Page - 430 - in Differential Geometrical Theory of Statistics

Entropy2016,18, 375 Weobtaina (conservative) connected, symmetricconfidencesetCV⊆CH bytestingζ∈CH with σ̂2max= supζ∈CH σ̂ 2 asacommonupperboundfor thevarianceperpendicular toanyζ∈CH.Note that σ̂2max canbeobtainedas thesolutionofEquation(9)with V˜n= sup ζ∈CH 1 n n ∑ k=1 ( Imζ−1Zk )2. Furthermore, we can shortenCV by iteratively redefining V˜n = supζ∈CV 1 n∑ n k=1 ( Imζ−1Zk )2 and recalculatingCV (seeAlgorithm1). TheresultingopeninganglewillbedenotedbyδV=arcsin sV |Z¯n|. Algorithm1:AlgorithmforcomputationofCV. Data: observationsZ1, . . . ,Zn∈S1; significance levelα; stopcriterion ε Result: anon-asymptoticconfidencesetCV for thecircularpopulationmean 1 compute theconfidencesetCH; 2 ifCH=S1 then 3 CV←S1 4 else 5 CV←CH; σ̂2max←1; 6 while supζ∈CV σ̂ 2< σ̂2max−εdo 7 σ̂2max← supζ∈CV σ̂2; 8 sV←w(α4, σ̂2); 9 CV← { ζ∈S1 : |Arg(ζ−1μˆn)|<arcsin sV|Z¯n| } 10 end 11 end Proposition2. LetZ1, . . . ,Zn berandomvariables takingvalueson theunit circleS1,and letα∈ (0,1). (i) The set CV resulting fromAlgorithm 1 is a (1−α)-confidence set for the circular populationmean set. Inparticular, ifEZ= 0, i.e., the circular populationmeanset equalsS1, then |Z¯n|> √ 2s0 with probabilityatmostα, so indeedCV=S1withprobabilityof at least1−α. (ii) sV is of ordern −12 . (iii) IfEZ = 0, i.e., if the circular populationmean is unique, then√nδV → 0 in probability, and the probability of obtaining a trivial confidence set, i.e., P(CH = S1) = P(|Z¯n| ≤ √ 2s0), goes to 0 exponentially fast. (iv) IfEZ =0, then limsup n→∞ δV δA ≤ √ −2ln α4 q 1−α2 a.s. withq1−α2 denoting the (1− α2)-quantile of the standardnormaldistributionN(0,1). Proof. Again, (i) followsbyconstruction,while (iii) is shownas inProposition1. For (ii),note that sV≤ s0 since theboundinEquation (10) forρ2= 1agreeswith theboundin Equation(6) for a=−1,b=1andv=0, thus sV andδV areat leastof theordern− 1 2. 430