推导1

$\gamma_t(i)$的推导

$\gamma_t(i)$代表t时刻状态为i的概率。

用公式表达为：

$$\begin{aligned} \gamma_t(i) & = P(i_t = q_i | O, \lambda) \\ & = \frac{P(i_t = q_i , O| \lambda)}{P(O|\lambda)} & \text{贝叶斯公式,}P(A|B) = \frac{P(A, B)}{P(B)} \\ & = \frac{P(i_t = q_i , O| \lambda)}{\sum_{j=1}^NP(i_t = q_i , O| \lambda)} \\ & = \frac{\alpha_t(i)\beta_t(i)}{\sum_{j=1}^N\alpha_t(i)\beta_t(i)} & \text{公式说明1} \end{aligned}$$

公式说明

1. 根据前向概率$\alpha_t(i)$和后向概率$\beta_t(i)$的字义，得：

   $$\begin{aligned} \alpha_t(i)\beta_t(i) & = P(o_1,o_2,\cdots,o_t,i_t=q_i|\lambda) * P(o_{t+1},o_{t+2},\cdots,o_T |i_t=q_i, \lambda) \\ & = P(o_1,o_2,\cdots,o_t|i_t=q_i, \lambda) * P(o_{t+1},o_{t+2},\cdots,o_T |i_t=q_i, \lambda) * P(i_t=q_i|\lambda) \\ & = P(O|i_t=q_i, \lambda) * P(i_t=q_i|\lambda) \\ & = P(O, i_t=q_i| \lambda) \end{aligned}$$

   我的推导方法可能比较笨。

$\xi_t(i,j)$的推导