From 55a275e705dcab1f255c14bac6670f9f6b7ebfae Mon Sep 17 00:00:00 2001 From: ")s" Date: Mon, 28 Oct 2024 19:45:26 +0800 Subject: [PATCH] Update chapter14.md to v2 --- docs/chapter14/chapter14.md | 88 ------------------------------------- 1 file changed, 88 deletions(-) diff --git a/docs/chapter14/chapter14.md b/docs/chapter14/chapter14.md index c762a99..cf25a0c 100644 --- a/docs/chapter14/chapter14.md +++ b/docs/chapter14/chapter14.md @@ -32,9 +32,7 @@ $y$ 的概率, 即根据 $\boldsymbol{x}$ "判别" $y$, 因此称为 "判别式 $$ - P\left(x_1, y_1, \ldots, x_n, y_n\right)=P\left(x_1, \ldots, x_n \mid y_1, \ldots, y_n\right) \cdot P\left(y_1, \ldots, y_n\right) - $$ @@ -43,14 +41,12 @@ $$ $$ - \begin{aligned} P\left(y_1, \ldots, y_n\right) & =P\left(y_n \mid y_1, \ldots, y_{n-1}\right) \cdot P\left(y_1, \ldots, y_{n-1}\right) \\ & =P\left(y_n \mid y_1, \ldots, y_{n-1}\right) \cdot P\left(y_{n-1} \mid y_1, \ldots, y_{n-2}\right) \cdot P\left(y_1, \ldots, y_{n-2}\right) \\ & =\ldots \ldots \\ & =P\left(y_n \mid y_1, \ldots, y_{n-1}\right) \cdot P\left(y_{n-1} \mid y_1, \ldots, y_{n-2}\right) \cdot \ldots \cdot P\left(y_2 \mid y_1\right) \cdot P\left(y_1\right) \end{aligned} - $$ @@ -59,13 +55,11 @@ $$ 决定; 基于这种依赖关系, 有 $$ - \begin{aligned} P\left(y_n \mid y_1, \ldots, y_{n-1}\right) & =P\left(y_n \mid y_{n-1}\right) \\ P\left(y_{n-1} \mid y_1, \ldots, y_{n-2}\right) & =P\left(y_{n-1} \mid y_{n-2}\right) \\ P\left(y_{n-2} \mid y_1, \ldots, y_{n-3}\right) & =P\left(y_{n-2} \mid y_{n-3}\right) \end{aligned} - $$ @@ -73,12 +67,10 @@ $$ 因此 $P\left(y_1, \ldots, y_n\right)$ 可化简为 $$ - \begin{aligned} P\left(y_1, \ldots, y_n\right) & =P\left(y_n \mid y_{n-1}\right) \cdot P\left(y_{n-1} \mid y_{n-2}\right) \cdot \ldots \cdot P\left(y_2 \mid y_1\right) \cdot P\left(y_1\right) \\ & =P\left(y_1\right) \prod_{i=2}^n P\left(y_i \mid y_{i-1}\right) \end{aligned} - $$ @@ -88,13 +80,11 @@ $$ 与其它状态变量及观测变量的取值无关。因此 $$ - \begin{aligned} P\left(x_1, \ldots, x_n \mid y_1, \ldots, y_n\right) & =P\left(x_1 \mid y_1, \ldots, y_n\right) \cdot \ldots \cdot P\left(x_n \mid y_1, \ldots, y_n\right) \\ & =P\left(x_1 \mid y_1\right) \cdot \ldots \cdot P\left(x_n \mid y_n\right) \\ & =\prod_{i=1}^n P\left(x_i \mid y_i\right) \end{aligned} - $$ @@ -102,13 +92,11 @@ $$ 综上所述, 可得 $$ - \begin{aligned} P\left(x_1, y_1, \ldots, x_n, y_n\right) & =P\left(x_1, \ldots, x_n \mid y_1, \ldots, y_n\right) \cdot P\left(y_1, \ldots, y_n\right) \\ & =\left(\prod_{i=1}^n P\left(x_i \mid y_i\right)\right) \cdot\left(P\left(y_1\right) \prod_{i=2}^n P\left(y_i \mid y_{i-1}\right)\right) \\ & =P\left(y_1\right) P\left(x_1 \mid y_1\right) \prod_{i=2}^n P\left(y_i \mid y_{i-1}\right) P\left(x_i \mid y_i\right) \end{aligned} - $$ @@ -150,9 +138,7 @@ $\psi_{A C}\left(x_A^{\prime}, x_C\right)$ 与变量 $x_B^{\prime}$ 无关, $$ - \sum_{x_A^{\prime}} \sum_{x_B^{\prime}} \psi_{A C}\left(x_A^{\prime}, x_C\right) \psi_{B C}\left(x_B^{\prime}, x_C\right)=\sum_{x_A^{\prime}} \psi_{A C}\left(x_A^{\prime}, x_C\right) \sum_{x_B^{\prime}} \psi_{B C}\left(x_B^{\prime}, x_C\right) - $$ @@ -162,13 +148,11 @@ $\mathbf{x}=\left\{x_1, x_2, x_3\right\}, \mathbf{y}=\left\{y_1, y_2, y_3\right\ 则 $$ - \begin{aligned} \sum_{i=1}^3 \sum_{j=1}^3 x_i y_j & =x_1 y_1+x_1 y_2+x_1 y_3+x_2 y_1+x_2 y_2+x_2 y_3+x_3 y_1+x_3 y_2+x_3 y_3 \\ & =x_1 \times\left(y_1+y_2+y_3\right)+x_2 \times\left(y_1+y_2+y_3\right)+x_3 \times\left(y_1+y_2+y_3\right) \\ & =\left(x_1+x_2+x_3\right) \times\left(y_1+y_2+y_3\right)=\left(\sum_{i=1}^3 x_i\right)\left(\sum_{j=1}^3 y_j\right) \end{aligned} - $$ @@ -221,9 +205,7 @@ $P(\mathbf{y} \mid \mathbf{x})$, 因此它 是一种判别式模型, 参见"西 $$ - P\left(y_{v} | \mathbf{x}, \mathbf{y}_{V \backslash\{v\}}\right)=P\left(y_{v} | \mathbf{x}, \mathbf{y}_{n(v)}\right) - $$ @@ -264,14 +246,12 @@ $$ 请一定理解并记住其含义。依次推导如下: $$ - \begin{aligned} & m_{12}\left(x_2\right)=\sum_{x_1} P\left(x_1\right) P\left(x_2 \mid x_1\right)=\sum_{x_1} P\left(x_2, x_1\right)=P\left(x_2\right) \\ & m_{23}\left(x_3\right)=\sum_{x_2} P\left(x_3 \mid x_2\right) m_{12}\left(x_2\right)=\sum_{x_2} P\left(x_3, x_2\right)=P\left(x_3\right) \\ & \left.m_{43}\left(x_3\right)=\sum_{x_4} P\left(x_4 \mid x_3\right) m_{23}\left(x_3\right)=\sum_{x_4} P\left(x_4, x_3\right)=P\left(x_3\right) \text { (这里与书中不一样 }\right) \\ & m_{35}\left(x_5\right)=\sum_{x_3} P\left(x_5 \mid x_3\right) m_{43}\left(x_3\right)=\sum_{x_3} P\left(x_5, x_3\right)=P\left(x_5\right) \end{aligned} - $$ 注意: 这里的过程与"西瓜书"中不太一样, 但本质一样, 因为 @@ -313,14 +293,12 @@ $n(3) \backslash 5=\{2,4\}$ （因为 $x_3$ 有邻接结点 2,4 和 5 )。 接下来, 仍然以图14.7 计算 $P\left(x_5\right)$ 为例: $$ - \begin{aligned} & m_{12}\left(x_2\right)=\sum_{x_1} \psi_{12}\left(x_1, x_2\right) \prod_{k \in n(1) \backslash 2} m_{k 1}\left(x_1\right)=\sum_{x_1} \psi_{12}\left(x_1, x_2\right) \\ & m_{23}\left(x_3\right)=\sum_{x_2} \psi_{23}\left(x_2, x_3\right) \prod_{k \in n(2) \backslash 3} m_{k 2}\left(x_2\right)=\sum_{x_1} \psi_{12}\left(x_1, x_2\right) m_{12}\left(x_2\right) \\ & m_{43}\left(x_3\right)=\sum_{x_4} \psi_{34}\left(x_3, x_4\right) \prod_{k \in n(4) \backslash 3} m_{k 4}\left(x_4\right)=\sum_{x_4} \psi_{34}\left(x_3, x_4\right) \\ & m_{35}\left(x_5\right)=\sum_{x_3} \psi_{35}\left(x_3, x_5\right) \prod_{k \in n(3) \backslash 5} m_{k 3}\left(x_3\right)=\sum_{x_3} \psi_{35}\left(x_3, x_5\right) m_{23}\left(x_3\right) m_{43}\left(x_3\right) \end{aligned} - $$ @@ -337,14 +315,12 @@ $$ $$ - \begin{aligned} \hat{f}&=\frac{1}{N} \sum_{j=1}^{M} f\left(x_{j}\right) \cdot m_j \\ &= \sum_{j=1}^{M} f\left(x_{j}\right)\cdot \frac{m_j}{N} \\ &\approx \sum_{j=1}^{M} f\left(x_{j}\right)\cdot p(x_j) \\ &\approx \int f(x) p(x) dx \end{aligned} - $$ @@ -403,11 +379,9 @@ $x_3$ 之间还有很多个结点呢? 使得 $$ - \begin{aligned} \boldsymbol{\pi} \mathbf{T}=\boldsymbol{\pi} \end{aligned} - $$ 其中, @@ -419,22 +393,18 @@ $\boldsymbol{\pi}$是一个是一个$n$维向量，代表$s_1,s_2,..,s_n$对应 事实上，转移矩阵只需要满足马尔可夫细致平稳条件 $$ - \begin{aligned} \pi_i \mathbf{T}_{ij}=\pi_j \mathbf{T}_{ji} \end{aligned} - $$ 即式(14.26)，这里采用的符号与西瓜书略有区别以便于理解. 证明如下 $$ - \begin{aligned} \boldsymbol{\pi} \mathbf{T}_{j\cdot} = \sum _i \pi_i\mathbf{T}_{ij} = \sum _i \pi_j\mathbf{T}_{ji} = \pi_j \end{aligned} - $$ @@ -467,14 +437,12 @@ $A\left(\mathbf{x}^* \mid \mathbf{x}^{t-1}\right)$ 的乘积表示。 式(14.27)等号左边将变为: $$ - \begin{aligned} & p\left(\mathbf{x}^{t-1}\right) Q\left(\mathbf{x}^* \mid \mathbf{x}^{t-1}\right) A\left(\mathbf{x}^* \mid \mathbf{x}^{t-1}\right) \\ = & p\left(\mathbf{x}^{t-1}\right) Q\left(\mathbf{x}^* \mid \mathbf{x}^{t-1}\right) \min \left(1, \frac{p\left(\mathbf{x}^*\right) Q\left(\mathbf{x}^{t-1} \mid \mathbf{x}^*\right)}{p\left(\mathbf{x}^{t-1}\right) Q\left(\mathbf{x}^* \mid \mathbf{x}^{t-1}\right)}\right) \\ = & \min \left(p\left(\mathbf{x}^{t-1}\right) Q\left(\mathbf{x}^* \mid \mathbf{x}^{t-1}\right), p\left(\mathbf{x}^{t-1}\right) Q\left(\mathbf{x}^* \mid \mathbf{x}^{t-1}\right) \frac{p\left(\mathbf{x}^*\right) Q\left(\mathbf{x}^{t-1} \mid \mathbf{x}^*\right)}{p\left(\mathbf{x}^{t-1}\right) Q\left(\mathbf{x}^* \mid \mathbf{x}^{t-1}\right)}\right) \\ = & \min \left(p\left(\mathbf{x}^{t-1}\right) Q\left(\mathbf{x}^* \mid \mathbf{x}^{t-1}\right), p\left(\mathbf{x}^*\right) Q\left(\mathbf{x}^{t-1} \mid \mathbf{x}^*\right)\right) \end{aligned} - $$ @@ -487,9 +455,7 @@ $\mathbf{x}^{t-1}$ 和 $\mathbf{x}^*$ 调换位置), 同理可得如上结果, $$ - A\left(\mathbf{x}^* \mid \mathbf{x}^{t-1}\right)=C \cdot p\left(\mathbf{x}^*\right) Q\left(\mathbf{x}^{t-1} \mid \mathbf{x}^*\right) - $$ @@ -520,9 +486,7 @@ $\frac{1}{p\left(\mathbf{x}^{t-1}\right) Q\left(\mathbf{x}^* \mid \mathbf{x}^{t- $$ - C=\min \left(\frac{1}{\cdot p\left(\mathbf{x}^*\right) Q\left(\mathbf{x}^{t-1} \mid \mathbf{x}^*\right)}, \frac{1}{p\left(\mathbf{x}^{t-1}\right) Q\left(\mathbf{x}^* \mid \mathbf{x}^{t-1}\right)}\right) - $$ @@ -575,9 +539,7 @@ $\mathbf{x}_{\bar{i}}^*=\mathbf{x}_{\bar{i}}^{t-1}$ ) $$ - \frac{p\left(\mathbf{x}^*\right) Q\left(\mathbf{x}^{t-1} \mid \mathbf{x}^*\right)}{p\left(\mathbf{x}^{t-1}\right) Q\left(\mathbf{x}^* \mid \mathbf{x}^{t-1}\right)}=\frac{p\left(x_i^* \mid \mathbf{x}_i^*\right) p\left(\mathbf{x}_i^*\right) p\left(x_i^{t-1} \mid \mathbf{x}_{\bar{i}}^*\right)}{p\left(x_i^{t-1} \mid \mathbf{x}_{\bar{i}}^{t-1}\right) p\left(\mathbf{x}_{\bar{i}}^{t-1}\right) p\left(x_i^* \mid \mathbf{x}_{\bar{i}}^{t-1}\right)}=1 - $$ @@ -620,9 +582,7 @@ $p(\mathbf{x}, \mathbf{z})=p(\mathbf{z} \mid \mathbf{x}) p(\mathbf{x})$, $$ - p(\mathbf{x})=\frac{p(\mathbf{x}, \mathbf{z})}{p(\mathbf{z} \mid \mathbf{x})} - $$ @@ -631,9 +591,7 @@ $$ $$ - p(\mathbf{x})=\frac{p(\mathbf{x}, \mathbf{z}) / q(\mathbf{z})}{p(\mathbf{z} \mid \mathbf{x}) / q(\mathbf{z})} - $$ @@ -642,9 +600,7 @@ $$ $$ - \ln p(\mathbf{x})=\ln \frac{p(\mathbf{x}, \mathbf{z}) / q(\mathbf{z})}{p(\mathbf{z} \mid \mathbf{x}) / q(\mathbf{z})}=\ln \frac{p(\mathbf{x}, \mathbf{z})}{q(\mathbf{z})}-\ln \frac{p(\mathbf{z} \mid \mathbf{x})}{q(\mathbf{z})} - $$ @@ -653,9 +609,7 @@ $$ $$ - \int q(\mathbf{z}) \ln p(\mathbf{x}) \mathrm{d} \mathbf{z}=\int q(\mathbf{z}) \ln \frac{p(\mathbf{x}, \mathbf{z})}{q(\mathbf{z})} \mathrm{d} \mathbf{z}-\int q(\mathbf{z}) \ln \frac{p(\mathbf{z} \mid \mathbf{x})}{q(\mathbf{z})} \mathrm{d} \mathbf{z} - $$ @@ -665,9 +619,7 @@ $$ $$ - \int q(\mathbf{z}) \ln p(\mathbf{x}) \mathrm{d} \mathbf{z}=\ln p(\mathbf{x}) \int q(\mathbf{z}) \mathrm{d} \mathbf{z}=\ln p(\mathbf{x}) - $$ @@ -677,9 +629,7 @@ $$ $$ - \ln p(\mathbf{x})=\int q(\mathbf{z}) \ln \frac{p(\mathbf{x}, \mathbf{z})}{q(\mathbf{z})} \mathrm{d} \mathbf{z}-\int q(\mathbf{z}) \ln \frac{p(\mathbf{z} \mid \mathbf{x})}{q(\mathbf{z})} \mathrm{d} \mathbf{z} - $$ @@ -693,9 +643,7 @@ $p(\mathbf{z} \mid \mathbf{x})$, 而 $\mathrm{KL}$ $$ - \min _{q(\mathbf{z})} \operatorname{KL}(q(\mathbf{z}) \| p(\mathbf{z} \mid \mathbf{x})) - $$ @@ -721,13 +669,11 @@ $q(\mathbf{z})=\prod_{i=1}^M q_i\left(\mathbf{z}_i\right)$, 将式(14.35)代入式(14.33), 得: $$ - \begin{aligned} \mathcal{L}(q) & =\int q(\mathbf{z}) \ln \frac{p(\mathbf{x}, \mathbf{z})}{q(\mathbf{z})} \mathrm{d} \mathbf{z}=\int q(\mathbf{z})\{\ln p(\mathbf{x}, \mathbf{z})-\ln q(\mathbf{z})\} \mathrm{d} \mathbf{z} \\ & =\int \prod_{i=1}^M q_i\left(\mathbf{z}_i\right)\left\{\ln p(\mathbf{x}, \mathbf{z})-\ln \prod_{i=1}^M q_i\left(\mathbf{z}_i\right)\right\} \mathrm{d} \mathbf{z} \\ & =\int \prod_{i=1}^M q_i\left(\mathbf{z}_i\right) \ln p(\mathbf{x}, \mathbf{z}) \mathrm{d} \mathbf{z}-\int \prod_{i=1}^M q_i\left(\mathbf{z}_i\right) \ln \prod_{i=1}^M q_i\left(\mathbf{z}_i\right) \mathrm{d} \mathbf{z} \triangleq \mathcal{L}_1(q)-\mathcal{L}_2(q) \end{aligned} - $$ @@ -737,9 +683,7 @@ $Q(\mathbf{x}, \mathbf{z})$, 则上式可变形为: $$ - \mathcal{L}(q)=\int Q(\mathbf{x}, \mathbf{z}) \mathrm{d} \mathbf{z}=\int \cdots \int Q(\mathbf{x}, \mathbf{z}) \mathrm{d} \mathbf{z}_1 \mathrm{~d} \mathbf{z}_2 \cdots \mathrm{d} \mathbf{z}_M - $$ @@ -750,9 +694,7 @@ $$ $$ - \mathcal{L}_1(q)=\int \prod_{i=1}^M q_i\left(\mathbf{z}_i\right) \ln p(\mathbf{x}, \mathbf{z}) \mathrm{d} \mathbf{z}=\int q_j\left\{\int \ln p(\mathbf{x}, \mathbf{z}) \prod_{i \neq j}^M\left(q_i\left(\mathbf{z}_i\right) \mathrm{d} \mathbf{z}_i\right)\right\} \mathrm{d} \mathbf{z}_j - $$ @@ -763,9 +705,7 @@ $\ln \tilde{p}\left(\mathbf{x}, \mathbf{z}_j\right)=\int \ln p(\mathbf{x}, \math $$ - \mathcal{L}_1(q)=\int q_j \ln \tilde{p}\left(\mathbf{x}, \mathbf{z}_j\right) \mathrm{d} \mathbf{z}_j - $$ @@ -773,12 +713,10 @@ $$ 对于第 2 项 $\mathcal{L}_2(q):$ $$ - \begin{aligned} \mathcal{L}_2(q) & =\int \prod_{i=1}^M q_i\left(\mathbf{z}_i\right) \ln \prod_{i=1}^M q_i\left(\mathbf{z}_i\right) \mathrm{d} \mathbf{z}=\int \prod_{i=1}^M q_i\left(\mathbf{z}_i\right) \sum_{i=1}^M \ln q_i\left(\mathbf{z}_i\right) \mathrm{d} \mathbf{z} \\ & =\sum_{i=1}^M \int \prod_{i=1}^M q_i\left(\mathbf{z}_i\right) \ln q_i\left(\mathbf{z}_i\right) \mathrm{d} \mathbf{z}=\sum_{i_1=1}^M \int \prod_{i_2=1}^M q_{i_2}\left(\mathbf{z}_{i_2}\right) \ln q_{i_1}\left(\mathbf{z}_{i_1}\right) \mathrm{d} \mathbf{z} \end{aligned} - $$ @@ -789,12 +727,10 @@ $$ 积分项，考虑当 $i_1=j$ 时: $$ - \begin{aligned} \int \prod_{i_2=1}^M q_{i_2}\left(\mathbf{z}_{i_2}\right) \ln q_j\left(\mathbf{z}_j\right) \mathrm{d} \mathbf{z} & =\int q_j\left(\mathbf{z}_j\right) \prod_{i_2 \neq j} q_{i_2}\left(\mathbf{z}_{i_2}\right) \ln q_j\left(\mathbf{z}_j\right) \mathrm{d} \mathbf{z} \\ & =\int q_j\left(\mathbf{z}_j\right) \ln q_j\left(\mathbf{z}_j\right)\left\{\int \prod_{i_2 \neq j} q_{i_2}\left(\mathbf{z}_{i_2}\right) \prod_{i_2 \neq j} \mathrm{~d} \mathbf{z}_{i_2}\right\} \mathrm{d} \mathbf{z}_j \end{aligned} - $$ @@ -806,9 +742,7 @@ $q_2\left(\mathbf{z}_2\right)$ 和 $q_3\left(\mathbf{z}_3\right)$, 即: $$ - \iiint q_1\left(\mathbf{z}_1\right) q_2\left(\mathbf{z}_2\right) q_3\left(\mathbf{z}_3\right) \mathrm{d} \mathbf{z}_1 \mathrm{~d} \mathbf{z}_2 \mathrm{~d} \mathbf{z}_3=\int q_1\left(\mathbf{z}_1\right) \int q_2\left(\mathbf{z}_2\right) \int q_3\left(\mathbf{z}_3\right) \mathrm{d} \mathbf{z}_3 \mathrm{~d} \mathbf{z}_2 \mathrm{~d} \mathbf{z}_1 - $$ @@ -819,9 +753,7 @@ $\int q_1\left(\mathbf{z}_1\right) \mathrm{d} \mathbf{z}_1=\int q_2\left(\mathbf $$ - \int \prod_{i_2=1}^M q_{i_2}\left(\mathbf{z}_{i_2}\right) \ln q_j\left(\mathbf{z}_j\right) \mathrm{d} \mathbf{z}=\int q_j\left(\mathbf{z}_j\right) \ln q_j\left(\mathbf{z}_j\right) \mathrm{d} \mathbf{z}_j - $$ @@ -829,12 +761,10 @@ $$ 进而第 2 项可化简为: $$ - \begin{aligned} \mathcal{L}_2(q) & =\sum_{i_1=1}^M \int q_{i_1}\left(\mathbf{z}_{i_1}\right) \ln q_{i_1}\left(\mathbf{z}_{i_1}\right) \mathrm{d} \mathbf{z}_{i_1} \\ & =\int q_j\left(\mathbf{z}_j\right) \ln q_j\left(\mathbf{z}_j\right) \mathrm{d} \mathbf{z}_j+\sum_{i_1 \neq j}^M \int q_{i_1}\left(\mathbf{z}_{i_1}\right) \ln q_{i_1}\left(\mathbf{z}_{i_1}\right) \mathrm{d} \mathbf{z}_{i_1} \end{aligned} - $$ @@ -844,9 +774,7 @@ $$ $$ - \mathcal{L}_2(q)=\int q_j\left(\mathbf{z}_j\right) \ln q_j\left(\mathbf{z}_j\right) \mathrm{d} \mathbf{z}_j+\text { const } - $$ @@ -868,7 +796,6 @@ $\ln \tilde{p}\left(\mathbf{x}, \mathbf{z}_j\right)$, 但该式却包 项, 即: $$ - \begin{aligned} & \int q_j\left\{\int \ln p(\mathbf{x}, \mathbf{z}) \prod_{i \neq j}^M\left(q_i\left(\mathbf{z}_i\right) \mathrm{d} \mathbf{z}_i\right)\right\} \mathrm{d} \mathbf{z}_j=\int q_j \mathbb{E}_{i \neq j}[\ln p(\mathbf{x}, \mathbf{z})] \mathrm{d} \mathbf{z}_j \\ & =\int q_j\left(\ln \tilde{p}\left(\mathbf{x}, \mathbf{z}_j\right)-\text { const }\right) \mathrm{d} \mathbf{z}_j \\ @@ -876,7 +803,6 @@ $$ & =\int q_j \ln \tilde{p}\left(\mathbf{x}, \mathbf{z}_j\right) \mathrm{d} \mathbf{z}_j-\text { const } \\ & \end{aligned} - $$ @@ -890,13 +816,11 @@ $$ 对于式(14.36), 可继续变形为: $$ - \begin{aligned} \mathcal{L}(q) & =\int q_j \ln \tilde{p}\left(\mathbf{x}, \mathbf{z}_j\right) \mathrm{d} \mathbf{z}_j-\int q_j \ln q_j \mathrm{~d} \mathbf{z}_j+\mathrm{const} \\ & =\int q_j \ln \frac{\tilde{p}\left(\mathbf{x}, \mathbf{z}_j\right)}{q_j} \mathrm{~d} \mathbf{z}_j+\mathrm{const} \\ & =-\mathrm{KL}\left(q_j \| \tilde{p}\left(\mathbf{x}, \mathbf{z}_j\right)\right)+\mathrm{const} \end{aligned} - $$ 注意, 在前面关于 "式(14.32) 式(14.34)的推导" 中提到, @@ -924,12 +848,10 @@ $\ln q_j=\mathbb{E}_{i \neq j}[\ln p(\mathbf{x}, \mathbf{z})]+\mathrm{const}$, 对式(14.39)两边同时取 $\exp (\cdot)$ 操作, 得 $$ - \begin{aligned} q_j^*\left(\mathbf{z}_j\right) & =\exp \left(\mathbb{E}_{i \neq j}[\ln p(\mathbf{x}, \mathbf{z})]+\text { const }\right) \\ & =\exp \left(\mathbb{E}_{i \neq j}[\ln p(\mathbf{x}, \mathbf{z})]\right) \cdot \exp (\text { const }) \end{aligned} - $$ 两边同时取积分 $\int(\cdot) \mathrm{d} \mathbf{z}_j$ @@ -938,33 +860,27 @@ $\int q_j^*\left(\mathbf{z}_j\right) \mathrm{d} \mathbf{z}_j=1$, 因此有 $$ - \begin{aligned} 1 & =\int \exp \left(\mathbb{E}_{i \neq j}[\ln p(\mathbf{x}, \mathbf{z})]\right) \cdot \exp (\text { const }) \mathrm{d} \mathbf{z}_j \\ & =\exp (\text { const }) \int \exp \left(\mathbb{E}_{i \neq j}[\ln p(\mathbf{x}, \mathbf{z})]\right) \mathrm{d} \mathbf{z}_j \end{aligned} - $$ 这里就是将常数拿到了积分号外面, 因此: $$ - \exp (\text { const })=\frac{1}{\int \exp \left(\mathbb{E}_{i \neq j}[\ln p(\mathbf{x}, \mathbf{z})]\right) \mathrm{d} \mathbf{z}_j} - $$ 代入刚开始的表达式, 可得本式: $$ - \begin{aligned} q_j^*\left(\mathbf{z}_j\right) & =\exp \left(\mathbb{E}_{i \neq j}[\ln p(\mathbf{x}, \mathbf{z})]\right) \cdot \exp (\text { const }) \\ & =\frac{\exp \left(\mathbb{E}_{i \neq j}[\ln p(\mathbf{x}, \mathbf{z})]\right)}{\int \exp \left(\mathbb{E}_{i \neq j}[\ln p(\mathbf{x}, \mathbf{z})]\right) \mathrm{d} \mathbf{z}_j} \end{aligned} - $$ 实际上, 本式的分母为归一化因子, 以保证 @@ -982,12 +898,10 @@ $q_j^*\left(\mathbf{z}_j\right)$ 为概率分布。 $$ - p(\boldsymbol W,\boldsymbol z,\boldsymbol \beta,\boldsymbol \theta | \boldsymbol \alpha,\boldsymbol \eta) = \prod_{t=1}^{T}p(\boldsymbol \theta_t | \boldsymbol \alpha) \prod_{k=1}^{K}p(\boldsymbol \beta_k | \boldsymbol \eta) (\prod_{n=1}^{N}P(w_{t,n} | z_{t,n}, \boldsymbol \beta_k)P( z_{t,n} | \boldsymbol \theta_t)) - $$ @@ -1016,9 +930,7 @@ $p(\mathbf{W}, \mathbf{z}, \boldsymbol{\beta}, \boldsymbol{\Theta} \mid \boldsym $$ - p_{\boldsymbol{\alpha}, \boldsymbol{\eta}}(\mathbf{z}, \boldsymbol{\beta}, \boldsymbol{\Theta} \mid \mathbf{W})=\frac{p_{\boldsymbol{\alpha}, \boldsymbol{\eta}}(\mathbf{W}, \mathbf{z}, \boldsymbol{\beta}, \boldsymbol{\Theta})}{p_{\boldsymbol{\alpha}, \boldsymbol{\eta}}(\mathbf{W})} - $$