《統計學習方法》讀書筆記 阿新 • • 發佈：2018-12-21 10.2.3 後向演算法 βt(i)=∑j=1Naijbj(ot+1)βt+1(j)\beta_t(i)=\displaystyle \sum_{j=1}^N a_{ij} b_j(o_{t+1}) \beta_{t+1}(j)βt(i)=j=1∑Naijbj(ot+1)βt+1(j)（10.20） 證明： ∑j=1Naijbj(ot+1)βt+1(j)=∑j=1NaijP(ot+1∣it+1=qj)P(ot+2,ot+3,⋯ ,oT∣it+1=qj,λ)=∑j=1NaijP(P(ot+1,ot+2,⋯ ,oT∣it+1=qj,λ) =∑j=1NP(it+1=qj∣it=qi)P(ot+1,ot+2,⋯ ,oT∣it+1=qj,λ)=∑j=1NP(it=qi∣it+1=qj)P(it+1=qj)P(it=qi)P(ot+1,ot+2,⋯ ,oT∣it+1=qj,λ)=1P(it=qi)∑j=1NP(ot+1,ot+2,⋯ ,oT,it=qi∣it+1=qj,λ)P(it+1=qj)=∑j=1NP(ot+1,ot+2,⋯ ,oT,it=qi,it+1=qj∣λ)P(it=qi)=P(ot+1,ot+2,⋯ , oT,it=qi∣λ)P(it=qi)=P(ot+1,ot+2,⋯ ,oT,∣it=qi,λ)=βt(i) \begin{aligned} &\displaystyle \sum_{j=1}^N a_{ij} b_j(o_{t+1}) \beta_{t+1}(j) \\ =&\displaystyle \sum_{j=1}^N a_{ij} P(o_{t+1}|i_{t+1}=q_j) P(o_{t+2},o_{t+3},\dotsb,o_{T}| i_{t+1}=q_j,\lambda) \\ =&\displaystyle \sum_{j=1}^N a_{ij} P(P(o_{t+1},o_{t+2},\dotsb,o_{T} | i_{t+1}=q_j,\lambda) \\ =&\displaystyle \sum_{j=1}^N P(i_{t+1}=q_j | i_t=q_i) P(o_{t+1},o_{t+2},\dotsb,o_{T} | i_{t+1}=q_j,\lambda) \\ =&\displaystyle \sum_{j=1}^N \dfrac {P(i_t=q_i | i_{t+1}=q_j)P(i_{t+1}=q_j)}{P(i_t=q_i)} P(o_{t+1},o_{t+2},\dotsb,o_{T} | i_{t+1}=q_j,\lambda)\\ =&\dfrac{1}{P(i_t=q_i)}\displaystyle \sum_{j=1}^N P(o_{t+1},o_{t+2},\dotsb,o_{T},i_t=q_i | i_{t+1}=q_j,\lambda) P(i_{t+1}=q_j)\\ =&\dfrac{\displaystyle \sum_{j=1}^NP(o_{t+1},o_{t+2},\dotsb,o_{T},i_t=q_i,i_{t+1}=q_j|\lambda)}{P(i_t=q_i)} \\ =&\dfrac{P(o_{t+1},o_{t+2},\dotsb,o_{T},i_t=q_i|\lambda)}{P(i_t=q_i)} \\ =&P(o_{t+1},o_{t+2},\dotsb,o_{T},|i_t=q_i, \lambda)\\ =&\beta_t(i) \end{aligned} =========j=1∑Naijbj(ot+1)βt+1(j)j=1∑NaijP(ot+1∣it+1=qj)P(ot+2,ot+3,⋯,oT∣it+1=qj,λ)j=1∑NaijP(P(ot+1,ot+2,⋯,oT∣it+1=qj,λ)j=1∑NP(it+1=qj∣it=qi)P(ot+1,ot+2,⋯,oT∣it+1=qj,λ)j=1∑NP(it=qi)P(it=qi∣it+1=qj)P(it+1=qj)P(ot+1,ot+2,⋯,oT∣it+1=qj,λ)P(it=qi)1j=1∑NP(ot+1,ot+2,⋯,oT,it=qi∣it+1=qj,λ)P(it+1=qj)P(it=qi)j=1∑NP(ot+1,ot+2,⋯,oT,it=qi,it+1=qj∣λ)P(it=qi)P(ot+1,ot+2,⋯,oT,it=qi∣λ)P(ot+1,ot+2,⋯,oT,∣it=qi,λ)βt(i) 10.2.4 一些概率與期望值的計算 P(it=qi,it+1=qj,O∣λ)=αt(i
