Page - 431 - in Differential Geometrical Theory of Statistics

Entropy2016,18, 375 For(iv),wewillusetheestimateinEquation(11). Recall that ln(1+x)= x− x22 +o(x2); therefore, for largenandhencesmall sV a.s. α 4 ≤ exp ( −nsV [( 1+ σ̂ 2 max sV )( sV σ̂2max − s2V 2(σ̂2max)2 +o(s2V) ) −1 ]) = exp (−ns2V/2σ̂2max+o(s2V)), thus sV≤ √ −2σ̂2max ln(α4) / n+o ( n− 1 2 ) .Additionally,arcsinx= x+o(x) forx close to0whichgives δV= sV /|Z¯n|+o(sV)≤√−2σ̂2max ln α4/(√n|Z¯n|)+o(n−12)a.s. Furthermore, σ̂2max→σ2 a.s. forn→∞, andweobtain limsup n→∞ δV δA ≤ √ −2ln α4 q1−α2 a.s. since δA= q1−α2√ n|Z¯n| √ 1 n n ∑ k=1 ( Im(μˆ−1n Zk) )2 ︸ ︷︷ ︸ → √ σ2 (seeEquation(5)). 4. SimulationandApplicationtoRealData Wewill compare theasymptotic confidencesetCA, theconfidencesetCH constructeddirectly usingHoeffding’s inequality inSection2,andtheconfidencesetCV resulting fromAlgorithm1by reportingtheircorrespondingopeninganglesδA,δH, andδV indegrees (◦)aswellas theircoverage frequencies insimulations. All computationshavebeenperformedusingourowncodebasedon the softwarepackageR (version2.15.3) [14] . 4.1. Simulation1: TwoPointsofEqualMassat±10◦ First,weconsiderarather favourablesituation:n=400 independentdrawsfromthedistribution with P(Z = exp(10πi/180)) = P(Z = exp(−10πi/180)) = 12. Then, we have |EZ| = EZ = cos(10πi/180)≈ 0.985, implying that thedata are highly concentrated, μ = 1 is unique, and the varianceofZ in thedirectionofμ is0; there isonlyvariationperpendicular toμ, i.e., in thedirectionof the imaginaryaxis (seeFigure4). 0 10◦ −10◦ EZ Figure4.Twopointsofequalmassat±10◦ andtheirEuclideanmean. 431