Tests of Concentration for Low-Dimensional and High-Dimensional Directional Data

Abstract

We consider asymptotic inference for the concentration of directional data. More precisely, we propose tests for concentration (1) in the low-dimensional case where the sample size n goes to infinity and the dimension p remains fixed, and (2) in the high-dimensional case where both n and p become arbitrarily large. To the best of our knowledge, the tests we provide are the first procedures for concentration that are valid in the (n, p)-asymptotic framework. Throughout, we consider parametric FvML tests, that are guaranteed to meet asymptotically the nominal level constraint under FvML distributions only, as well as “pseudo-FvML” versions of such tests, that meet asymptotically the nominal level constraint within the whole class of rotationally symmetric distributions. We conduct a Monte-Carlo study to check our asymptotic results and to investigate the finite-sample behavior of the proposed tests.

Appendix

Proof of Theorem 1

 (i) All expectations and variances when proving Part (i) of the theorem are taken under $\mathcal{R}_{p}^{(n)}(\boldsymbol{\theta },F)$ and all stochastic convergences are taken as n → ∞ under $\mathcal{R}_{p}^{(n)}(\boldsymbol{\theta },F)$. Since

$$\displaystyle{ n^{1/2}(\bar{\mathbf{X}}_{ n} - e_{10}\boldsymbol{\theta }) = O_{\mathrm{P}}(1), }$$

(5)

the delta method (applied to the mapping x ↦ x∕ ∥ x ∥ ) yields

$$\displaystyle{ n^{1/2}(\mathbf{Y}_{ n}-\boldsymbol{\theta }) = e_{10}^{-1}[\mathbf{I}_{ p} -\boldsymbol{\theta }\boldsymbol{\theta }^{{\prime}}]n^{1/2}(\bar{\mathbf{X}}_{ n} - e_{10}\boldsymbol{\theta }) + o_{\mathrm{P}}(1), }$$

(6)

where we wrote $\mathbf{Y}_{n}:=\bar{ \mathbf{X}}_{n}/\|\bar{\mathbf{X}}_{n}\|$. This, and the fact that

$$\displaystyle{\mathbf{S}_{n}\stackrel{\mathrm{P}}{\rightarrow }\mathrm{E}[\mathbf{X}_{1}\mathbf{X}_{1}^{{\prime}}] = \mathrm{E}[(\mathbf{X}_{ 1}^{{\prime}}\boldsymbol{\theta })^{2}]\boldsymbol{\theta }\boldsymbol{\theta }' + \frac{1 -\mathrm{E}[(\mathbf{X}_{1}^{{\prime}}\boldsymbol{\theta })^{2}]} {p - 1} \,(\mathbf{I}_{p} -\boldsymbol{\theta }\boldsymbol{\theta }^{{\prime}}),}$$

where I _p denotes the p-dimensional identity matrix, readily implies that

$$\displaystyle{ \hat{\sigma }_{n}^{2}:= \frac{\bar{\mathbf{X}}_{n}^{{\prime}}\mathbf{S}_{ n}\bar{\mathbf{X}}_{n}} {\|\bar{\mathbf{X}}_{n}\|^{2}} - e_{10}^{2} = \mathbf{Y}_{ n}^{{\prime}}\mathbf{S}_{ n}\mathbf{Y}_{n} - e_{10}^{2}\stackrel{\mathrm{P}}{\rightarrow }\mathrm{E}[(\mathbf{X}_{ 1}^{{\prime}}\boldsymbol{\theta })^{2}] - e_{ 10}^{2} = \mathrm{Var}[\mathbf{X}_{ 1}^{{\prime}}\boldsymbol{\theta }]. }$$

(7)

Now, write

$$\displaystyle{ \frac{n^{1/2}(\|\bar{\mathbf{X}}_{n}\| - e_{10})} {\hat{\sigma }_{n}} = \frac{n^{1/2}\bar{\mathbf{X}}_{n}^{{\prime}}(\mathbf{Y}_{n}-\boldsymbol{\theta })} {\hat{\sigma }_{n}} + \frac{n^{1/2}(\bar{\mathbf{X}}_{n}^{{\prime}}\boldsymbol{\theta }- e_{10})} {\hat{\sigma }_{n}} =: S_{1n} + S_{2n}, }$$

(8)

say. It directly follows from (5) to (7) that S _1n  = o _P(1) as n → ∞. As for S _2n , the central limit theorem and Slutsky’s lemma yield that S _2n is asymptotically standard normal. This readily implies that

$$\displaystyle{T_{\mathrm{WJm}}^{(n)} =\bigg (\frac{n^{1/2}(\|\bar{\mathbf{X}}_{ n}\| - e_{10})} {\hat{\sigma }_{n}} \bigg)^{2}\stackrel{\mathcal{L}}{\rightarrow }\chi _{ 1}^{2}.}$$

(ii) In view of the derivations above, the continuous mapping theorem implies that, for any $\boldsymbol{\theta }\in \mathcal{S}^{p-1}$ and $F \in \mathcal{F}_{0}$,

$$\displaystyle{T_{\mathrm{WJm}}^{(n)} = \frac{n(\|\bar{\mathbf{X}}_{n}\| - e_{10})^{2}} {\mathrm{Var}[\mathbf{X}_{1}^{{\prime}}\boldsymbol{\theta }]} + o_{\mathrm{P}}(1)}$$

as n → ∞ under $\mathcal{R}_{p}^{(n)}(\boldsymbol{\theta },F)$. The result then follows from the fact that, under $\mathcal{R}_{p}^{(n)}(\boldsymbol{\theta },F_{p,\kappa _{0}})$, with κ ₀ = h _p ⁻¹(e ₁₀), $\mathrm{Var}[\mathbf{X}_{1}^{{\prime}}\boldsymbol{\theta }] = 1 -\frac{p-1} {\kappa _{0}} e_{10} - e_{10}^{2};$ see, e.g., Lemma S.2.1 from [7]. □ 

Proof of Proposition 1

From Lemma S.2.1 in [7], we have that, under $\mathcal{R}_{p_{n}}^{(n)}(\boldsymbol{\theta }_{n},F_{p_{n},\kappa _{n}})$,

$$\displaystyle{e_{n1} = \frac{I_{p_{n}/2}(\kappa _{n})} {I_{p_{n}/2-1}(\kappa _{n})}\ \text{ and }\ \tilde{e}_{n2} = 1 -\frac{p_{n} - 1} {\kappa _{n}} \,e_{n1} - e_{n1}^{2}.}$$

The result then readily follows from

$$\displaystyle{ \frac{z} {\nu +1 + \sqrt{z^{2 } + (\nu +1)^{2}}} \leq \frac{I_{\nu +1}(z)} {I_{\nu }(z)} \leq \frac{z} {\nu +\sqrt{z^{2 } +\nu ^{2}}} }$$

(9)

for any ν, z > 0; see (9) in [1]. □ 

Proof of Theorem 2

Writing $e_{n2}:= \mathrm{E}[(\mathbf{X}_{n1}^{{\prime}}\boldsymbol{\theta }_{n})^{2}]$, Theorem 5.1 in [7] entails that, under $\mathcal{R}_{p_{n}}^{(n)}(\boldsymbol{\theta }_{n},F_{p_{n},\kappa _{n}})$, where (κ _n ) is an arbitrary sequence in (0, ∞),

$$\displaystyle{ \frac{\sqrt{p_{n}}\big(n\|\bar{\mathbf{X}}_{n}\|^{2} - 1 - (n - 1)e_{n1}^{2}\big)} {\sqrt{2}\left (\,p_{n}\tilde{e}_{n2}^{2} + 2np_{n}e_{n1}^{2}\tilde{e}_{n2} + (1 - e_{n2})^{2}\right )^{1/2}}}$$

converges weakly to the standard normal distribution as n → ∞. The result then follows from the fact that, under $\mathcal{R}_{p_{n}}^{(n)}(\boldsymbol{\theta }_{n},F_{p_{n},\kappa _{n}})$, where the sequence (κ _n ) is such that, for any n, e _n1 = e ₁₀ under $\mathcal{R}_{p_{n}}^{(n)}(\boldsymbol{\theta }_{n},F_{p_{n},\kappa _{n}})$, one has

$$\displaystyle{e_{n2} = 1 -\frac{p_{n} - 1} {\kappa _{n}} \,e_{10},\quad \tilde{e}_{n2} = 1 -\frac{p_{n} - 1} {\kappa _{n}} \,e_{10} - e_{10}^{2},\quad \text{and}\quad \kappa _{ n}/p_{n} \rightarrow c_{0}\ \text{ as }n \rightarrow \infty;}$$

see Proposition 1(ii). □ 

The proof of Theorem 3 requires the three following preliminary results:

Lemma 1

Let Z be a random variable such that P [|Z|≤ 1] = 1. Then Var [Z ² ] ≤ 4 Var [Z].

Lemma 2

Let the assumptions of Theorem  3 hold. Write $\hat{e}_{n1} =\|\bar{ \mathbf{X}}_{n}\|$ and $\hat{e}_{n2}:=\bar{ \mathbf{X}}_{n}^{{\prime}}\mathbf{S}_{n}\bar{\mathbf{X}}_{n}/\|\bar{\mathbf{X}}_{n}\|^{2}$ . Then, as n →∞ under  $\mathcal{R}_{p_{n}}^{(n)}(\boldsymbol{\theta }_{n},F_{p_{n},\kappa _{n}})$ , (i) $(\hat{e}_{n1}^{2} - e_{10}^{2})/(e_{n2} - e_{10}^{2}) = o_{\mathrm{P}}(1)$ and (ii) $(\hat{e}_{2n} - e_{n2})/(e_{n2} - e_{10}^{2}) = o_{\mathrm{P}}(1)$.

Lemma 3

Let the assumptions of Theorem  3 hold. Write σ _n ² := p _n (e _n2 − e ₁₀ ² ) ² + 2np _n e ₁₀ ² (e _n2 − e ₁₀ ² ) + (1 − e _n2 ) ² and $\hat{\sigma }_{n}^{2}:= p_{n}(\hat{e}_{n2} -\hat{ e}_{n1}^{2})^{2} + 2np_{n}e_{10}^{2}(\hat{e}_{n2} -\hat{ e}_{n1}^{2}) + (1 -\hat{ e}_{n2})^{2}$ . Then $(\hat{\sigma }_{n}^{2} -\sigma _{n}^{2})/\sigma _{n}^{2} = o_{\mathrm{P}}(1)$ as n →∞ under  $\mathcal{R}_{p_{n}}^{(n)}(\boldsymbol{\theta }_{n},F_{p_{n},\kappa _{n}})$.

Proof of Lemma 1

Let Z _a and Z _b be mutually independent and identically distributed with the same distribution as Z. Since | x ² − y ² | ≤ 2 | x − y | for any x, y ∈ [−1, 1], we have that

$$\displaystyle{\mathrm{Var}[Z^{2}] = \frac{1} {2}\,\mathrm{E}[(Z_{a}^{2} - Z_{ b}^{2})^{2}] \leq 2\,\mathrm{E}[(Z_{ a} - Z_{b})^{2}] = 4\,\mathrm{Var}[Z],}$$

which proves the result. □ 

Proof of Lemma 2

All expectations and variances in this proof are taken under the sequence of hypotheses $\mathcal{R}_{p_{n}}^{(n)}(\boldsymbol{\theta }_{n},F_{n})$ considered in the statement of Theorem 3, and all stochastic convergences are taken as n → ∞ under the same sequence of hypotheses. (i) Proposition 5.1 from [7] then yields

$$\displaystyle{ \mathrm{E}[\hat{e}_{n1}^{2}] = \mathrm{E}[\|\bar{\mathbf{X}}_{ n}\|^{2}] = \frac{n - 1} {n} \,e_{10}^{2} + \frac{1} {n} }$$

(10)

and

$$\displaystyle\begin{array}{rcl} \mathrm{Var}[\hat{e}_{n1}^{2}]& =& \mathrm{Var}[\|\bar{\mathbf{X}}_{ n}\|^{2}] = \frac{2(n - 1)} {n^{3}} \,\tilde{e}_{2n}^{2} + \frac{4(n - 1)^{2}} {n^{3}} \,e_{10}^{2}\tilde{e}_{ n2} + \frac{2(n - 1)} {n^{3}(\,p_{n} - 1)}(1 - e_{n2}^{2})^{2} \\ & =& \frac{4} {n}\,e_{10}^{2}\tilde{e}_{ n2} + O(n^{-2}) {}\end{array}$$

(11)

as n → ∞. In view of Condition (i) in Theorem 3, this readily implies

$$\displaystyle\begin{array}{rcl} \mathrm{E}\Big[\Big(\frac{\hat{e}_{n1}^{2} - e_{10}^{2}} {\tilde{e}_{n2}} \Big)^{2}\Big]& =& \mathrm{Var}\Big[\frac{\hat{e}_{n1}^{2} - e_{ 10}^{2}} {\tilde{e}_{n2}} \Big] +\Big (\mathrm{E}\Big[\frac{\hat{e}_{n1}^{2} - e_{10}^{2}} {\tilde{e}_{n2}} \Big]\Big)^{2} {}\\ & =& \frac{4e_{10}^{2}} {n\tilde{e}_{n2}} + O\Big( \frac{1} {n^{2}\tilde{e}_{n2}^{2}}\Big) +\Big (\frac{1 - e_{10}^{2}} {n\tilde{e}_{n2}} \Big)^{2} = o(1) {}\\ \end{array}$$

as n → ∞, which establishes Part (i) of the result.

(ii) Write

$$\displaystyle{\frac{\hat{e}_{n2} - e_{n2}} {\tilde{e}_{n2}} = \frac{1} {\tilde{e}_{n2}}\bigg(\Big( \frac{1} {\hat{e}_{n1}^{2}} - \frac{1} {e_{10}^{2}}\Big)\bar{\mathbf{X}}_{n}^{{\prime}}\mathbf{S}_{ n}\bar{\mathbf{X}}_{n} + \frac{1} {e_{10}^{2}}\,\bar{\mathbf{X}}_{n}^{{\prime}}\mathbf{S}_{ n}\bar{\mathbf{X}}_{n} - e_{n2}\bigg).}$$

Part (i) of the result shows that $(\hat{e}_{n1}^{2} - e_{10}^{2})/\tilde{e}_{n2}$ is o _P(1) as n → ∞. Since (10) and (11) yield that $\hat{e}_{n1}$ converges in probability to e ₁₀( ≠ 0), this implies that $(\hat{e}_{n1}^{-2} - e_{10}^{-2})/\tilde{e}_{n2}$ is o _P(1) as n → ∞. This, and the fact that $\bar{\mathbf{X}}_{n}^{{\prime}}\mathbf{S}_{n}\bar{\mathbf{X}}_{n} = O_{\mathrm{P}}(1)$ as n → ∞, readily yields

$$\displaystyle{ \frac{\hat{e}_{n2} - e_{n2}} {\tilde{e}_{n2}} = \frac{1} {\tilde{e}_{n2}}\bigg( \frac{1} {e_{10}^{2}}\,\bar{\mathbf{X}}_{n}^{{\prime}}\mathbf{S}_{ n}\bar{\mathbf{X}}_{n} - e_{n2}\bigg) + o_{\mathrm{P}}(1) }$$

(12)

as n → ∞. Since

$$\displaystyle{ \frac{1} {e_{10}^{2}}\,\bar{\mathbf{X}}_{n}^{{\prime}}\mathbf{S}_{ n}\bar{\mathbf{X}}_{n} = \frac{1} {e_{10}^{2}}\,(\bar{\mathbf{X}}_{n} - e_{10}\boldsymbol{\theta })^{{\prime}}\mathbf{S}_{ n}(\bar{\mathbf{X}}_{n} - e_{10}\boldsymbol{\theta }) + \frac{2} {e_{10}}\,(\bar{\mathbf{X}}_{n} - e_{10}\boldsymbol{\theta })^{{\prime}}\mathbf{S}_{ n}\boldsymbol{\theta } +\boldsymbol{\theta } ^{{\prime}}\mathbf{S}_{ n}\boldsymbol{\theta },}$$

the result follows if we can prove that

$$\displaystyle\begin{array}{rcl} & & A_{n}:= \frac{1} {\tilde{e}_{n2}}(\bar{\mathbf{X}}_{n} - e_{10}\boldsymbol{\theta })^{{\prime}}\mathbf{S}_{ n}(\bar{\mathbf{X}}_{n} - e_{10}\boldsymbol{\theta }),\ \ B_{n}:= \frac{1} {\tilde{e}_{n2}}(\bar{\mathbf{X}}_{n} - e_{10}\boldsymbol{\theta })^{{\prime}}\mathbf{S}_{ n}\boldsymbol{\theta }, {}\\ & & \quad \text{ and }\ \ C_{n}:= \frac{1} {\tilde{e}_{n2}}(\boldsymbol{\theta }^{{\prime}}\mathbf{S}_{ n}\boldsymbol{\theta } - e_{n2}) {}\\ \end{array}$$

all are o _P(1) as n → ∞.

Starting with A _n , (10) yields

$$\displaystyle{ \mathrm{E}[\vert A_{n}\vert ] \leq \frac{1} {\tilde{e}_{n2}}\,\mathrm{E}[\|\bar{\mathbf{X}}_{n} - e_{10}\boldsymbol{\theta }\|^{2}] = \frac{1} {\tilde{e}_{n2}}\,\Big(\frac{n - 1} {n} \,e_{10}^{2} + \frac{1} {n} - e_{10}^{2}\Big) = \frac{1 - e_{10}^{2}} {n\tilde{e}_{n2}} = o(1) }$$

(13)

as n → ∞. Since convergence in L ₁ is stronger than convergence in probability, this implies that A _n  = o _P(1) as n → ∞. Turning to B _n , the Cauchy–Schwarz inequality and (13) provide

$$\displaystyle{\mathrm{E}[\vert B_{n}\vert ] \leq \frac{1} {\tilde{e}_{n2}}\,\mathrm{E}[\|\bar{\mathbf{X}}_{n} - e_{10}\boldsymbol{\theta }\|^{2}] = o(1),}$$

as n → ∞, so that B _n is indeed o _P(1) as n → ∞. Finally, it follows from Lemma 1 that

$$\displaystyle\begin{array}{rcl} \mathrm{E}[C_{n}^{2}] = \frac{1} {\tilde{e}_{n2}^{2}}\mathrm{E}[(\boldsymbol{\theta }^{{\prime}}\mathbf{S}_{ n}\boldsymbol{\theta } - e_{n2})^{2}]& =& \frac{1} {n\tilde{e}_{n2}^{2}}\mathrm{Var}[(\mathbf{X}_{n1}^{{\prime}}\boldsymbol{\theta })^{2}] \leq \frac{4} {n\tilde{e}_{n2}} = o(1) {}\\ \end{array}$$

as n → ∞, so that C _n is also o _P(1) as n → ∞. This establishes the result. □ 

Proof of Lemma 3

As in the proof of Lemma 2, all expectations and variances in this proof are taken under the sequence of hypotheses $\mathcal{R}_{p_{n}}^{(n)}(\boldsymbol{\theta }_{n},F_{n})$ considered in the statement of Theorem 3, and all stochastic convergences are taken as n → ∞ under the same sequence of hypotheses.

Let then $\tilde{\sigma }_{n}^{2}:= 2np_{n}e_{10}^{2}(e_{n2} - e_{10}^{2})$. Since Condition (i) in Theorem 3 directly entails that $\sigma _{n}^{2}/\tilde{\sigma }_{n}^{2} \rightarrow 1$ as n → ∞, it is sufficient to show that $(\hat{\sigma }_{n}^{2} -\sigma _{n}^{2})/\tilde{\sigma }_{n}^{2}$ is o _P(1) as n → ∞. To do so, write

$$\displaystyle{ \hat{\sigma }_{n}^{2} -\sigma _{ n}^{2} = A_{ n} + B_{n} + C_{n}, }$$

(14)

where

$$\displaystyle{A_{n}:= p_{n}\left ((\hat{e}_{n2} -\hat{ e}_{n1}^{2})^{2} - (e_{ n2} - e_{10}^{2})^{2}\right ),\quad B_{ n}:= 2np_{n}e_{10}^{2}\left (\hat{e}_{ n2} -\hat{ e}_{n1}^{2} - e_{ n2} + e_{10}^{2}\right ),}$$

and

$$\displaystyle{C_{n}:= (1 -\hat{ e}_{n2})^{2} - (1 - e_{ n2})^{2}.}$$

Since

$$\displaystyle{\frac{\vert A_{n}\vert } {\tilde{\sigma }_{n}^{2}} \leq \frac{p_{n}} {\tilde{\sigma }_{n}^{2}} = \frac{1} {2ne_{10}^{2}(e_{n2} - e_{10}^{2})}\quad \text{ and }\quad \frac{\vert C_{n}\vert } {\tilde{\sigma }_{n}^{2}} \leq \frac{1} {\tilde{\sigma }_{n}^{2}} = \frac{1} {2np_{n}e_{10}^{2}(e_{n2} - e_{10}^{2})},}$$

almost surely, Condition (i) in Theorem 3 implies that $A_{n}/\tilde{\sigma }_{n}^{2}$ and $C_{n}/\tilde{\sigma }_{n}^{2}$ are o _P(1) as n → ∞. The result then follows from the fact that, in view of Lemma 2,

$$\displaystyle{\frac{B_{n}} {\tilde{\sigma }_{n}^{2}} = \frac{(\hat{e}_{n2} - e_{n2}) - (\hat{e}_{n1}^{2} - e_{10}^{2})} {e_{n2} - e_{10}^{2}} }$$

is also o _P(1) as n → ∞. □ 

Proof of Theorem 3

Decompose Q _CPVm ⁽ⁿ⁾ into

$$\displaystyle{ Q_{\mathrm{CPVm}}^{(n)} = \frac{\sigma _{n}} {\hat{\sigma }_{n}} \times \frac{\sqrt{p_{n}}\left (n\|\bar{\mathbf{X}}_{n}\|^{2} - 1 - (n - 1)e_{10}^{2}\right )} {\sqrt{2}\,\sigma _{n}} =: \frac{\sigma _{n}} {\hat{\sigma }_{n}} \times V _{n}, }$$

(15)

say. Theorem 5.1 in [7] entails that, under the sequence of hypotheses $\mathcal{R}_{p_{n}}^{(n)}(\boldsymbol{\theta }_{n},F_{n})$ considered in the statement of the theorem, V _n is asymptotically standard normal as n → ∞. The result therefore follows from Lemma 3 and the Slutsky’s lemma. □