本文參考的視頻鏈接

首先要知道什么式序列（Series），什么是集合（Set）

時間序列模型 Discrete Dynamic Model: Hidden Markov Model

P ( X t ∣ X t ? 1 , X t ? 2 … . X 1 ) = P ( X t ∣ X t ? 1 ) (1) \begin{aligned} & P\left(X_{t} | X_{t-1}, X_{t-2} \ldots . X_{1}\right) \\ =& P\left(X_{t} \mid X_{t-1}\right)\tag{1} \end{aligned} =?P(Xt?∣Xt?1?,Xt?2?….X1?)P(Xt?∣Xt?1?)?(1)

1. 馬爾可夫程序簡介

在我們知道一系列隱狀態之后，我們的觀測都是獨立的

股市中的箭頭的數值指的就是式(1)的概率值

為了所有的符號一致，現在把所有的隱狀態記為q：

• p ( q t ∣ q t ? 1 ) p(q_t|q_{t-1}) p(qt?∣qt?1?)→transition probability（轉移概率，在HMM里面，一定是離散的）
• p ( y t ∣ q t ) p(y_t|q_t) p(yt?∣qt?)→emission/measurement probability（發射概率，并不一定是離散的）

這兩個概率決定了HMM模型

在語音里面的應用如下所示（音標是隱變數）

HMM圖模型

知道隱狀態之后，觀測都是獨立的！

2. {A、B、 π \pi π}

在HMM里面，transition probability用一個矩陣（k×k）來表示：

• 我們用一個 A k × k A_{k×k} Ak×k?的矩陣代表 p ( q t ∣ q t ? 1 ) p(q_t|q_{t-1}) p(qt?∣qt?1?)
• 假設 p ( y t ∣ q t ) p(y_t|q_t) p(yt?∣qt?)是離散的，我們用一個 B k × L B_{k×L} Bk×L?來表示

現在思考，是否只有 λ = { A , B } \lambda=\{A,B\} λ={A,B}便可以描述HMM模型

所以請思考，怎么計算股票觀測到 P ( y 1 = u p , y 2 = u p , y 3 = d o w n ) P(y_1=up,y_2=up,y_3=down) P(y1?=up,y2?=up,y3?=down)的概率？

我們知道： P ( x ) = ∫ y P ( x , y ) d y P(x)=\int_{y} P(x, y) d y P(x)=∫y?P(x,y)dy
所以原概率可以轉化為：
P ( y 1 , y 2 , y 3 ) = ∑ q 1 = 1 k ∑ q 2 = 1 k ∑ q 3 = 1 k p ( y 1 , y 2 , y 3 , q 1 , q 2 , q 3 ) = ∑ q 1 = 1 k ∑ q 2 = 1 k ∑ q 3 = 1 k p ( y 3 ∣ y 1 , y 2 , q 1 , q 2 , q 3 ) × p ( y 1 , y 2 , q 1 , q 2 , q 3 ) = ∑ q 1 = 1 k ∑ q 2 = 1 k ∑ q 3 = 1 k p ( y 3 ∣ q 3 ) × p ( y 1 , y 2 , q 1 , q 2 , q 3 ) = ∑ q 1 = 1 k ∑ q 2 = 1 k ∑ q 3 = 1 k p ( y 3 ∣ q 3 ) × p ( q 3 ∣ y 1 , y 2 , q 1 , q 2 ) × ( y 1 , y 2 , q 1 , q 2 ) = ∑ q 1 = 1 k ∑ q 2 = 1 k ∑ q 3 = 1 k p ( y 3 ∣ q 3 ) × p ( q 3 ∣ q 2 ) × ( y 1 , y 2 , q 1 , q 2 ) = ∑ q 1 = 1 k ∑ q 2 = 1 k ∑ q 3 = 1 k p ( y 3 ∣ q 3 ) × p ( q 3 ∣ q 2 ) × p ( y 2 ∣ q 2 ) × p ( q 2 ∣ q 1 ) × p ( y 1 ∣ q 1 ) × p ( q 1 ) \begin{aligned} P\left(y_{1}, y_{2}, y_{3}\right)&=\sum_{q_1=1}^{k} \sum_{q_2=1}^{k} \sum_{q_{3}=1}^{k} p\left(y_{1}, y_{2}, y_{3}, q_{1}, q_{2}, q_{3}\right)\\ &=\sum_{q_1=1}^{k} \sum_{q_2=1}^{k} \sum_{q_{3}=1}^{k}p(y_3|y_{1}, y_{2}, q_{1}, q_{2}, q_{3})\times p(y_{1}, y_{2}, q_{1}, q_{2}, q_{3})\\ &=\sum_{q_1=1}^{k} \sum_{q_2=1}^{k} \sum_{q_{3}=1}^{k}p(y_3|q_{3})\times p(y_{1}, y_{2}, q_{1}, q_{2}, q_{3})\\ &=\sum_{q_1=1}^{k} \sum_{q_2=1}^{k} \sum_{q_{3}=1}^{k}p(y_3|q_{3})\times p(q_{3}|y_{1}, y_{2}, q_{1}, q_{2})\times ({y_{1}, y_{2}, q_{1}, q_{2}})\\ &=\sum_{q_1=1}^{k} \sum_{q_2=1}^{k} \sum_{q_{3}=1}^{k}p(y_3|q_{3})\times p(q_{3}|q_{2})\times ({y_{1}, y_{2}, q_{1}, q_{2}})\\ &=\sum_{q_1=1}^{k} \sum_{q_2=1}^{k} \sum_{q_{3}=1}^{k}p(y_3|q_{3})\times p(q_{3}|q_{2})\times p(y_2|q_{2})\times p(q_{2}|q_{1})\times p(y_1|q_{1})\times p(q_{1}) \end{aligned} P(y1?,y2?,y3?)?=q1?=1∑k?q2?=1∑k?q3?=1∑k?p(y1?,y2?,y3?,q1?,q2?,q3?)=q1?=1∑k?q2?=1∑k?q3?=1∑k?p(y3?∣y1?,y2?,q1?,q2?,q3?)×p(y1?,y2?,q1?,q2?,q3?)=q1?=1∑k?q2?=1∑k?q3?=1∑k?p(y3?∣q3?)×p(y1?,y2?,q1?,q2?,q3?)=q1?=1∑k?q2?=1∑k?q3?=1∑k?p(y3?∣q3?)×p(q3?∣y1?,y2?,q1?,q2?)×(y1?,y2?,q1?,q2?)=q1?=1∑k?q2?=1∑k?q3?=1∑k?p(y3?∣q3?)×p(q3?∣q2?)×(y1?,y2?,q1?,q2?)=q1?=1∑k?q2?=1∑k?q3?=1∑k?p(y3?∣q3?)×p(q3?∣q2?)×p(y2?∣q2?)×p(q2?∣q1?)×p(y1?∣q1?)×p(q1?)?
所以現在還差了一個 p ( q 1 ) p(q_1) p(q1?)，所以現在需要一個初始狀態的概率，所以我們還需要一個引數 π \pi π

HMM三個主要的操作如下：
Evaluate p ( Y ∣ λ ) λ MLE = arg ? max ? λ p ( Y ∣ λ ) arg ? max ? Q p ( Y ∣ Q , λ ) \begin{aligned} &\text { Evaluate } p(Y \mid \lambda) \\ &\lambda_{\text {MLE }}=\underset{\lambda}{\arg \max } p(Y \mid \lambda) \\ &\underset{Q}{\arg \max } p(Y \mid Q, \lambda) \end{aligned} ? Evaluate p(Y∣λ)λMLE ?=λargmax?p(Y∣λ)Qargmax?p(Y∣Q,λ)?
下面首先討論Evaluation

如何應用HMM，以語音識別為例，例如找50個人說同樣的cat或者dog…

λ c a t = arg?max ? λ l o g P ( y ( 1 ) ( 1 ) , y ( 2 ) ( 2 ) . . . ∣ λ ) \lambda_{cat} = \argmax_{\lambda} logP(y_{(1)}^{(1)},y_{(2)}^{(2)}...|\lambda) λcat?=λargmax?logP(y(1)(1)?,y(2)(2)?...∣λ)
之后我們想要干嘛？語音識別，有個人說了一段從來沒有聽說過的錄音，現在可以去評估說哪個單詞的概率最高，即Evaluate

下面來看Evaluation，即有了 λ \lambda λ來估計觀測值：
P ( y 1 . . . y T ∣ λ ) P\left(y_{1} ... y_{T} \mid \lambda\right) P(y1?...yT?∣λ)

= ∑ q 1 = 1 k ∑ q 2 = 1 k … ∑ q T = 1 k P ( y 1 … . y T , q 1 … q T ) ? =\sum_{q_{1}=1}^{k} \sum_{q_{2}=1}^{k} \ldots \sum_{q_{T}=1}^{k} \underbrace{P\left(y_{1} \ldots . y_{T}, q_{1} \ldots q_{T}\right)} =q1?=1∑k?q2?=1∑k?…qT?=1∑k? P(y1?….yT?,q1?…qT?)?

通常計算的方法如下所示：通常計算的方法如下所示：
p ( Y ∣ λ ) = ∑ Q [ p ( Y , Q ∣ λ ) ] = ∑ q 1 = 1 k … , ∑ q T = 1 k [ p ( y 1 , … , y T , q 1 , … q T ∣ λ ) ] = ∑ q 1 = 1 k … , ∑ q T = 1 k [ p ( y 1 , … , y T , q 0 , q 1 , … q T ∣ λ ) ] = ∑ q 1 = 1 k … , ∑ q T = 1 k p ( q 1 ) p ( y 1 ∣ q 1 ) p ( q 2 ∣ q 1 ) … p ( q t ∣ q t ? 1 ) p ( y t ∣ q t ) = ∑ q 1 = 1 k … , ∑ q T = 1 k π ( q 1 ) ∏ t = 2 T a q t ? 1 , q t ∏ t = 1 T b q t ( y t ) \begin{aligned} p(Y \mid \lambda) &=\sum_{Q}[p(Y, Q \mid \lambda)]=\sum_{q_{1}=1}^{k} \ldots, \sum_{q_{T}=1}^{k}\left[p\left(y_{1}, \ldots, y_{T}, q_{1}, \ldots q_{T} \mid \lambda\right)\right] \\ &=\sum_{q_{1}=1}^{k} \ldots, \sum_{q_{T}=1}^{k}\left[p\left(y_{1}, \ldots, y_{T}, q_{0}, q_{1}, \ldots q_{T} \mid \lambda\right)\right] \\ &=\sum_{q_{1}=1}^{k} \ldots, \sum_{q_{T}=1}^{k} p\left(q_{1}\right) p\left(y_{1} \mid q_{1}\right) p\left(q_{2} \mid q_{1}\right) \ldots p\left(q_{t} \mid q_{t-1}\right) p\left(y_{t} \mid q_{t}\right) \\ &=\sum_{q_{1}=1}^{k} \ldots, \sum_{q_{T}=1}^{k} \pi\left(q_{1}\right) \prod_{t=2}^{T} a_{q_{t-1}, q_{t}} \prod_{t=1}^{T}b_{q_{t}}\left(y_{t}\right) \end{aligned} p(Y∣λ)?=Q∑?[p(Y,Q∣λ)]=q1?=1∑k?…,qT?=1∑k?[p(y1?,…,yT?,q1?,…qT?∣λ)]=q1?=1∑k?…,qT?=1∑k?[p(y1?,…,yT?,q0?,q1?,…qT?∣λ)]=q1?=1∑k?…,qT?=1∑k?p(q1?)p(y1?∣q1?)p(q2?∣q1?)…p(qt?∣qt?1?)p(yt?∣qt?)=q1?=1∑k?…,qT?=1∑k?π(q1?)t=2∏T?aqt?1?,qt??t=1∏T?bqt??(yt?)?

其中，定義轉移概率為轉移概率矩陣的第 i i i行，第 j j j列： p ( q t = j ∣ q t ? 1 = i ) ≡ a i , j p\left(q_{t}=j \mid q_{t-1}=i\right) \equiv a_{i, j} p(qt?=j∣qt?1?=i)≡ai,j?，同時定義測量概率為： p ( y t ∣ q t = j ) ≡ b j ( y t ) p\left(y_{t} \mid q_{t}=j\right) \equiv b_{j}\left(y_{t}\right) p(yt?∣qt?=j)≡bj?(yt?)，注意到這里有 k T k^T kT個可能的 Q Q Q值，所以我們需要更簡單的方法，

3. 引入 α , β \alpha,\beta α,β便于Evaluate

存在的問題，運算量太大了！所以這里假設一個 α \alpha α和 β \beta β

聯合概率：
α i ( t ) = p ( y 1 , y 2 , … y t , q t = i ) \alpha_{i}(t)=p\left(y_{1}, y_{2}, \ldots y_{t}, q_{t}=i \right) αi?(t)=p(y1?,y2?,…yt?,qt?=i)
按照以上概率：
α i ( 1 ) = p ( y 1 , q 1 = i ) = p ( y 1 ∣ q 1 = i ) × p ( q 1 = i ) = b i ( y 1 ) ? π ( q 1 ) \alpha_{i}(1)=p\left(y_{1}, q_{1}=i \right)=p(y_1|q_{1}=i)\times p(q_{1}=i)=b_i(y_1) \cdot \pi(q_1) αi?(1)=p(y1?,q1?=i)=p(y1?∣q1?=i)×p(q1?=i)=bi?(y1?)?π(q1?)

對于 α i ( 2 ) \alpha_i(2) αi?(2)，由于 P ( y 2 ∣ q 2 = j ) P\left(y_{2} \mid q_{2}=j\right) P(y2?∣q2?=j)一項與 i i i沒有關系：

α i ( 2 ) = p ( y 1 , y 2 , q 2 = j ) = ∑ i = 1 k p ( y 1 , y 2 , q 1 = i , q 2 = j ) = ∑ i = 1 k p ( y 2 ∣ q 2 = j ) ? p ( q 1 = j ∣ q 1 = i ) ? p ( y 1 , q 1 = i ) ? α i ( 1 ) = p ( y 2 ∣ q 2 = j ) ? b j ( y 2 ) ∑ i = 1 k p ( q 1 = j ∣ q 1 = i ) ? a i , j α i ( 1 ) = b j ( y 2 ) ∑ i = 1 k a i , j α i ( 1 ) . . . α j ( t + 1 ) = [ ∑ i = 1 k α i ( t ) a i , j ] b j ( y t + 1 ) . . . α i ( T ) = b j ( y T ) [ ∑ i = 1 k a i , j α i ( T ? 1 ) ] \begin{aligned} \alpha_{i}(2)&=p(y_1,y_2,q_2=j)\\ &=\sum_{i=1}^{k}p(y_1,y_2,q_1=i,q_2=j)\\ &=\sum_{i=1}^{k}p\left(y_{2} \mid q_{2}=j\right) \cdot p\left(q_{1}=j \mid q_1=i\right) \cdot \underbrace{p\left(y_{1}, q_{1}=i\right)}_{\alpha_{i}(1)}\\ &=\underbrace{p\left(y_{2} \mid q_{2}=j\right)}_{b_j(y_2)}\sum_{i=1}^{k}\underbrace{p\left(q_{1}=j \mid q_1=i\right)}_{a_{i,j}}\alpha_{i}(1)\\ &=b_j(y_2)\sum_{i=1}^{k}a_{i,j}\alpha_{i}(1)\\ &...\\ \alpha_{j}(t+1)&=\left[\sum_{i=1}^{k} \alpha_{i}(t) a_{i, j}\right] b_{j}\left(y_{t+1}\right)\\ &...\\ \alpha_{i}(T)&=b_{j}\left(y_{T}\right)\left[\sum_{i=1}^{k} a_{i, j} \alpha_{i}(T-1)\right] \end{aligned} αi?(2)αj?(t+1)αi?(T)?=p(y1?,y2?,q2?=j)=i=1∑k?p(y1?,y2?,q1?=i,q2?=j)=i=1∑k?p(y2?∣q2?=j)?p(q1?=j∣q1?=i)?αi?(1) p(y1?,q1?=i)??=bj?(y2?) p(y2?∣q2?=j)??i=1∑k?ai,j? p(q1?=j∣q1?=i)??αi?(1)=bj?(y2?)i=1∑k?ai,j?αi?(1)...=[i=1∑k?αi?(t)ai,j?]bj?(yt+1?)...=bj?(yT?)[i=1∑k?ai,j?αi?(T?1)]?

現在只有 k × T k\times T k×T個運算量而不是 K T K^T KT了，所以到此為止，我們定義了前向程序：
α i ( t ) = p ( y 1 , y 2 , … y t , q t = i ∣ λ ) ? p ( Y ∣ λ ) = ∑ i = 1 k α i ( T ) \alpha_{i}(t)=p\left(y_{1}, y_{2}, \ldots y_{t}, q_{t}=i \mid \lambda\right) \Longrightarrow p(Y \mid \lambda)=\sum_{i=1}^{k} \alpha_{i}(T) αi?(t)=p(y1?,y2?,…yt?,qt?=i∣λ)?p(Y∣λ)=i=1∑k?αi?(T)

這是最后在t時處于狀態i部分序列 y 1 , … , y t y_1,\dots,y_t y1?,…,yt?的概率，

現在的運算量只有KT而不是K^T：
P ( y 1 … y T ) = ∑ j = 1 k α j ( T ) {P\left(y_{1} \ldots y_{T}\right)}{=\sum_{j=1}^{k} \alpha_{j}(T)} P(y1?…yT?)=j=1∑k?αj?(T)

以上講解的都是Evaluate，現在講怎么把 λ \lambda λ學出來

4. EM演算法引數學習

EM演算法：

θ ( g + 1 ) = argmax ? θ ∫ Q log ? P ( Y , Q ) ? P ( Q ∣ Y , θ ( g ) ) ? d Q \theta^{(g+1)}=\underset{\theta}{\operatorname{argmax}} \int_{Q} \log P(Y, Q) \cdot P\left(Q \mid Y ,\theta^{(g)}\right) \cdot d Q θ(g+1)=θargmax?∫Q?logP(Y,Q)?P(Q∣Y,θ(g))?dQ
因為 P ( Q , Y ∣ θ ( g ) ) P\left(Q ,Y \mid\theta^{(g)}\right) P(Q,Y∣θ(g))= P ( Q ∣ Y , θ ( g ) ) ? P ( Y ∣ θ ( g ) ) P\left(Q \mid Y ,\theta^{(g)}\right) \cdot P(Y \mid \theta^{(g)}) P(Q∣Y,θ(g))?P(Y∣θ(g))，由于最后一項與 θ \theta θ沒有關系，與 Q Q Q沒有關系，這是一個常數，乘不乘對結果沒有影響，所以我們寫成：
θ ( g + 1 ) = argmax ? θ ∫ Q log ? P ( Y , Q ) ? P ( Q , Y ∣ θ ( g ) ) ? d Q \theta^{(g+1)}=\underset{\theta}{\operatorname{argmax}} \int_{Q} \log P(Y, Q) \cdot P\left(Q ,Y \mid\theta^{(g)}\right) \cdot d Q θ(g+1)=θargmax?∫Q?logP(Y,Q)?P(Q,Y∣θ(g))?dQ
之前我們已經知道：
p ( Y ∣ λ ) = ∑ Q [ p ( Y , Q ∣ λ ) ] = ∑ q 1 = 1 k … , ∑ q T = 1 k [ p ( y 1 , … , y T , q 1 , … q T ∣ λ ) ] = ∑ q 1 = 1 k … , ∑ q T = 1 k [ p ( y 1 , … , y T , q 0 , q 1 , … q T ∣ λ ) ] = ∑ q 1 = 1 k … , ∑ q T = 1 k p ( q 1 ) p ( y 1 ∣ q 1 ) p ( q 2 ∣ q 1 ) … p ( q t ∣ q t ? 1 ) p ( y t ∣ q t ) = ∑ q 1 = 1 k … , ∑ q T = 1 k π ( q 1 ) ∏ t = 2 T a q t ? 1 , q t b q t ( y t ) \begin{aligned} p(Y \mid \lambda) &=\sum_{Q}[p(Y, Q \mid \lambda)]=\sum_{q_{1}=1}^{k} \ldots, \sum_{q_{T}=1}^{k}\left[p\left(y_{1}, \ldots, y_{T}, q_{1}, \ldots q_{T} \mid \lambda\right)\right] \\ &=\sum_{q_{1}=1}^{k} \ldots, \sum_{q_{T}=1}^{k}\left[p\left(y_{1}, \ldots, y_{T}, q_{0}, q_{1}, \ldots q_{T} \mid \lambda\right)\right] \\ &=\sum_{q_{1}=1}^{k} \ldots, \sum_{q_{T}=1}^{k} p\left(q_{1}\right) p\left(y_{1} \mid q_{1}\right) p\left(q_{2} \mid q_{1}\right) \ldots p\left(q_{t} \mid q_{t-1}\right) p\left(y_{t} \mid q_{t}\right) \\ &=\sum_{q_{1}=1}^{k} \ldots, \sum_{q_{T}=1}^{k} \pi\left(q_{1}\right) \prod_{t=2}^{T} a_{q_{t-1}, q_{t}} b_{q_{t}}\left(y_{t}\right) \end{aligned} p(Y∣λ)?=Q∑?[p(Y,Q∣λ)]=q1?=1∑k?…,qT?=1∑k?[p(y1?,…,yT?,q1?,…qT?∣λ)]=q1?=1∑k?…,qT?=1∑k?[p(y1?,…,yT?,q0?,q1?,…qT?∣λ)]=q1?=1∑k?…,qT?=1∑k?p(q1?)p(y1?∣q1?)p(q2?∣q1?)…p(qt?∣qt?1?)p(yt?∣qt?)=q1?=1∑k?…,qT?=1∑k?π(q1?)t=2∏T?aqt?1?,qt??bqt??(yt?)?
所以前一個式子即為：
θ ( g + 1 ) = argmax ? θ ∫ Q log ? [ p ( q 1 ) ∏ t = 2 T a q t ? 1 , q t ∏ t = 1 T b q t ( y t ) ] ? P ( Q , Y ∣ θ ( g ) ) ? d Q = ∑ q 1 = 1 k … , ∑ q T = 1 k [ log ? π ( q 1 ) + ∑ t = 2 T log ? a q t ? 1 , q t + ∑ t = 1 T log ? b q t ( y t ) ] ? P ( Q , Y ∣ θ ( g ) ) \begin{aligned} \theta^{(g+1)}&=\underset{\theta}{\operatorname{argmax}} \int_{Q} \log [ p\left(q_{1}\right) \prod_{t=2}^{T} a_{q_{t-1}, q_{t}} \prod_{t=1}^{T}b_{q_{t}}\left(y_{t}\right)] \cdot P\left(Q ,Y \mid\theta^{(g)}\right) \cdot d Q\\ &=\sum_{q_{1}=1}^{k} \ldots, \sum_{q_{T}=1}^{k} [\log \pi\left(q_{1}\right) +\sum_{t=2}^{T} \log a_{q_{t-1}, q_{t}}+ \sum_{t=1}^{T}\log b_{q_{t}}\left(y_{t}\right)] \cdot P\left(Q ,Y \mid\theta^{(g)}\right)\\ \end{aligned} θ(g+1)?=θargmax?∫Q?log[p(q1?)t=2∏T?aqt?1?,qt??t=1∏T?bqt??(yt?)]?P(Q,Y∣θ(g))?dQ=q1?=1∑k?…,qT?=1∑k?[logπ(q1?)+t=2∑T?logaqt?1?,qt??+t=1∑T?logbqt??(yt?)]?P(Q,Y∣θ(g))?

對于每一個term，可以分別由觀測資料學習其對應引數，例如對于 π \pi π：
Q term ? 1 = ∑ q 0 = 1 k ? ∑ q T = 1 k ln ? π q 0 p ( q , Y ∣ λ ( g ) ) = ∑ i = 1 k ln ? π i p ( q 0 = i , Y ∣ λ ( g ) ) \mathcal{Q}^{\operatorname{term} 1}=\sum_{q_{0}=1}^{k} \cdots \sum_{q_{T}=1}^{k} \ln \pi_{q_{0}} p\left(q, Y \mid \lambda^{(g)}\right)=\sum_{i=1}^{k} \ln \pi_{i} p\left(q_{0}=i, Y \mid \lambda^{(g)}\right) Qterm1=q0?=1∑k??qT?=1∑k?lnπq0??p(q,Y∣λ(g))=i=1∑k?lnπi?p(q0?=i,Y∣λ(g))
約束條件：
arg ? max ? ( Q term 1 ) with ∑ i = 1 k π i = 1 \arg \max \left(\mathcal{Q}^{\text {term } 1}\right) \text { with } \sum_{i=1}^{k} \pi_{i}=1 argmax(Qterm 1) with i=1∑k?πi?=1
使用拉格朗日中值定理：
L M term 1 = ∑ i = 1 k ln ? π i p ( q 0 = i , Y ∣ λ ( g ) ) + τ ( ∑ i = 1 k π i ? 1 ) \mathbb{L M}^{\text {term } 1}=\sum_{i=1}^{k} \ln \pi_{i} p\left(q_{0}=i, Y \mid \lambda^{(g)}\right)+\tau\left(\sum_{i=1}^{k} \pi_{i}-1\right) LMterm 1=i=1∑k?lnπi?p(q0?=i,Y∣λ(g))+τ(i=1∑k?πi??1)
對兩項求導，令其等于0：
? L M term 1 ? π i = p ( q , Y ∣ λ ( g ) ) π j + τ = 0 ? L M term 1 ? τ = ∑ i = 1 k π i ? 1 = 0 \frac{\partial \mathbb{L} \mathbb{M}^{\text {term } 1}}{\partial \pi_{i}}=\frac{p\left(q, Y \mid \lambda^{(g)}\right)}{\pi_{j}}+\tau=0 \quad \frac{\partial \mathbb{L} \mathbb{M}^{\text {term } 1}}{\partial \tau}=\sum_{i=1}^{k} \pi_{i}-1=0 ?πi??LMterm 1?=πj?p(q,Y∣λ(g))?+τ=0?τ?LMterm 1?=i=1∑k?πi??1=0
p ( q 0 = i , Y ∣ λ ( g ) ) = ? τ π i p\left(q_{0}=i, Y \mid \lambda^{(g)}\right)=-\tau \pi_{i} p(q0?=i,Y∣λ(g))=?τπi?
為了使用到約束條件，兩邊相加：
∑ i = 1 k p ( q 0 = i , Y ∣ λ ( g ) ) = ? τ ∑ i = 1 k π i = ? τ \sum_{i=1}^{k} p\left(q_{0}=i, Y \mid \lambda^{(g)}\right)=-\tau \sum_{i=1}^{k} \pi_{i}=-\tau i=1∑k?p(q0?=i,Y∣λ(g))=?τi=1∑k?πi?=?τ
代入可得：
π i = p ( q 0 = i , Y ∣ λ ( g ) ) ? τ ? π i = p ( q 0 = i , Y ∣ λ ( g ) ) ∑ i = 1 k p ( q 0 = i , Y ∣ λ ( g ) ) \pi_{i}=\frac{p\left(q_{0}=i, Y \mid \lambda^{(g)}\right)}{-\tau} \Longrightarrow \pi_{i}=\frac{p\left(q_{0}=i, Y \mid \lambda^{(g)}\right)}{\sum_{i=1}^{k} p\left(q_{0}=i, Y \mid \lambda(g)\right)} πi?=?τp(q0?=i,Y∣λ(g))??πi?=∑i=1k?p(q0?=i,Y∣λ(g))p(q0?=i,Y∣λ(g))?

小結

基本算是入了一個門，主要在聽思路，三個典型問題的第三個這個視頻沒有講，應該需要用維特比(Viterbi)演算法進行遞回，這個不會；EM演算法公式怎么推的很不會，