Page - 244 - in Differential Geometrical Theory of Statistics

Entropy2016,18, 383 which is similar to theright-handsideof (23). Thus, theproof isparallel toTheorem2.Namely,by inductionontherank, it suffices toshowthat:∫ PV e−(x|ξ)ΔVs (x)dx =(2π)q1/2Γ(s1+1+ q1 2 )(φ1(ξ) −1)s1+1+q1/211 (detψ1(ξ)) −sgn(q1)/2 × ∫ PV′ e−(x ′|ξ)ΔV ′ s′ (x ′)dx′ (36) thanks to (33). Whenq1=0,wehave (x|ξ)= x11ξ11+(x′|ξ′)andΔVs (x)= xs111ΔV ′ s (x′). Thus:∫ PV e−(x|ξ)ΔVs (x)dx= ∫ ∞ 0 e−x11ξ11xs111dx11× ∫ PV′ e−(x ′|ξ)ΔV ′ s′ (x ′)dx′. Since ∫∞ 0 e −x11ξ11xs111dx11=Γ(s1+1)ξ −s1−1 11 ,weget (36). When q1 > 1, we use the change of variable (6). Since x˜′ = x′ − x−111UtU, we have ΔVs (x) = x s1 11Δ V′ s′ (x˜ ′) by (31). Therefore, by the sameGaussian integral formula as in the proof of Theorem2, the integral ∫ PV e −(x|ξ)ΔVs (x)dxequals:∫ ∞ 0 ∫ W ∫ PV′ e−x11(ξ11− tvψ(ξ′)−1v)e−x11 t(u˜+ψ(ξ′)−1v)ψ(ξ′)(u˜+ψ(ξ′)−1v)e−(x˜ ′|ξ′)xs111Δ V′ s′ (x˜ ′) ×2q1/2xq111dx11du˜dx˜′ =(2π)q1/2(detψ(ξ))−1/2 ∫ ∞ 0 e−x11(ξ11− tvψ(ξ′)−1v)xs1+q1/211 dx11 × ∫ PV′ e−(x˜ ′|ξ′)ΔV ′ s′ (x˜ ′)dx˜′ =(2π)q1/2(detψ(ξ))−1/2Γ(sk+1+ q1 2 )(ξ11− tvψ(ξ′)−1v)−sk−1−qk/2 × ∫ PV′ e−(x˜ ′|ξ′)ΔV ′ s′ (x˜ ′)dx˜′. Hence,weget (36)by(18). Weshallobtainan integral formulaoverP∗V as follows. Theorem4. When sk> qk/2 for k=1,. . . ,r, onehas:∫ P∗V e−(x|ξ)δVs (ξ)ϕV(ξ)dξ=CVΓV(s)ΔV−s(x) (x∈PV), (37) whereΓV(s) :=(2π)(N−r)/2∏rk=1Γ(sk−qk/2). Proof. Using(24), (31)and(33),werewrite (37)as:∫ P∗V e−(x|ξ)(φ1(ξ)−1) −s1+1+q1/2 11 (detψ1(ξ)) −sgn(q1)/2δV ′ s′ (ξ ′)ϕV′(ξ′)dξ =CV′(2π)q1/2Γ(s1−q1/2)ΓV′(s′)x−s111 ΔV ′ −s′(x˜ ′), (38) where: x˜′ := { x′ (q1=0), x′−x−111UtU (q1>0). 244