Novel Test for the Equality of Continuous Curves with Homoscedastic or Heteroscedastic Measurement Errors

Abstract

Testing equality of two curves occurs often in functional data analysis. In this paper, we develop procedures for testing if two curves measured with either homoscedastic or heteroscedastic errors are equal. The method is applicable to a general class of curves. Compared with existing tests, ours does not require repeated measurements to obtain the variances at each of the explanatory values. Instead, our test calculates the overall variances by pooling all of the data points. The null distribution of the test statistic is derived and an approximation formula to calculate the p value is developed when the heteroscedastic variances are either known or unknown. Simulations are conducted to show that this procedure works well in the finite sample situation. Comparisons with other test procedures are made based on simulated data sets. Applications to our motivating example from an environmental study will be illustrated. An R package was created for ease of general applications.

Appendices

Appendix

Appendix A. Proof of Lemma 4.1

Let $X_1=\sum _{i=1}^{n_1}W_i^2,$ where $W_i\sim _{ iid} N(0,1).$ Since $EX_1=n_1, Var(X_1)=2n_1,$ we can approximate $X_1$ by $X_1 {\mathop {=}\limits ^{d}} n_1+\sqrt{2n_1} Z_1 + o(\sqrt{n_1}),$ $\;\;X_2$ by $X_2{\mathop {=}\limits ^{d}} n_2+\sqrt{2n_2} Z_2 + o(\sqrt{n_2}),$ where $Z_1$ and $Z_2$ are independent standard normal random variables and notation ${\mathop {=}\limits ^{d}}$ denotes equality in distribution. Therefore,

$$\begin{aligned} G_i=&( {\frac{X_i }{X_1+X_2 }} )/ ({ \frac{n_i}{n_1+n_2}}) \\ {\mathop {=}\limits ^{d}}&\frac{n_i+\sqrt{2n_i} Z_i + o(\sqrt{n_i})}{n_1+n_2 +\sqrt{2n_1} Z_1 +\sqrt{2n_2} Z_2 + o(\sqrt{n_2})+ o(\sqrt{n_1})}/({ \frac{n_i}{n_1+n_2}}) \\ =&\frac{ 1+\frac{\sqrt{2n_i} Z_i + o(\sqrt{n_i})}{n_i}}{1 +\frac{\sqrt{2n_1} Z_1 +\sqrt{2n_2} Z_2 + o(\sqrt{n_2})+ o(\sqrt{n_1})}{n_1+n_2}} {\mathop {\sim }\limits ^{d}} _{approx} 1, \end{aligned}$$

as $n_i \rightarrow \infty$ for $i=1,2.$ QED

Appendix B. Proof of Lemma 4.2

Clearly by $S_i/\nu _i \sim \chi ^2_{\nu _i},$ we have $ES_i=1, ~~Var(S_i)=2/\nu _i$, and there exists a sequence of i.i.d. standard normal random variables $W_{i,k},$ for $k=1, 2, \dots , \nu _i, i=1,2$ such that $S_i= (1/\nu _i) {\sum _{k=1}^{\nu _i} W_{i,k}}$ by definition. A large sample theory gives that $S_i{\mathop {=}\limits ^{d}} 1+\sqrt{\frac{2}{\nu _i}} Z_i+o(\frac{1}{\sqrt{\nu _i}})$, where $Z_i \sim N(0,1)$.

Expanding $Y=f(S_1, S_2)= (X_1+X_2 )/(X_1/S_1+X_2/S_2)$ around (1, 1),  we have:

$$\begin{aligned} Y=f(S_1, S_2)=&1+\sqrt{ \frac{2}{\nu _1} }Z_1 \frac{ X_1}{X_1+X_2 }+ \sqrt{\frac{2}{\nu _2} }Z_2 \frac{ X_2}{X_1+X_2 }+ o(\frac{1}{\sqrt{\nu _1}}+\frac{1}{\sqrt{\nu _2}}). \end{aligned}$$

It is easy to find that $EY\rightarrow 1.$ Conditioning on $X_1, X_2$, we have

$$\begin{aligned} Var(Y) =&Var(E(Y|X_1, X_2)) +E (Var(Y|X_1, X_2)) \\ =&E (Var(Y|X_1, X_2)) +o(\frac{1}{\sqrt{\nu _1}}+\frac{1}{\sqrt{\nu _2}})\\ \approx&E \left\{ Var( \sqrt{\frac{2}{\nu _1}}Z_1 \frac{ X_1}{X_1+X_2 } + \sqrt{\frac{2}{\nu _2}}Z_2 \frac{ X_2}{X_1+X_2 } |X_1, X_2)\right\} \\ =&E \left\{ \frac{2}{\nu _1} (\frac{ X_1}{X_1+X_2 })^2 + \frac{2}{\nu _2} (\frac{ X_2}{X_1+X_2 })^2 \right\} \\ =&\frac{2}{\nu _1} E (\frac{ X_1}{X_1+X_2 })^2 + \frac{2}{\nu _2} E(\frac{ X_2}{X_1+X_2 })^2 \\ \approx&\frac{2}{\nu _1} (\frac{n_1}{n_1+n_2})^2 + \frac{2}{\nu _2} (\frac{ n_2}{ n_1+n_2})^2 \qquad \qquad \qquad {\text {(by Lemma (4.1))}} \\ =&\frac{2n_1^2\nu _2 +2n_2^2\nu _1}{(n_1+n_2)^2\nu _1\nu _2}. \end{aligned}$$

Thus by comparing $Var{(\chi ^2_\nu /\nu )}=2/\nu$ with $Var(Y)=(2n_1^2\nu _2 +2n_2^2\nu _1)/((n_1+n_2)^2\nu _1\nu _2)$, we have estimate $\nu$ in formula (4.3). QED

Appendix C. Proof of Theorem 5.1

Z(t) in formula (5.6) can be written as $~~Z(t)=\left\langle u_1(t), \; \xi _1 \right\rangle$ $- \left\langle u_2(t),\; \xi _2 \right\rangle$ $=\left\langle \mathcal {M}(t),\; \xi \right\rangle ,$ where $\mathcal {M}(t)= \left( u_1(t), -u_2(t)\right) ' \in \mathcal {S}^{n-1}$ and $\xi = \left( \xi _1, \xi _2\right) ' = \left( \varepsilon _1/\sigma , \varepsilon _2/\sigma \right) '\in \mathcal {R}^n$ is standard multivariate normal. Conditioning on $\Vert \xi \Vert ,$ the probability can be written as

$$\begin{aligned} Pr(T\ge t_0) =&Pr(\sup _{t\in \mathcal {T}} |\left\langle \mathcal {M}(t),\;\xi \right\rangle |\ge t_0) \nonumber \\ =&\int _{t_0}^{\infty } Pr(\sup _{t\in \mathcal {T}}\left| \left\langle \mathcal {M}(t),\; \frac{\xi }{\Vert \xi \Vert }\right\rangle \right| \ge \frac{t_0}{y}\; \;|\;\; \Vert \xi \Vert =y)g(y, n)dy \end{aligned}$$

(10.1)

where g(y, n) is the probability density function (pdf) of the square root of a $\chi ^2$ random variable with n degrees of freedom. Since $U={\xi }/{\Vert \xi \Vert } \sim uniform (\mathcal {S}^{n-1})$ is independent of ${\Vert \xi \Vert },$ we can drop the condition in the probability. Let $T=\{x\in S^{n-1}:\; \sup _{t\in \mathcal {T}}\left| <\mathcal {M}(t),\; x>\right| \ge ( {t_0}/ {y})\}$, we then have tubes around curve $\mathcal {M}(t)$ and curve $-\mathcal {M}(t)$ embedded in $S^{n-1}$, with radius $r =\sqrt{2-2{t_0}/{y}}$ (See relation (4.4)). The probability inside the integral of (10.1) can be calculated by ${Vol(T)} /{Vol(S^{n-1})}.$ We then plug-in the tube formula (4.5) to get result (5.9). See also [22] [ Proposition 1, p. 1330]. Result (5.10) is obtained by replacing g(y, n) of the pdf of ${\Vert \xi \Vert } =\sqrt{\varepsilon '\varepsilon / \sigma ^2}$ by the pdf of $\Vert \hat{\xi }\Vert =\sqrt{\varepsilon '\varepsilon / \hat{\sigma }^2},$ where $\Vert \hat{\xi }\Vert ^2/n \sim F_{n, \nu }$. For details, see the above citation. QED

Appendix D. Proof of Theorem 5.2

Suppose $\sigma _1^2$ and $\sigma _2^2$ are known. Let $\mathcal {M}(t)=\left( u_1(t), \;\; u_2(t)\right) ' \in \mathcal {S}^{n-1} \subseteq \mathcal {R}^n$ for $n=n_1+n_2,\; t\in \mathcal {T},$ as before. Let $\xi =\left( \xi _1, \; \; \xi _2 \right) '$ and $U= { \xi }/{ \Vert \xi \Vert }.$ Then U is uniformly distributed (over $\mathcal {S}^{n-1}$), independent of $\Vert \xi \Vert$. Since $\sigma _1$ and $\sigma _2$ are known, $\Vert \xi \Vert ^2$ will follow a $\chi ^2_n$ distribution.

Conditioning on value $\Vert \xi \Vert ,$ making use of the fact that $\varepsilon _1$ and $\varepsilon _2$ are independent, we have

$$\begin{aligned} Pr(T>t_0)=&Pr(\sup _{t\in {\mathcal {T}}}\Vert Z(t)\Vert \ge t_0 ) =Pr(\sup _{t\in {\mathcal {T}}}\langle {\mathcal {M}}(t), \xi \rangle \;\; \ge t_0)\\ =&Pr(\sup _{t\in {\mathcal {T}}}\langle {\mathcal {M}}(t), \frac{\xi }{ \Vert \xi \Vert } \rangle \ge \frac{t_0 }{ \Vert \xi \Vert })\\ =&\int _{y\ge t_0}Pr(\sup _{t\in {\mathcal {T}}}\langle {\mathcal {M}}(t), U \rangle \ge \frac{t_0 }{y} \mid \Vert \xi \Vert =y) f_{\Vert \xi \Vert }(y)dy\\ =&\int _{y\ge t_0}Pr(\sup _{t\in {\mathcal {T}}}\langle {\mathcal {M}}(t), U \rangle \ge \frac{t_0 }{ y}) f_{\Vert \xi \Vert }(y)dy, \end{aligned}$$

where $f_{\Vert \xi \Vert }(y)$ denotes the pdf of $\Vert \xi \Vert ,$ whose square follows a $\chi ^2_n$ distribution. The last equation holds because of the fact that $\xi /\Vert \xi \Vert$ and $\Vert \xi \Vert$ are independent.

When $t_0$ is large, the following tube formula [cf: Lemma 4.5] can be plugged into the last equation

$$Pr(\sup _{t\in {\mathcal {T}}}\Vert \langle {\mathcal {M}}(t), U \rangle \Vert \ge c)\approx 2*\left\{ \frac{\kappa _0}{2\pi } (1-c^2)^{n/2-1}+\frac{E}{4} Pr(\beta _{\{1/2, (n-1)/2\}}\ge c^2\right\} $$

where $\beta _{\{1/2, (n-1)/2\}}$ denotes a Beta random variable with parameters 1/2 and (n–1)/2. The factor 2 corresponds to the two curves satisfying the probability condition: one is ${\mathcal {M}}(t)$, another $-{\mathcal {M}}(t).$ This gives formula (5.17).

Suppose we don’t know both $\sigma ^2_i,i=1,2, \;$ but they are estimated as in formula (5.5). Let

$$\begin{aligned}&X_i= \frac{\varepsilon '_i\varepsilon _i }{\sigma _i^2},&S_i= \frac{\hat{\sigma }^2_i }{\sigma ^2_i},\; \;\;\;\;\; i=1,2, \end{aligned}$$

(10.2)

$$\begin{aligned}&Y= \frac{X_1}{S_1} +\frac{X_2}{S_2},&X=\frac{X_1+X_2 }{ X_1/S_1+X_2/S_2 }=\frac{X_1+X_2 }{Y}. \end{aligned}$$

(10.3)

Then the requirements of Lemma 4.2 are satisfied, hence we have $X\sim _{approx}\chi ^2_{\nu }/\nu ,$ with degrees of freedom $\nu$ estimated in formula (4.3). Therefore, $Y= ({X_1+X_2 })/{X}\rightarrow (\chi ^2_n)/(\chi ^2_{\nu }/\nu )\sim nF_{n,\nu }.$

Now let

$$\begin{aligned}&{\hat{\mathcal {M}}(t) }=\frac{(\hat{\sigma }_{1}{\mathbf {l}}_1(t) , -\hat{\sigma }_{2}{\mathbf {l}}_2(t) )}{\sqrt{\hat{\sigma }_{1}^2 \left\| {\mathbf {l}}_1(t) \right\| ^2+ \hat{\sigma }_{2}^2 \left\| {\mathbf {l}}_2(t) \right\| ^2}} \in \mathcal {S}^{n-1},&U=\frac{\hat{\xi }}{ \Vert {\hat{\xi }} \Vert }, \;\;\;\;\; \\&~~~~~~\hat{\xi }= (\frac{\varepsilon _1}{\hat{\sigma }_{1}}, \frac{\varepsilon _2}{\hat{\sigma }_{2}})\in \mathcal {R}^n,&\hat{Z}(t)=<{\hat{\mathcal {M}}(t) }, \hat{\xi }>. \end{aligned}$$

Then

$$\begin{aligned} Pr(T>t_0)=&Pr(\sup _{t\in {\mathcal {T}}}\Vert \hat{Z}(t)\Vert \ge t_0 ) =Pr(\sup _{t\in {\mathcal {T}}}\langle {\hat{\mathcal {M}}(t) },\hat{\xi }\rangle \;\; \ge t_0)\\ =&Pr(\sup _{t\in {\mathcal {T}}}\langle {\hat{\mathcal {M}}(t) }, \frac{\hat{\xi }}{\Vert {\hat{\xi }} \Vert } \rangle \ge \frac{t_0 }{\Vert {\hat{\xi }} \Vert })\\ =&\int _{y\ge t_0}Pr(\sup _{t\in {\mathcal {T}}}\langle \hat{\mathcal {M}}(t), U \rangle \ge \frac{t_0 }{y} \mid \Vert {\hat{\xi }} \Vert =y) f_{\Vert {\hat{\xi }} \Vert }(y)dy\\ =&\int _{y\ge t_0}Pr(\sup _{t\in {\mathcal {T}}}\langle \hat{\mathcal {M}}(t), U \rangle \ge \frac{t_0 }{ y}) f_{\Vert {\hat{\xi }} \Vert }(y)dy, \end{aligned}$$

where ${Y}=:\Vert {\hat{\xi }} \Vert ^2\sim _{approx} \chi ^2_{\nu }/\nu ,$ as Y was defined in (4.2).

Let the $\sigma _i^2$ be known values in ${\hat{\mathcal {M}}(t) }$ in order to estimate $\kappa _0,$ so that we can use the Tube formula (4.5) for the estimation of the probability inside the integral. After some calculation, we get result (5.18). QED

Appendix E. Software

Our R package curvetest is freely available at http://www.r-project.org/. This package tests the equality of curves as described in this paper. The main function curvetest has the following parameters:  

formula::

specified the regression formula.

data1::

data.frame representing the first (discretized) curve.

data2::

a data frame representing the second curve. If it is NULL, then the test is test $f(t)==0.$ data1 and data2 must have two columns with same column names, that can be retrieved by calls on the formula.

equal.var::

logic value, specifies if equal.variances are assumed. Default=TRUE.

plotit::

logic, asks if curvetest should generate the scatter plots and smoothing curves. It is useful to plot it to select the window width bw below. Default=F.

bw::

Window bandwidth for both curves.

alpha::

Smoothing parameter. Default=0.5.

nn::

number of points used to smooth the curves. The points are equally spaced between the domains that appeared in the two data set. Default=100.

myx::

x (or t) values to estimate the curves. Default= NULL. This will put n points specified by nn in the data range. If myx is non-null, papameter nn will be suppressed.

bcorrect::

method to use for boundary correction. Default=‘simple’. Other options are: ’none’=no corrections.

Conf.level::

the $\alpha$ value for the type I error level. Default=.05.

kernel::

kernel function to choose for smoothing. Users can choose one of ‘Trio’, ‘Gaussian’, ‘Uniform’, ‘Triweight’, ‘Triangle’, ‘Epanechnikov’, ‘Quartic’. See the definitions of them in Table 2.

 

Table 2 List of kernel functions

Usage: