# 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$$\alpha\_t(i)$$](https://windmising.gitbook.io/lihang-tongjixuexifangfa/hmm/3)和[后向概率$$\beta\_t(i)$$](https://windmising.gitbook.io/lihang-tongjixuexifangfa/hmm/4)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


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