Tests for Patterned Alternatives Using Logarithmic Quantile Estimation

Abstract

We investigate the logarithmic quantile estimation (LQE) method using fully nonparametric rank statistics to test for known trend and umbrella patterns in the main effects of three widely used designs: a fixed-effect two-factor model, a mixed-effect repeated measures model, and a mixed-effect cross-classification model. We also test for patterned alternatives in the interaction between the main effect and time in the repeated measures model. We determine the level and power of the test statistics using LQE with simulated and real data. The LQE procedure uses only the data to estimate quantiles of test statistics and does not require the estimation of the asymptotic variance nor the Satterthwaite–Smith degrees of freedom estimation. Our results show that the LQE method commonly yields conservative tests with high power when testing for patterned alternatives.

Appendix

We sketch the proofs of Propositions 1 and 2 following the ideas from [13, 14]. For convenience, we present here the formulas from [13] that we use in our derivations (see Appendix, page 2948). [13] showed that a simple linear rank statistic based on n independent random vectors of possibly unequal lengths satisfies the almost sure central limit theorem. In our case, we use n independent random vectors of equal length ${{{\mathbf{X_1}}}},\dots ,{{{\mathbf{X_n}}}}$, having components ${{{\mathbf{X_u}}}}=(X_{u1},$ $\dots ,$ $X_{um})$ ($u=1,\dots ,n$) and $N(n)=nm$ total number of observations. We now define the simple linear rank statistic (see [14], formulas (6)-(10))

$$\begin{aligned} T_n=\frac{1}{N(n)}\sum _{u=1}^{n}\sum _{v=1}^{m}\lambda _{uv}^{(n)}\frac{R_{uv}(n)}{N(n)+1}-\int _{-\infty }^{\infty }HdF_n, \end{aligned}$$

(31)

which has the following term that appears in its Taylor decomposition

$$\begin{aligned} B_n=\int _{-\infty }^{\infty }Hd({\hat{F}}_n-F_n)+\int _{-\infty }^{\infty }({\hat{H}}_n-H)dF_n, \text { and} \end{aligned}$$

(32)

$$\begin{aligned} \sigma _n^2=N^2(n)Var(B_n). \end{aligned}$$

(33)

$R_{uv}(n)$ denotes the rank of observation $X_{uv}\backsim F_{uv}$ among all N(n) random variables, and $\lambda _{uv}^{(n)}$ are known regression constants with maximum absolute value of one. We define the distribution function weighted by the regression constants for the n independent random vectors

$$\begin{aligned} F_n(x)=\frac{1}{mn}\sum _{u=1}^{n}\sum _{v=1}^{m}\lambda _{uv}^{(n)}F_{uv}(x), \end{aligned}$$

(34)

and its empirical form

$$\begin{aligned} {\hat{F}}_n(x)=\frac{1}{mn}\sum _{u=1}^{n}\sum _{v=1}^{m}\lambda _{uv}^{(n)}{\mathbb {I}}(X_{uv}\le x). \end{aligned}$$

(35)

In the proofs equations (31) through (35) are specified for each model and used to prove the almost sure central limit theorems. The almost sure weak convergence of the linear rank statistic $T_n$ in equation (31) can be obtained by showing the almost sure weak convergence of $B_n$ in equation (32). By verifying the conditions in Theorem 3.1 in [19], we obtain the almost sure weak convergence of $B_n$ in (32) (see Appendix in [13]).

Proof of Proposition 1

In the two-way fixed-effects model introduced in Section 2.1, $X_{ijk} \backsim F_{ij}$ are independent random variables, and $N= abn$ is the total number of subjects. We define independent random vectors ${{{\mathbf{X}}}_{{{\mathbf{ik}}}}}= (X_{i1k}, \dots , X_{ibk})'$, $1\le i\le a$, $1\le k \le n$. Let $1\le l\le a$ denote the $l^{th}$ level of factor A and define

$$\begin{aligned} \lambda _{ij}^{(n)}= {\left\{ \begin{array}{ll} 1, &{} i=l \text {, }j=1,\dots ,b\\ 0, &{} otherwise. \end{array}\right. } \end{aligned}$$

(36)

Equations (34) and (35) become

$$\begin{aligned} F^{(l)}_n(x)= & {} \frac{1}{ab}\sum _{j=1}^{b}F_{lj}(x),\text { and} \\ {\hat{F}}^{(l)}_n(x)= & {} \frac{1}{abn}\sum _{j=1}^{b}\sum _{k=1}^{n}{\mathbb {I}}(X_{ljk}\le x), \end{aligned}$$

which are the average distribution function across the levels of factor B (groups) for a fixed time ($i=l$) of factor A, and its empirical analog, respectively. The linear rank statistic is

$$\begin{aligned} T^{(l)}_n=\frac{1}{abn(abn+1)}\sum _{j=1}^{b}\sum _{k=1}^{n}R_{ljk}-\frac{1}{ab}\sum _{j=1}^{b}\int _{-\infty }^{\infty }H(x)dF_{lj}(x), \end{aligned}$$

where H(x) is defined in (1), and

$$\begin{aligned} B^{(l)}_n= & {} \frac{1}{N}\sum _{k=1}^{n} \sum _{j=1}^{b}H(X_{ljk})-\frac{2}{N}\sum _{k=1}^{n}\sum _{j=1}^{b}\int _{-\infty }^{\infty }H(x)dF_{lj}(x) \\&+\frac{1}{N}\sum _{k=1}^{n}\sum _{s=1}^{a}\sum _{j=1}^{b}\int _{-\infty }^{\infty }{\mathbb {I}}(X_{sjk}\le x)d\left( \frac{1}{ab}\sum _{t=1}^{b}F_{lt}(x)\right) . \end{aligned}$$

We note that under $H_0^F(A)$ in (10) the test statistic $P_N^{(\text {fix})}(A)$ in (13) can be expressed as

$$\begin{aligned} P_N^{(\text {fix})}(A)=\frac{a(N+1)}{\sqrt{N}}\sum _{i=1}^{a}(w_i-{\bar{w}})T^{(i)}_n. \end{aligned}$$

(37)

The almost sure weak convergence of the vector

$$\begin{aligned} {{\tilde{{{\mathbf{T}}}}}_{{\mathbf{n}}}}=\left( \frac{a(N+1)}{\sqrt{N}}(w_i-{\bar{w}})T^{(i)}_n\right) _{1\le i \le a} \end{aligned}$$

(38)

can be obtained by showing the almost sure weak convergence of the vector

$$\begin{aligned} {{\tilde{{\mathbf{B}}}}_{{\mathbf{n}}}}=\left( \frac{a(N+1)}{\sqrt{N}}(w_i-{\bar{w}})B^{(i)}_n\right) _{1\le i \le a}. \end{aligned}$$

(39)

As in [13], we can express ${{\tilde{{\mathbf{B}}}}_{{\mathbf{n}}}}$ as a sum of a dimensional independent random vectors ${{{\mathbf{Z_k}}}}$, which are bounded with $E{{{\mathbf{Z_k}}}}={{{\mathbf{0}}}}$

$$\begin{aligned} {{\tilde{{{\mathbf{B}}}}}_{{\mathbf{n}}}}=\sqrt{\frac{a}{b}}\left( \frac{abn+1}{abn}\right) \frac{1}{\sqrt{n}}\sum _{k=1}^{n}{{{\mathbf{Z_k}}}}, \end{aligned}$$

(40)

where ${\mathbf{Z_k}}$ has components

$$\begin{aligned} Z_{ki}=\\ (w_i-{\bar{w}})\sum _{j=1}^{b}\left[ H(X_{ijk})-2\int HdF_{ij}+\frac{1}{ab}\sum _{s=1}^{a}\int {\mathbb {I}}\left( X_{sjk}\le x\right) d\left( \sum _{t=1}^{b}F_{it}\right) \right] . \end{aligned}$$

By the multivariate central limit theorem (MVCLT) and under Assumption 1, the vector

$$\begin{aligned} {{\tilde{{\mathbf{B}}}}_{{\mathbf{n}}}}\overset{d}{\longrightarrow }{\mathscr {N}}\left( {{{\mathbf{0}}}},\frac{a}{b} {\varvec{{\varSigma }}} \right) ,&\text { as }n\rightarrow \infty . \end{aligned}$$

(41)

It can be verified using Theorem 3.1 of [19] that

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{1}{\log n}\sum _{k=1}^{n}\frac{1}{k}{\mathbb {I}}\left( \frac{1}{\sqrt{k}}\sum _{l=1}^{k}{{{\mathbf{Z_l}}}}\le {{{\mathbf{x}}}}\right) =G_X({{{\mathbf{x}}}}),\text { }a.s.,\text { }\forall {{{\mathbf{x}}}}\in {\mathbb {R}}^a, \end{aligned}$$

(42)

where $G_X$ is the distribution function of ${{{\mathbf{X}}}}\backsim {\mathscr {N}}({{{\mathbf{0}}}},{\varvec{{\varSigma }}})$. Lemma 2.2 of [16] implies

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{1}{\log n}\sum _{k=1}^{n}\frac{1}{k}{\mathbb {I}}\left( {{\tilde{{{\mathbf{B}}}}}_{{\mathbf{k}}}}\le {{\mathbf{x}}}\right) =G_X\left( {{\mathbf{x}}}\sqrt{\frac{a}{b}}\right) ,\text { }a.s.,\text { }\forall {{\mathbf{x}}}\in {\mathbb {R}}^a. \end{aligned}$$

(43)

For the continuous function $f:{\mathbb {R}}^a\longrightarrow {\mathbb {R}}$, $f(x_1,\dots ,x_a)=\sum _{i=1}^{a}x_i$, and by [19] and the techniques in [24] equation (43) is equivalent to

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{1}{\log n}\sum _{k=1}^{n}\frac{1}{k}{\mathbb {I}}\left( f({{\tilde{{\mathbf{T}}}}_{{\mathbf{k}}}})\le x\right) =G_{f(X)}\left( x\sqrt{\frac{a}{b}}\right) ,\text { }a.s.,\text { }\forall x\in {\mathbb {R}}. \end{aligned}$$

(44)

Hence, the almost sure weak convergence of the test statistic (37) follows. $\square$

Proof of Proposition 2

Recall that each of the bn subjects in the partial hierarchical model of Section 2.2 can be expressed as independent random vectors ${{{\mathbf{X}}}_{{{\mathbf{jk}}}}}$ $=$ $\left( X_{1jk},\dots ,X_{ajk}\right) '$, $1\le j\le 2, 1\le k\le n$, where $X_{ijk}\backsim F_{ij}$. For fixed l,v such that $1\le l \le a$, $1 \le v \le b$ let

$$\begin{aligned} \lambda _{ij}^{(n)}={\left\{ \begin{array}{ll} 1, &{} i=l, \text { } j=v \\ 0, &{} otherwise. \end{array}\right. } \end{aligned}$$

(45)

Then, the linear rank statistic has the form

$$\begin{aligned} T^{(l,v)}_n=\frac{1}{N}\sum _{k=1}^{n}\frac{R_{lvk}}{N+1}-\frac{n}{N}\int _{-\infty }^{\infty }H(x)dF_{lv}(x), \end{aligned}$$

(46)

and

$$\begin{aligned}&B_n^{(l,v)} \nonumber \\&\quad =\frac{1}{N}\sum _{k=1}^{n}\left[ H(X_{lvk})-2\int HdF_{lv} +\frac{1}{N}\sum _{s=1}^{a}\sum _{t=1}^{b}\sum _{u=1}^{n}\int {\mathbb {I}}\left( X_{stu}\le x\right) dF_{lv}\right] . \end{aligned}$$

(47)

The test statistic in (16) can be written

$$\begin{aligned} P_{bn}^{(ph)}(A)=\sqrt{\frac{a}{b}}\text { } \frac{(abn+1)}{\sqrt{n}}\sum _{i=1}^{a}\sum _{j=1}^{b}(w_i-{\bar{w}}) T_n^{(i,j)}. \end{aligned}$$

(48)

Consider the vectors

$$\begin{aligned} {{\tilde{{\mathbf{T}}}}_{{\mathbf{n}}}}=\left( \frac{\sqrt{a}(N+1)(w_i-{\bar{w}})}{\sqrt{bn}}T_n^{(i,j)}\right) _{1\le i\le a, 1\le j \le b}, \end{aligned}$$

(49)

and

$$\begin{aligned} {{\tilde{{\mathbf{B}}}}_{{\mathbf{n}}}}=\frac{\sqrt{a}(abn+1)}{\sqrt{bn}}\left( w_i-{\bar{w}}) B_n^{(i,j)}\right) _{1\le i \le a, 1\le j \le b}. \end{aligned}$$

(50)

${{\tilde{{{\mathbf{B}}}}}_{{\mathbf{n}}}}$ can be expressed as a sum of a dimensional independent random vectors

$$\begin{aligned} {{\tilde{{\mathbf{B}}}}_{{\mathbf{n}}}}=\sqrt{\frac{a}{b}}\text { } \frac{(abn+1)}{abn}\frac{1}{\sqrt{n}}\sum _{k=1}^{n} {{{\mathbf{Z_k}}}}, \end{aligned}$$

(51)

where components of the vector ${{{\mathbf{Z_k}}}}$ are

$$\begin{aligned} {{{\mathbf{Z}}}_{{{\mathbf{ki}}}}}=\left( (w_i-{\bar{w}})\left[ H(X_{ijk})+\frac{1}{N}\sum _{s=1}^{a}\sum _{t=1}^{b}\sum _{u=1}^{n}\int {\mathbb {I}}(X_{stu}\le x)dF_{ij}\right] \right) _{1 \le j \le b}, \end{aligned}$$

(52)

for $1\le i \le a$. By the multivariate central limit theorem (MVCLT) and under Assumption 1, the vector

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{k=1}^{n}{{{\mathbf{Z_k}}}}\rightarrow {\mathscr {N}}({{{\mathbf{0}}}},{\varvec{{\varSigma }}}),&\text { as }n\rightarrow \infty , \end{aligned}$$

(53)

and hence,

$$\begin{aligned} {{\tilde{{\mathbf{B}}}}_{{\mathbf{n}}}}\rightarrow {\mathscr {N}}\left( {{{\mathbf{0}}}},\frac{a}{b}{\varvec{{\varSigma }}}\right) ,&\text { as }n\rightarrow \infty . \end{aligned}$$

(54)

The proof follows as in Proposition 1 by taking the function $f:{\mathbb {R}}^{ab}\rightarrow {\mathbb {R}}$, $f(x_1,\dots ,x_{ab})=\sum _{i=1}^{ab}x_i$. $\square$