Seite - 427 - in Differential Geometrical Theory of Statistics

Entropy2016,18, 375 Now, for any ζ ∈ S1, we will test the hypothesis that ζ is a circular population mean. Thishypothesis isequivalent tosayingthat there is someλ∈ [0,1] suchthatEZ=λζ.Multiplication byζ−1 thenrotatesEZonto thenon-negativerealaxis:Eζ−1Z=λ≥0. Now,fixζ andconsiderXk=Re(ζ−1Zk),Yk= Im(ζ−1Zk) fork=1,. . . ,nwhichmaybeviewed as theprojectionofZ1, . . . ,Zk onto the line in thedirectionof ζ andonto the lineperpendicular to it. Botharesequencesof independent randomvariables takingvalues in [−1,1]withEXk=λand EYk=0under thehypothesis. Theythus fulfill theconditions forHoeffding’s inequalitywith a=−1, b=1andν=λor0, respectively. Wewillfirst consider thecaseofnon-uniquenessof thecircularmean, i.e.,μ=S1,orequivalently λ = 0. Then, the critical value s0 = t(α4,0,−1,1) is well-defined for any α4 > 2−n, andwe get P(X¯n ≥ s0) ≤ α4, and also, by considering−X1, . . . ,−Xn, thatP(−X¯n ≥ s0) ≤ α4. Analogously, P(|Y¯n|≥ s0)≤2α4 = α2.Weconcludethat P (|Z¯n|≥√2s0)=P(|X¯n|2+ |Y¯n|2≥2s20)≤P(|X¯n|2≥ s20)+P(|Y¯n|2≥ s20)≤α. Rejecting thehypothesisμ= S1, i.e.,EZ= 0, if |Z¯n| ≥ √ 2s0 thus leads toa testwhoseprobability of falserejection isatmostα (seeFigure1).Ofcourse,onemayworkwith |X¯n|2≥ s20 and |Y¯n|2≥ s20 as criterions for rejection;however,wepreferworkingwith |Z¯n| ≥ √ 2s0 since it is independentof thechosenζ. 0 s0 s0 P(Re Z¯n≥ s0)≤ α4P(Re Z¯n≤−s0)≤ α4 P(Im Z¯n≤−s0)≤ α4 P(Im Z¯n≥ s0)≤ α4 Figure1.Theconstructionfor the testof thehypothesisμ=S1,orequivalentlyEZ=0. In the case of uniqueness of the circular mean, i.e., for the hypothesis λ > 0, we use the monotonicityofν+ t(γ,ν,a,b) inνandobtain P ( X¯n≤−s0 ) =P (−X¯n≥ t(α4,0,−1,1))≤P(−X¯n≥−λ+ t(α4,−λ,−1,1))≤ α4 aswell. Forthedirectionperpendiculartothedirectionofζ (seeFigure2),however,wemaynowwork with 38α, sofor sp= t( 3 8α,0,−1,1)—whichiswell-definedwhenever s0 issince 38α> α4 >2−n—weobtain P ( Y¯n≥ sp ) +P ( Y¯n≤−sp )≤2 · 38α. Rejecting if X¯n≤−s0 or |Y¯n| ≥ sp, then,willhappenwithprobabilityatmost α4+2 · 38α= αunder thehypothesisμ = ζ. In case thatwealready rejected thehypothesisμ = S1, i.e., if |Z¯n| ≥ √ 2s0, ζ will not be rejected if and only if X¯n > s0 > 0 and |Y¯n|< sp < s0 which is then equivalent to |Arg(ζ−1Z¯n)|=arcsin(|Y¯n|/|Z¯n|)<arcsin(sp/|Z¯n|)= δH (seeFigure3). DefineCH asallζwhichwecouldnot reject, i.e., CH= { S1, ifα≤2−n+2 or |Z¯n|≤ √ 2s0,{ ζ∈S1 : |Arg(ζ−1μˆn)|< δH } otherwise. (8) 427