)^{j} \binom{k}{j} \bigl[\bigl\{k G_1(t)-j\bigr \}_+\bigr]^{k-1} \,\mathrm{I}\bigl\{k G_1(t)\ge j\bigr\} \frac{dG_1(t)}{dt}. \end{aligned}$$

Applying L’Hospital’s rule and the fact lim _t→0+ G ₁(t)=0,

$$ \lim_{t\to0+} \frac{G_1(t)}{t} = \lim_{t\to0+} \frac{d G_1(t)}{dt}= \lim_{t\to0+}\frac{f_1(F_0^{-1}(1-t))}{f_0(F_0^{-1}(1-t))} = \lim _{x\to x_0-}\frac{f_1(x)}{f_0(x)} = \infty, $$

(A.3)

where $x = F_{0}^{-1}(1-t)$. Note that $\widehat{\mathrm{FDR}_{L}}^{\infty}(t)$ is a decreasing function of $G_{1}^{*}(t)/G_{0}^{*}(t)$. Using $\lim_{t\to0+} G_{1}^{*}(t)=0$, $\lim_{t\to0+} G_{0}^{*}(t)=0$, and (3.3),

$$\begin{aligned} \lim_{t\to0+} \frac{G_1^*(t)}{G_0^*(t)} =& \lim_{t\to0+} \frac{\frac{dG_1^*(t)}{dt}}{\frac{dG_0^*(t)}{dt}} \\ =& \lim_{t\to0+} \frac{\sum_{j=0}^{k} (-1)^{j} \binom{k}{j} [\{k G_1(t)-j\}_+]^{k-1} \,\mathrm{I}\{k G_1(t)\ge j\}}{ \sum_{j=0}^{k} (-1)^{j} \binom{k}{j} \{(k t-j)_+\}^{k-1} \,\mathrm {I}(k t\ge j)} \frac{dG_1(t)}{dt} \\ =& \lim_{t\to0+} \biggl\{\frac{G_1(t)}{t} \biggr\}^{k-1} \lim_{t\to 0+} \frac{dG_1(t)}{dt} \end{aligned}$$

(A.4)

which together with (A.3) shows $\lim_{t\to0+} G_{1}^{*}(t)/G_{0}^{*}(t) = \lim_{t\to0+} \{G_{1}(t)/t\}^{k} = \infty$. Thus, $\sup_{0 < t \le1} G_{1}^{*}(t)/G_{0}^{*}(t)=\infty$, that is, $\alpha_{\infty}^{\mathrm{FDR}_{L}} = 0$ for the FDR _L procedure.

Part II. Following $\widehat{\mathrm{FDR}_{L}}^{\infty}(t)$, we conclude that $\alpha_{\infty}^{\mathrm{FDR}_{L}} \ne0$ if

$$ \sup_{0 < t \le1}G_1^*(t)/G_0^*(t)<\infty . $$

(A.5)

We first verify (A.5) for the FDR _L procedure. Assume (A.5) fails, i.e., $\sup_{0<t \le 1} G_{1}^{*}(t) / G_{0}^{*}(t)=\infty$. Note that for any δ>0, the function $G_{1}^{*}(t)/G_{0}^{*}(t)$, for t∈[δ,1], is continuous and bounded away from ∞, thus, $\sup_{0<t \le1}G_{1}^{*}(t)/G_{0}^{*}(t)=\infty$ only if there exists a sequence t ₁>t ₂>⋯>0, such that lim _m→∞ t _m =0 and $\lim_{m\to\infty} G_{1}^{*}(t_{m})/G_{0}^{*}(t_{m}) = \infty$. For each m, recall that both $G_{1}^{*}(t)$ and $G_{0}^{*}(t)$ are continuous on [0,t _m ], and differentiable on (0,t _m ). Applying Cauchy’s mean-value theorem, there exists ξ _m ∈(0,t _m ) such that $G_{1}^{*}(t_{m})/G_{0}^{*}(t_{m}) = \{G_{1}^{*}(t_{m})-G_{1}^{*}(0)\}/{\{G_{0}^{*}(t_{m})-G_{0}^{*}(0)\}} = \frac{d G_{1}^{*}(t)/dt}{d G_{0}^{*}(t)/dt}|_{t=\xi_{m}}$. Since $\lim_{m\to\infty} G_{1}^{*}(t_{m})/G_{0}^{*}(t_{m}) = \infty$, it follows that $\limsup_{t\to0+}\frac{d G_{1}^{*}(t)/dt}{d G_{0}^{*}(t)/dt} = \infty$, which combined with (A.4) implies

$$ \limsup_{t\to0+}\frac{d G_1(t)}{d t} = \infty. $$

(A.6)

On the other hand, the condition $\limsup_{x\to x_{0}-} {f_{1}(x)}/{f_{0}(x)} < \infty$ indicates that

$$ \limsup_{t\to0+} \frac{d G_1(t)}{dt} = \limsup _{t\to0+}\frac{f_1(F_0^{-1}(1-t))}{f_0(F_0^{-1}(1-t))} = \limsup_{x\to x_0-} \frac{f_1(x)}{f_0(x)} <\infty, $$

(A.7)

where $x = F_{0}^{-1}(1-t)$. Clearly, (A.7) contradicts (A.6). The proof is completed. □

Proof of Theorem 2

We first show Lemma 1.

Lemma 1

Let B(t) be the C.D.F. of the p.d.f. corresponding to (3.7) with k≥3. Then I. for t∈(0,0.5), B(t)/t is a strictly increasing function and B(t)<t; II. for t∈(0.5,1), B(t)>t; III. for t ₁∈(0,0.5] and t ₂∈[t ₁,1], B(t ₁)/t ₁≤B(t ₂)/t ₂.

Proof

Since (3.7) is symmetric with respect to 0.5, we deduce that B(t)=1−B(1−t) and B′(t)=B′(1−t), i.e. the p.d.f. B′(t) is symmetric with respect to 0.5. It follows that

$$ B''(t)=(-1) B''(1-t), $$

(A.8)

namely, B″(t) is antisymmetric with respect to 0.5. More precisely, it is easy to see from (3.7) that

$$ B''(t)= \frac{k^2 (k-1)}{2(k-1)!}\sum _{j=0}^{k} (-1)^{j} \binom {k}{j} (k t-j)^{k-2} \mathrm{sign}(k t - j). $$

Hence from (3.4) and (A.8), the possible roots of B″(t) are at {0,0.5,1}. For positive t close to 0, B″(t) is a positive polynomial function of degree k−2.

To show part I, define F₁(t)=B(t)/t. Then $\mathrm{F}_{1}'(t)=\{B'(t)t-B(t)\}/t^{2}$, where d{B′(t)t−B(t)}/dt=B″(t)t. For t∈(0,0.5), (A.8) and the above analysis indicate B″(t)>0, i.e., B′(t)t−B(t) is strictly increasing, implying B′(t)t−B(t)>B′(0)0−B(0)=0. Hence for t∈(0,0.5), B(t)/t is strictly increasing and therefore B(t)/t<B(0.5)/0.5=1.

For part II, define F₂(t)=B(t)−t. Then $\mathrm{F}_{2}''(t)=B''(t)$. By (A.8), B″(t)<0 for t∈(0.5,1), thus F₂(t) is strictly concave, giving F₂(t)>max{F₂(0.5),F₂(1)}=0.

Last, we show part III. For t ₂∈[t ₁,0.5], part I indicates that B(t ₁)/t ₁≤B(t ₂)/t ₂; for t ₂∈[0.5,1], part II indicates that B(t ₂)/t ₂≥1 which, combined with B(t ₁)/t ₁≤1 from part I, yields B(t ₁)/t ₁≤B(t ₂)/t ₂. □

We now prove Theorem 2. It suffices to show that

$$\begin{aligned} \bigl\{1-G_1(\lambda)\bigr\}/{(1-\lambda)} \ge& \bigl \{1-G_1^*(\lambda)\bigr\}/\bigl\{1-G_0^*(\lambda)\bigr \}, \end{aligned}$$

(A.9)

$$\begin{aligned} \sup_{0< t\le1} {G_1(t)}/{t} \le& \sup _{0<t \le1} {G_1^*(t)}/{G_0^*(t)}. \end{aligned}$$

(A.10)

To verify (A.9), it suffices to show that $G_{1}(\lambda) \le G_{1}^{*}(\lambda)$ and $\lambda\ge G_{0}^{*}(\lambda)$. Following (A.2), for 0≤t≤1,

$$ G_1^*(t) =G_0^*\bigl(G_1(t)\bigr). $$

(A.11)

Applying (A.11), (A.1), $1-F_{0}(F_{1}^{-1}(0.5))\le\lambda$ and part II of Lemma 1 yields $G_{1}(\lambda) \le G_{1}^{*}(\lambda)$; applying λ≤0.5 and part I of Lemma 1 implies $\lambda \ge G_{0}^{*}(\lambda)$. This shows (A.9).

To verify (A.10), let M=sup_0<t≤1 G ₁(t)/t. Since G ₁(1)/1=1, we have M≥1 which will be discussed in two cases. Case 1: if M=1, then

$$ \sup_{0<t \le1} \frac{G_1^*(t)}{G_0^*(t)} \ge\frac{G_1^*(1)}{G_0^*(1)} = 1 = \sup _{0< t\le1} \frac{G_1(t)}{t}. $$

(A.12)

Case 2: if M>1, then there exist t ₀∈[0,1] and t _n ∈(0,1) such that lim _n→∞ t _n =t ₀, and

$$ \lim_{n\to\infty} \bigl\{G_1(t_n)/t_n \bigr\} =\sup_{0<t\le1} \bigl\{G_1(t)/t\bigr\} = M>1. $$

(A.13)

Thus, there exists N ₁ such that for all n>N ₁,

$$ G_1(t_n)>t_n. $$

(A.14)

Cases of t ₀=1, t ₀=0 and t ₀∈(0,1) will be discussed separately. First, if t ₀=1, then M=lim _n→∞{G ₁(t _n )/t _n }=lim _n→∞ G ₁(t _n )≤1, which contradicts (A.13). Thus t ₀<1. Second, if t ₀=0, then there exists N ₂ such that t _n <0.5 for all n>N ₂. Thus for all n>N≡max{N ₁,N ₂}, applying (A.11), (A.14) and part III of Lemma 1, we have

$$ {G_1^*(t_n)}/{G_1(t_n)} = {G_0^*\bigl(G_1(t_n)\bigr)}/{G_1(t_n)} \ge {G_0^*(t_n)}/{t_n}. $$

This together with (A.13) shows

$$ \sup_{0<t\le1}\frac{G_1^*(t)}{G_0^*(t)}\ge\limsup_{n\to\infty} \frac{G_1^*(t_n)}{G_0^*(t_n)} \ge \lim_{n\to\infty}\frac{G_1(t_n)}{t_n} = M =\sup _{0<t\le 1}\frac{G_1(t)}{t}. $$

(A.15)

Third, for t ₀∈(0,1), since both F ₀ and F ₁ are differentiable and f ₀ is supported in a single interval, $G_{1}(t)/t=\{1-F_{1}(F_{0}^{-1}(1-t))\}/t$ is differentiable in (0,1). Thus,

$$ \sup_{0<t\le1} {G_1(t)}/{t} = {G_1(t_0)} / {t_0} = M, $$

(A.16)

and ${d\{G_{1}(t)/t\}}/{dt}|_{t=t_{0}} = 0$. Notice

$$ \frac{d\{G_1(t)/t\}}{dt} \bigg|_{t=t_0}= \frac{\frac{dG_1(t)}{dt}|_{t=t_0}-G_1(t_0)/t_0}{t_0} = \frac{\frac {dG_1(t)}{dt}|_{t=t_0}-M}{t_0} = 0. $$

(A.17)

If t ₀>0.5, then $F_{0}^{-1}(1-t_{0})\le F_{0}^{-1}(0.5)$. By (A.3) and the assumption on f ₀ and f ₁, ${dG_{1}(t)}/{dt}|_{t=t_{0}} = f_{1}(F_{0}^{-1}(1-t_{0}))/f_{0}(F_{0}^{-1}(1-t_{0})) \le1$, which contradicts (A.17). Thus, 0<t ₀≤0.5. This together with (A.11), (A.16), and part III of Lemma 1 gives

$$ {G_1^*(t_0)}/{G_1(t_0)} = {G_0^*\bigl(G_1(t_0)\bigr)}/{G_1(t_0)} \ge {G_0^*(t_0)}/{t_0}. $$

This together with (A.16) shows,

$$ \sup_{0<t\le1}\frac{G_1^*(t)}{G_0^*(t)} \ge \frac{G_1^*(t_0)}{G_0^*(t_0)} \ge \frac{G_1(t_0)}{t_0} = M = \sup_{0<t\le1}\frac{G_1(t)}{t}. $$

(A.18)

Combining (A.12), (A.15) and (A.18) completes the proof.

Calculation of $\alpha_{\infty}^{\mathrm{FDR}_{L}}$ of the mean filter in Table 3 of Sect. 4.1

From (A.1) and the conditions given in Sect. 4.1,

$$ G_0(t) = t \quad\mbox{for}\ t\in[0,1], \quad\mbox{and} \quad G_1(t) = \begin{cases} te^C, &\mbox{if}\ t\in[0, e^{-C}], \\ 1, &\mbox{if}\ t\in(e^{-C}, 1]. \end{cases} $$

(A.19)

Now we compute $\alpha_{\infty}^{\mathrm{FDR}_{L}}$ of the mean filter. Recall that the distribution $G_{0}^{*}(t)$ is that of $\overline{\mathrm{U}}_{k}=\sum_{i=1}^{k} \mathrm{U}_{i}/k$, where $\{\mathrm{U}_{i}\}_{i=1}^{k} \stackrel {\text{i.i.d.}}{\sim} \mathrm{Unif}(0,1)$, and the distribution G ^∗(t) is also that of $\overline{\mathrm{U}}_{k}$. Similarly, by (A.2), the distribution $G_{1}^{*}(t)$ is that of $\overline{\mathrm{U}}_{k}/e^{C}$.

Hence, $\widehat{\mathrm{FDR}_{L}}^{\infty}(t) = [{\pi_{0}+\pi_{1}\{1-G_{1}^{*}(\lambda)\}/\{1- G_{0}^{*}(\lambda)\}}] /\{ \pi_{0}+\pi_{1} G_{1}^{*}(t)/G_{0}^{*}(t)\}$, yielding

$$ \alpha_\infty^{\mathrm{FDR}_{L}} = \frac{\pi_{0}+\pi_{1}\{ 1-G_1^*(\lambda)\}/\{1- G_0^*(\lambda)\}}{\pi_{0}+\pi_{1} \sup_{0<t\le1} G_1^*(t)/G_0^*(t)}. $$

Note that

$$ \frac{G_1^*(t)}{G_0^*(t)} = \frac{\mathrm{P}(\overline{\mathrm{U}}_k \le te^C)}{\mathrm {P}(\overline{\mathrm{U}}_k \le t)} = \begin{cases} \frac{\mathrm{P}(\overline{ \mathrm{U}}_k \le te^C)}{\mathrm {P}(\overline {\mathrm{U}}_k \le t)}, & \mbox{if}\ 0 \le t \le e^{-C}, \\ \frac{1}{\mathrm{P}(\overline{ \mathrm{U}}_k \le t)}, & \mbox{if}\ e^{-C}<t \le1. \end{cases} $$

Note that ${1}/{\mathrm{P}(\overline{ \mathrm{U}}_{k} \le t)}$ is decreasing in t. By a graphical approach for k=5, ${\mathrm{P}(\overline{ \mathrm{U}}_{k} \le te^{C})}/{\mathrm {P}(\overline {\mathrm{U}}_{k} \le t)}$ is also decreasing in t. Thus, $\sup_{t\in(0, 1]}G_{1}^{*}(t)/G_{0}^{*}(t) = \lim_{t\to0+} G_{1}^{*}(t)/G_{0}^{*}(t) = \lim_{t\to0+} \{G_{1}(t)/t\}^{k} = e^{Ck}$ from (A.4) and (A.19). So $\alpha_{\infty}^{\mathrm{FDR}_{L}} = [{\pi_{0}+\pi_{1}\{ 1-G_{1}^{*}(\lambda)\}/\{1- G_{0}^{*}(\lambda)\}}] / (\pi_{0}+\pi_{1} e^{Ck})$, where $G_{0}^{*}(\cdot)$ can be calculated from (3.5). The completes the derivation.