Page - 106 - in Differential Geometrical Theory of Statistics

Entropy2016,18, 386 Combiningthis relationwith thatof 1 λ(x) ,wecandeduce thatλ(x)∂ 2Φ(θ) ∂θ2 =1andasλ(x)>0 then ∂2Φ(θ) ∂θ2 >0. Fréchet emphasizes at this step [141], anotherway to approach the problem. We can select arbitrarilyh(x)and l(x)andthenΦ(θ) isdeterminedby: +∞ −∞ e ∂Φ(θ) ∂θ [h(x)−θ]+Φ(θ)+ (x)dx=1 (A25) Thatcouldberewritten: eθ. ∂Φ(θ) ∂θ −Φ(θ) = +∞ −∞ e ∂Φ(θ) ∂θ h(x)+ (x)dx (A26) Ifwethenfixedarbitrarilyh(x)and l(x)andlet sanarbitraryvariable, the followingfunction willbeanexplicitpositive functiongivenby eΨ(s): +∞ −∞ es.h(x)+ (x)dx= eΨ(s) (A27) Fréchetobtainedfinally the functionΦ(θ)as solutionof the equation [141]: Φ(θ)= θ · ∂Φ(θ) ∂θ −Ψ ( ∂Φ(θ) ∂θ ) (A28) Fréchetnoted that this is theAlexisClairaut equation [141]. Thecase ∂Φ(θ) ∂θ = cstewouldreduce thedensity toa functionthatwouldbe independentofθ, andsoΦ(θ) isgivenbyasingular solutionof thisClairautequation,which isuniqueandcouldbe computedbyeliminatingthevariable sbetween: Φ= θ ·s−Ψ(s) andθ= ∂Ψ(s) ∂s (A29) Orbetween: eθ·s−Φ(θ) = +∞ −∞ es·h(x)+ (x)dxand +∞ −∞ es·h(x)+ (x) [h(x)−θ]dx=0 (A30) Φ(θ)=−log +∞ −∞ es·h(x)+ (x)dx+θ ·swhere s isgiven implicitlyby +∞ −∞ es·h(x)+ (x) [h(x)−θ]dx=0. Thenweknowthedistinguishedfunction,H′ amongfunctionsH(X1,...,Xn)verifyingEθ [H]= θ andsuchthatσH reaches foreachvalueofθ, anabsoluteminimum,equal to 1√ nσA . For thepreviousequation: h(x)= θ+ ∂logpθ(x) ∂θ +∞ −∞ [ ∂pθ(x) ∂θ ]2 dx pθ(x) (A31) 106