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@@ -25,7 +25,7 @@ $$
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\end{aligned}
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\tag{3}
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$$
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-假设采样得到的序列为$x_1,x_2,..,x_{t-1},x_t$,则可以使用$MH$算法来使得$x_{t-1}$(假设为状态$s_i$)转移到$x_t$(假设为状态$s_j$)的概率满足式$(2)$.
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+假设采样得到的序列为$x_1,x_2,..,x_{t-1},x_t$,则可以使用$MH$算法来使得$x_{t-1}$(假设为状态$s_i$)转移到$x_t$(假设为状态$s_j$)的概率满足式$(2)$.
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## 14.28
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@@ -127,3 +127,34 @@ $$
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\end{aligned}
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$$
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这个就是$14.36$右边的积分部分。
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+
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+## 14.40
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+
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+$$
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+\begin{aligned}
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+q_j^*(\mathbf{z}_j) = \frac{ \exp\left ( \mathbb{E}_{i\neq j}[\ln (p(\mathbf{x},\mathbf{z}))] \right ) }{\int \exp\left ( \mathbb{E}_{i\neq j}[\ln (p(\mathbf{x},\mathbf{z}))] \right ) \mathrm{d}\mathbf{z}_j}
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+\end{aligned}
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+$$
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+
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+[推导]:由$14.39$去对数直接可得
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+$$
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+\begin{aligned}
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+ \int q_j^*(\mathbf{z}_j)\mathrm{d}\mathbf{z}_j &=\int \exp\left ( \mathbb{E}_{i\neq j}[\ln (p(\mathbf{x},\mathbf{z}))] \right )\cdot\exp(const) \, \mathrm{d}\mathbf{z}_j \\
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+ &=\exp(const) \int \exp\left ( \mathbb{E}_{i\neq j}[\ln (p(\mathbf{x},\mathbf{z}))] \right ) \, \mathrm{d}\mathbf{z}_j \\
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+ &= 1
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+ \end{aligned}
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+ \tag{7}
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+$$
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+所以
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+$$
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+\exp(const) = \dfrac{1}{\int \exp\left ( \mathbb{E}_{i\neq j}[\ln (p(\mathbf{x},\mathbf{z}))] \right ) \, \mathrm{d}\mathbf{z}_j} \\
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+\tag{8}
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+$$
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+
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+$$
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+\begin{aligned}
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+ q_j^*(\mathbf{z}_j)\mathrm{d}\mathbf{z}_j &= \exp\left ( \mathbb{E}_{i\neq j}[\ln (p(\mathbf{x},\mathbf{z}))] \right )\cdot\exp(const) \, \mathrm{d}\mathbf{z}_j \\
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+ &= \frac{ \exp\left ( \mathbb{E}_{i\neq j}[\ln (p(\mathbf{x},\mathbf{z}))] \right ) }{\int \exp\left ( \mathbb{E}_{i\neq j}[\ln (p(\mathbf{x},\mathbf{z}))] \right ) \mathrm{d}\mathbf{z}_j}
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+ \end{aligned}
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+ \tag{9}
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+$$
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