Page - 241 - in Differential Geometrical Theory of Statistics

Entropy2016,18, 383 Therefore,weget: ϕV(ξ)=(2π)q1/2(detψ(ξ′))−1/2 ∫ ∞ 0 e−x11(ξ11− tvψ(ξ′)−1v)xq1/211 dx11 ∫ PV′ e−(x˜ ′|ξ′)dx˜′ =(2π)q1/2(detψ1(ξ))−1/2Γ(1+ q1 2 )(ξ11− tvψ(ξ′)−1v)−1−qk/2ϕV′(ξ′), whichtogetherwith (18) leadsus to (24). Example2. LetV={Vlk}1≤k<l≤3 beas inExample1. For: ξ= ⎛⎜⎜⎜⎝ ξ1 0 ξ4 0 0 ξ1 0 ξ5 ξ4 0 ξ2 ξ6 0 ξ5 ξ6 ξ3 ⎞⎟⎟⎟⎠∈ZV, (25) wehave: φ1(ξ)= ⎛⎜⎝ξ1 ξ4 ξ5ξ4 ξ2 0 ξ5 0 ξ3 ⎞⎟⎠ , φ2(ξ)= ( ξ2 ξ6 ξ6 ξ3 ) , φ3(ξ)= ξ3, ψ1(ξ)= ( ξ2 0 0 ξ3 ) , ψ2(ξ)= ξ3. TheconeP∗V isdescribedas: P∗V= ⎧⎪⎨⎪⎩ξ∈ZV ; ∣∣∣∣∣∣∣ ξ1 ξ4 ξ5 ξ4 ξ2 0 ξ5 0 ξ3 ∣∣∣∣∣∣∣>0, ∣∣∣∣∣ξ2 ξ6ξ6 ξ3 ∣∣∣∣∣>0, ξ3>0 ⎫⎪⎬⎪⎭ , and itsKoszul–Vinbergcharacteristic functionϕV is expressedas: ϕV(ξ)=CV ⎧⎪⎨⎪⎩ ∣∣∣∣∣∣∣ ξ1 ξ4 ξ5 ξ4 ξ2 0 ξ5 0 ξ3 ∣∣∣∣∣∣∣/(ξ2ξ3) ⎫⎪⎬⎪⎭ −2{∣∣∣∣∣ξ2 ξ6ξ6 ξ3 ∣∣∣∣∣/ξ3 }−3/2 ξ−13 ·(ξ2ξ3)−1/2(ξ3)−1/2 =CV ∣∣∣∣∣∣∣ ξ1 ξ4 ξ5 ξ4 ξ2 0 ξ5 0 ξ3 ∣∣∣∣∣∣∣ −2 ∣∣∣∣∣ξ2 ξ6ξ6 ξ3 ∣∣∣∣∣ −3/2 ξ3/22 ξ 3/2 3 , whereCV=(2π)3/2Γ(2)Γ(3/2)Γ(1)= √ 2π2. Suppose that the conePV is homogeneous. Then,P∗V, aswell asPV, is a homogeneous cone of rank3, so that theKoszul–Vinbergcharacteristic functionofP∗V hasatmost three irreducible factors (see [8]).However, wehave seen that thereare four irreducible factors in the functionϕV. Therefore,weconclude thatneitherPV, norP∗V ishomogeneous. 241