EM算法Q函数推导过程详解

Q函数

$$\begin{array}{rl}Q\left(\theta ,{\theta }^{(i)}\right)& =E_Z\left[\log P(Y,Z\mid \theta )\mid Y,{\theta }^{(i)}\right]\\ & =\sum _Z\log P(Y,Z\mid \theta )\cdot P\left(Z\mid Y,{\theta }^{(i)}\right)\end{array}$$

Q函数是EM算法中的一个重要函数，全称为“期望完全数据对数似然函数”。它的作用是在E步中计算出完全数据的对数似然函数的期望值，以便在M步中求出模型参数的最大似然估计值。

在之前的一篇文章（EM算法求解三硬币模型参数推导）中，为大家介绍了李航教授《统计学习方法》中求解三硬币模型的参数推导过程，其中使用的EM算法是从一个Q函数直接展开求解的，限于篇幅，文章并未展示证明过程，本篇文章作为上一篇文章以及《统计学习方法-第九章-179页》推导的补充，详细推导Q函数的由来。

Q函数推导证明

我们已知关于参数$\theta$的似然函数
$$\begin{gathered} L(\theta )=\log P(Y\mid \theta ) \\ =\log \frac{P(Y,\theta )}{P(\theta )} \\ =\log \frac{{\sum }_{Z}P(Y,\theta ,Z)}{P(\theta )} \\ =\log \sum _Z\frac{P(Y,\theta ,Z)}{P(\theta )}=\log \sum _{Z}P(Y,Z\mid \theta ) \\ =\log \sum _Z\frac{P(Y,Z,\theta )}{P(Z,\theta )}\cdot \frac{P(Z,\theta )}{P(\theta )} \end{gathered}$$
即
$$L(\theta )=\log \sum _{Z}P(Y\mid Z,\theta )\cdot P(Z\mid \theta )$$
假设第i次参数取 ${\theta }^{(i)}$ ，我们希望优化后$L(\theta )>L({\theta }^{(i)})$
于是可以作差

即
$$L(\theta )-L\left({\theta }^{(i)}\right)=\log {\Sigma }_{Z}P(Y\mid Z,\theta )\cdot P(Z\mid \theta )-\log P\left(Y\mid {\theta }^{(i)}\right)$$
第一项可以凑一个分式出来
$$L(\theta )-L\left({\theta }^{(i)}\right)=\log \left({\Sigma }_{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot \frac{P(Y\mid Z,\theta )\cdot P(Z\mid \theta )}{P\left(Z\mid Y,{\theta }^{(i)}\right)}\right)-\log P\left(Y\mid {\theta }^{(i)}\right)$$
利用${\sum }_{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)=1$的特性，第二项乘以这一串，可以得到
$$L(\theta )-L\left({\theta }^{(i)}\right)=\log \left({\Sigma }_{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot \frac{P(Y\mid Z,\theta )\cdot P(Z\mid \theta )}{P\left(Z\mid Y,{\theta }^{(i)}\right)}\right)-\log P\left(Y\mid {\theta }^{(i)}\right)\cdot {\Sigma }_{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)$$
利用$Jensen$不等式
$log{\sum }_j{\lambda }_j\cdot y_j⩾{\sum }_j{\lambda }_j\cdot log{y}_j$，其中$\lambda ⩾0,{\sum }_j{\lambda }_j=1$

可知
$$⩾\sum _{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot \log \frac{P(Y\mid Z,\theta )\cdot P(Z\mid \theta )}{P\left(Z\mid Y,{\theta }^{(i)}\right)}-\log P\left(Y\mid {\theta }^{(i)}\right)\cdot {\Sigma }_{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)$$

$$=\sum _{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot \left[\log \frac{P(Y\mid Z,\theta )\cdot P(Z\mid \theta )}{p\left(Z\mid Y,{\theta }^{(i)}\right)}-\log P\left(Y\mid {\theta }^{(i)}\right)\right]$$
$$={\Sigma }_{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot \log \frac{P(Y\mid Z,\theta )\cdot P(Z\mid \theta )}{P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot P\left(Y\mid {\theta }^{(i)}\right)}$$
即此时 $$L(\theta )-L\left({\theta }^{(i)}\right)⩾{\Sigma }_{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot \log \frac{P(Y\mid Z,\theta )\cdot P(Z\mid \theta )}{P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot P\left(Y\mid {\theta }^{(i)}\right)}$$
即 $$L(\theta )⩾L\left({\theta }^{(i)}\right)+{\Sigma }_{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot \log \frac{P(Y\mid Z,\theta )\cdot P(Z\mid \theta )}{P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot P\left(Y\mid {\theta }^{(i)}\right)}$$
令$$B\left(\theta ,{\theta }^{(i)}\right)=L\left({\theta }^{(i)}\right)+{\Sigma }_{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot \log \frac{P(Y\mid Z,\theta )\cdot P(Z\mid \theta )}{P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot P\left(Y\mid {\theta }^{(i)}\right)}$$
此时 $B\left(\theta ,{\theta }^{(i)}\right)$ 是 $L(\theta )$ 的下界，使 $B\left(\theta ,{\theta }^{(i)}\right)$ 最大化的 $\theta$ 也可使 $L\left(\theta \right)$ 最大

于是我们的目标是 ${\theta }^{(i+1)}=\underset{\theta }{\mathrm{argmax}}B\left(\theta ,{\theta }^{(i)}\right)$
也即
$${\theta }^{(i+1)}=\underset{\theta }{\mathrm{argmax}}\left[L\left({\theta }^{(i)}\right)+{\Sigma }_{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot \log \frac{P(Y\mid Z,\theta )\cdot P(Z\mid \theta )}{P(Z\mid Y,{\theta }^{(i)})\cdot P\left(Y\mid {\theta }^{(i)}\right)}\right]$$

可把 $L\left({\theta }^{(i)}\right)、P\left(Z\mid Y,{\theta }^{(i)}\right)、P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot P\left(Y\mid {\theta }^{(i)}\right)$ 三项视为常数
且已知 $P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot P\left(Y\mid {\theta }^{(i)}\right)>0$ ，这一项从分母去掉，不影响求最大值，注意这里的 $P(Z\mid Y,{\theta }^{(i)})$ 不能省略，因为它是$\sum$ 后面中的每一项的系数

于是
$${\theta }^{(i+1)}=\underset{\theta }{\mathrm{argmax}}\left[{\Sigma }_{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot \log P(Y\mid Z,\theta )\cdot P(Z\mid \theta )\right]$$

我们令 $Q\left(\theta ,{\theta }^{(i)}\right)={\Sigma }_{Z}P\left(Z\mid Y,{\theta }^{(i)}\right)\cdot \log P(Y\mid Z,\theta )\cdot P(Z\mid \theta )$
即
$${\theta }^{(i+1)}=\underset{\theta }{\mathrm{argmax}}Q\left(\theta ,{\theta }^{(i+1)}\right)$$

其中 $Q\left(\theta ,{\theta }^{(i)}\right)$ 就是所谓的 $Q$ 函数

参考资料

[1].EM算法求解三硬币模型参数推导