GMM

解决高斯分布的单峰性，引入多个高斯模型加权平均拟合数据

几何角度

• $$p(x)=\sum _{k=1}^{K} \alpha _k N( \mu _k, \Sigma _k), \sum \alpha = 1$$

生成模型角度

• 利用离散变量 z 来选择来自哪一个高斯分布

• $$p(x)=\sum\limits_zp(x,z)=\sum\limits_{k=1}^Kp(x,z=k)=\sum\limits_{k=1}^Kp(z=k)p(x|z=k)$$

• 得到 $$p(x)=\sum\limits_{k=1}^Kp_k\mathcal{N}(x|\mu_k,\Sigma_k)$$
  [[MLE]] 求解

• 求解上面的 px

• $$\theta_{MLE}=\mathop{argmax}\limits_{\theta}\log p(X)=\mathop{argmax}_{\theta}\sum\limits_{i=1}^N\log p(x_i)=\mathop{argmax}_\theta\sum\limits_{i=1}^N\log \sum\limits_{k=1}^Kp_k\mathcal{N}(x_i|\mu_k,\Sigma_k)$$

• log 中有连加号存在，无法求出解析解。
  [[EM]] 求解

• EM 目标求解：$\theta ^{(t+1)} = \arg \max _{\theta} \int _Z \log [p(X,Z|\theta)]p(Z|X, \theta ^{(t)})dZ = \arg \max _{\theta} E _{z|x,{\theta ^t}} [\log p(X,Z|\theta)]$

• E-Step

  • $$Q(\theta,\theta^t)=\sum\limits_z[\log\prod\limits_{i=1}^Np(x_i,z_i|\theta)]\prod \limits_{i=1}^Np(z_i|x_i,\theta^t)\\$$

+ 对第一个累加号展开，第一项为： + $$\sum\limits_z\log p(x_1,z_1|\theta)\prod\limits_{i=1}^Np(z_i|x_i,\theta^t)=\sum\limits_z\log p(x_1,z_1|\theta)p(z_1|x_1,\theta^t)\prod\limits_{i=2}^Np(z_i|x_i,\theta^t)\\ =\sum\limits_{z_1}\log p(x_1,z_1|\theta) p(z_1|x_1,\theta^t)\sum\limits_{z_2,\cdots,z_K}\prod\limits_{i=2}^Np(z_i|x_i,\theta^t)\\ =\sum\limits_{z_1}\log p(x_1,z_1|\theta)p(z_1|x_1,\theta^t)

  + 后面与 1 无关求和结果为1

+ $$Q(\theta,\theta^t)=\sum\limits_{i=1}^N\sum\limits_{z_i}\log p(x_i,z_i|\theta)p(z_i|x_i,\theta^t)$$

  + $$p(x,z|\theta)=p(z|\theta)p(x|z,\theta)=p_z\mathcal{N}(x|\mu_z,\Sigma_z)$$

  + $$p(z|x,\theta^t)=\frac{p(x,z|\theta^t)}{p(x|\theta^t)}=\frac{p_z^t\mathcal{N}(x|\mu_z^t,\Sigma_z^t)}{\sum\limits_kp_k^t\mathcal{N}(x|\mu_k^t,\Sigma_k^t)}$$

+ $$Q=\sum\limits_{i=1}^N\sum\limits_{z_i}\log p_{z_i}\mathcal{N(x_i|\mu_{z_i},\Sigma_{z_i})}\frac{p_{z_i}^t\mathcal{N}(x_i|\mu_{z_i}^t,\Sigma_{z_i}^t)}{\sum\limits_kp_k^t\mathcal{N}(x_i|\mu_k^t,\Sigma_k^t)}$$

• M-Step

  • $$Q=\sum\limits_{k=1}^K\sum\limits_{i=1}^N[\log p_k + \log \mathcal{N}(x_i|\mu_k,\Sigma_k)]p(z_i=k|x_i,\theta^t)$$

  • $$p_k^{t+1}=\mathop{argmax}_{p_k}\sum\limits_{k=1}^K\sum\limits_{i=1}^N[\log p_k+\log \mathcal{N}(x_i|\mu_k,\Sigma_k)]p(z_i=k|x_i,\theta^t)\ s.t.\ \sum\limits_{k=1}^Kp_k=1$$

    • 化简 $$p_k^{{t+1}=\mathop{argmax}_{p_k}\sum\limits_{k=1}}K\sum\limits_{i=1}^N\log p_kp(z_i=k|x_i,\theta^{t) s.t. \sum\limits_{k=1}}Kp_k=1$$

    • [[Lagrange Multiplier]] $$L(p_k,\lambda)=\sum\limits_{k=1}^{K\sum\limits_{i=1}}N\log p_kp(z_i=k|x_i,\theta^{t)-\lambda(1-\sum\limits_{k=1}}Kp_k)$$

    • [[Lagrange]] $$L(p_k,\lambda)=\sum\limits_{k=1}^{K\sum\limits_{i=1}}N\log p_kp(z_i=k|x_i,\theta^{t)-\lambda(1-\sum\limits_{k=1}}Kp_k)$$

    • $$\frac{\partial}{\partial p_k}L=\sum\limits_{i=1}^N\frac{1}{p_k}p(z_i=k|x_i,\theta^t)+\lambda=0\\$$

\Rightarrow\lambda=-N$$

  + $$p_k^{t+1}=\frac{1}{N}\sum\limits_{i=1}^Np(z_i=k|x_i,\theta^t)$$

+ $$\mu_k,\Sigma_k$$ 无约束参数，直接求导

对比 MLE 和 EM 在求解 GMM 问题上的区别

• $\theta_{MLE}=\mathop{argmax}\limits_{\theta}\log p(X)=\mathop{argmax}_{\theta}\sum\limits_{i=1}^N\log p(x_i)=\mathop{argmax}_\theta\sum\limits_{i=1}^N\log \sum\limits_{k=1}^Kp_k\mathcal{N}(x_i|\mu_k,\Sigma_k)$

• $Q=\sum\limits_{k=1}^K\sum\limits_{i=1}^N[\log p_k + \log \mathcal{N}(x_i|\mu_k,\Sigma_k)]p(z_i=k|x_i,\theta^t)$

• 极大似然需要计算 $$P(X)$$，EM 计算 $$P(X,Z)$$。$$P(X) = \sum _Z P(X,Z)$$，每次单独考虑 $$P(X,Z)$$，避免在 log 中求和。

E 和 M 具体的含义还没有整理！