間違いしかありません.コメントにてご指摘いただければ幸いです(気が付いた点を特に断りなく頻繁に書き直していますのでご注意ください).

尤度凾数の平均情報量(離散/標本からの)

尤度凾数の平均情報量(離散/標本からの)

p(x)θq(x;θ)θ()(θx)Θ0p(x)q(x;θ)θ0p(x)q(x;θ)(θ0Θ0)q(x;θ)(xθ(xθ))logq(x;θ)logq(x;θ)()
Ln(θ)=E[log(1q(Xi;θ))]()=E[log(q(Xi;θ)1)]1A=A1=E[log(q(Xi;θ))]logAB=BlogA=E[log(q(Xi;θ))]E[cX]=cE[X]=1ni=1nlog(q(Xi;θ))n=1ni=1nlog(q(Xi;θ)q(Xi;θ0)q(Xi;θ0))θ0:=1ni=1nlog(q(Xi;θ0)q(Xi;θ)q(Xi;θ0))=1ni=1n{log(q(Xi;θ0))+log(q(Xi;θ)q(Xi;θ0))}logAB=logA+logB=1n{i=1nlog(q(Xi;θ0))+i=1nlog(q(Xi;θ)q(Xi;θ0))}(A+B)=A+B=1ni=1nlog(q(Xi;θ0))1ni=1nlog(q(Xi;θ)q(Xi;θ0))=1ni=1nlog(q(Xi;θ0))+1ni=1nlog((q(Xi;θ)q(Xi;θ0))1)logx=logx1=1ni=1nlog(q(Xi;θ0))+1ni=1nlog(q(Xi;θ0)q(Xi;θ))(AB)1=1AB=BA=Ln(θ0)+Kn(θ)Ln(θ0)=1ni=1nlog(q(Xi;θ0)),Kn=1ni=1nlog(q(Xi;θ0)q(Xi;θ))
f(Xi,θ0,θ)=logq(Xi;θ0)q(Xi;θ)()Kn(θ)=1ni=1nf(Xi,θ0,θ)KLnKn(θ)=i=1nf(Xi,θ0,θ)Ln(θ)=Ln(θ0)+Kn(θ)Kn(θ)>0KLKn(θ)=0θΘ0exp(f(Xi,θ0,θ))=q(Xi;θ0)q(Xi;θ)A=log(B),exp(A)=Bq(Xi;θ)=q(Xi;θ0)1exp(f(Xi,θ0,θ))=q(Xi;θ0)exp(f(Xi,θ0,θ))

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