9.2 证明引理 9.2.
引理 9.2 若P~θ(Z)=P(Z∣Y,θ)\widetilde P_\theta(Z)=P(Z|Y,\theta)Pθ(Z)=P(Z∣Y,θ),则
F(P~,θ)=logP(Y∣θ) F(\widetilde P, \theta)=logP(Y|\theta)F(P,θ)=logP(Y∣θ).
证明:
F(P~,θ)=EP~[logP(Y,Z∣θ)]+H(P~)=EP~[logP(Y,Z∣θ)]−EP~logP~(Z)=∑ZlogP(Y,Z∣θ)P~θ(Z)−∑ZlogP~(Z)P~(Z)=∑ZlogP(Y,Z∣θ)P(Z∣Y,θ)−∑ZlogP(Z∣Y,θ)P(Z∣Y,θ)=∑ZP(Z∣Y,θ)(logP(Y,Z∣θ)−logP(Z∣Y,θ))=∑ZP(Z∣Y,θ)logP(Y,Z∣θ)P(Z∣Y,θ)=∑ZP(Z∣Y,θ)logP(Y∣θ)=logP(Y∣θ)∑ZP(Z∣Y,θ)=logP(Y∣θ)⋅1=logP(Y∣θ)
\begin{aligned}
F(\widetilde P, \theta) &= E_{\widetilde P}[logP(Y,Z|\theta)]+H(\widetilde P) \\
&= E_{\widetilde P}[logP(Y,Z|\theta)]-E_{\widetilde P}log\widetilde P(Z) \\
&= \sum_ZlogP(Y,Z|\theta)\widetilde P_\theta(Z) - \sum_Zlog\widetilde P(Z)\widetilde P(Z) \\
&= \sum_ZlogP(Y,Z|\theta)P(Z|Y,\theta)- \sum_ZlogP(Z|Y,\theta)P(Z|Y,\theta) \\
&= \sum_ZP(Z|Y,\theta)(logP(Y,Z|\theta)-logP(Z|Y,\theta)) \\
&= \sum_ZP(Z|Y,\theta)log\dfrac{P(Y,Z|\theta)}{P(Z|Y,\theta)} \\
&= \sum_ZP(Z|Y,\theta)logP(Y|\theta) \\
&= logP(Y|\theta)\sum_ZP(Z|Y,\theta) \\
&= logP(Y|\theta)\cdot1 \\
&= logP(Y|\theta)
\end{aligned}
F(P,θ)=EP[logP(Y,Z∣θ)]+H(P)=EP[logP(Y,Z∣θ)]−EPlogP(Z)=Z∑logP(Y,Z∣θ)Pθ(Z)−Z∑logP(Z)P(Z)=Z∑logP(Y,Z∣θ)P(Z∣Y,θ)−Z∑logP(Z∣Y,θ)P(Z∣Y,θ)=Z∑P(Z∣Y,θ)(logP(Y,Z∣θ)−logP(Z∣Y,θ))=Z∑P(Z∣Y,θ)logP(Z∣Y,θ)P(Y,Z∣θ)=Z∑P(Z∣Y,θ)logP(Y∣θ)=logP(Y∣θ)Z∑P(Z∣Y,θ)=logP(Y∣θ)⋅1=logP(Y∣θ)
证毕.
李航-统计学习方法-习题-第九章
最新推荐文章于 2025-05-10 10:08:38 发布