李航-统计学习方法-习题-第九章

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9.2 证明引理 9.2.
引理 9.2 P ~ θ ( Z ) = P ( Z Y , θ ) \widetilde P_\theta(Z)=P(Z|Y,\theta) ,则
F ( P ~ , θ ) = l o g P ( Y θ ) F(\widetilde P, \theta)=logP(Y|\theta) .
证明:
F ( P ~ , θ ) = E P ~ [ l o g P ( Y , Z θ ) ] + H ( P ~ ) = E P ~ [ l o g P ( Y , Z θ ) ] E P ~ l o g P ~ ( Z ) = Z l o g P ( Y , Z θ ) P ~ θ ( Z ) Z l o g P ~ ( Z ) P ~ ( Z ) = Z l o g P ( Y , Z θ ) P ( Z Y , θ ) Z l o g P ( Z Y , θ ) P ( Z Y , θ ) = Z P ( Z Y , θ ) ( l o g P ( Y , Z θ ) l o g P ( Z Y , θ ) ) = Z P ( Z Y , θ ) l o g P ( Y , Z θ ) P ( Z Y , θ ) = Z P ( Z Y , θ ) l o g P ( Y θ ) = l o g P ( Y θ ) Z P ( Z Y , θ ) = l o g P ( Y θ ) 1 = l o g P ( 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}
证毕.

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