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| Original file line number | Diff line number | Diff line change |
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| ## 14.26 | ||
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| $$p(x^t)T(x^{t-1}|x^t)=p(x^{t-1})T(x^t|x^{t-1})$$ | ||
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| [解析]：假设变量$x$所在的空间有$n$个状态($s_1,s_2,..,s_n$), 定义在该空间上的一个转移矩阵$T(n\times n)$如果满足一定的条件则该马尔可夫过程存在一个稳态分布$\pi$, 使得 | ||
| $$ | ||
| \begin{aligned} | ||
| \pi T=\pi | ||
| \end{aligned} | ||
| \tag{1} | ||
| $$ | ||
| 其中, $\pi$是一个是一个$n$维向量，代表$s_1,s_2,..,s_n$对应的概率. 反过来, 如果我们希望采样得到符合某个分布$\pi$的一系列变量$x_1,x_2,..,x_t$, 应当采用哪一个转移矩阵$T(n\times n)$呢？ | ||
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| 事实上，转移矩阵只需要满足马尔可夫细致平稳条件 | ||
| $$ | ||
| \begin{aligned} | ||
| \pi (i)T(i,j)=\pi (j)T(j,i) | ||
| \end{aligned} | ||
| \tag{2} | ||
| $$ | ||
| 即公式$14.26$，这里采用的符号与西瓜书略有区别以便于理解. 证明如下 | ||
| $$ | ||
| \begin{aligned} | ||
| \pi T(j) = \sum _i \pi (i)T(i,j) = \sum _i \pi (j)T(j,i) = \pi(j) | ||
| \end{aligned} | ||
| \tag{3} | ||
| $$ | ||
| 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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| $$A(x^* | x^{t-1}) = \min\left ( 1,\frac{p(x^*)Q(x^{t-1} | x^*) }{p(x^{t-1})Q(x^* | x^{t-1})} \right )$$ | ||
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| [推导]：这个公式其实是拒绝采样的一个trick，因为基于式$14.27$只需要 | ||
| $$ | ||
| \begin{aligned} | ||
| A(x^* | x^{t-1}) &= p(x^*)Q(x^{t-1} | x^*) \\ | ||
| A(x^{t-1} | x^*) &= p(x^{t-1})Q(x^* | x^{t-1}) | ||
| \end{aligned} | ||
| \tag{4} | ||
| $$ | ||
| 即可满足式$14.26$，但是实际上等号右边的数值可能比较小，比如各为0.1和0.2，那么好不容易才到的样本只有百分之十几得到利用，所以不妨将接受率设为0.5和1，则细致平稳分布条件依然满足，样本利用率大大提高, 所以可以将$(4)$改进为 | ||
| $$ | ||
| \begin{aligned} | ||
| A(x^* | x^{t-1}) &= \frac{p(x^*)Q(x^{t-1} | x^*)}{norm} \\ | ||
| A(x^{t-1} | x^*) &= \frac{p(x^{t-1})Q(x^* | x^{t-1}) }{norm} | ||
| \end{aligned} | ||
| \tag{5} | ||
| $$ | ||
| 其中 | ||
| $$ | ||
| \begin{aligned} | ||
| norm = \max\left (p(x^{t-1})Q(x^* | x^{t-1}),p(x^*)Q(x^{t-1} | x^*) \right ) | ||
| \end{aligned} | ||
| \tag{6} | ||
| $$ | ||
| 即教材的$14.28$. | ||
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| ## 14.32 | ||
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| $${\rm ln}p(x)=\mathcal{L}(q)+{\rm KL}(q \parallel p)$$ | ||
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| [推导] | ||
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| 根据条件概率公式$p(x,z)=p(z|x)*p(x)$，可以得到$p(x)=\frac{p(x,z)}{p(z|x)}$ | ||
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| 然后两边同时作用${\rm ln}$函数，可得${\rm ln}p(x)={\rm ln}\frac{p(x,z)}{p(z|x)}$ (1) | ||
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| 因为$q(z)$是概率密度函数，所以$1=\int q(z)dz$ | ||
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| 等式两边同时乘以${\rm ln}p(x)$，因为${\rm ln}p(x)$是不关于变量$z$的函数，所以${\rm ln}p(x)$可以拿进积分里面，得到${\rm ln}p(x)=\int q(z){\rm ln}p(x)dz$ | ||
| $$ | ||
| \begin{align} | ||
| {\rm ln}p(x)&=\int q(z){\rm ln}p(x) \\ | ||
| &=\int q(z){\rm ln}\frac{p(x,z)}{p(z|x)}\qquad(带入公式(1))\\ | ||
| &=\int q(z){\rm ln}\bigg\{\frac{p(x,z)}{q(z)}\cdot\frac{q(z)}{p(z|x)}\bigg\} \\ | ||
| &=\int q(z)\bigg({\rm ln}\frac{p(x,z)}{q(z)}-{\rm ln}\frac{p(z|x)}{q(z)}\bigg) \\ | ||
| &=\int q(z){\rm ln}\bigg\{\frac{p(x,z)}{q(z)}\bigg\}-\int q(z){\rm ln}\frac{p(z|x)}{q(z)} \\ | ||
| &=\mathcal{L}(q)+{\rm KL}(q \parallel p)\qquad(根据\mathcal{L}和{\rm KL}的定义) | ||
| \end{align} | ||
| $$ | ||
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| ## 14.36 | ||
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| $$ | ||
| \begin{align} | ||
| \mathcal{L}(q)&=\int \prod_{i}q_{i}\bigg\{ {\rm ln}p({\rm \mathbf{x},\mathbf{z}})-\sum_{i}{\rm ln}q_{i}\bigg\}d{\rm\mathbf{z}} \\ | ||
| &=\int q_{j}\bigg\{\int p(x,z)\prod_{i\ne j}q_{i}d{\rm\mathbf{z_{i}}}\bigg\}d{\rm\mathbf{z_{j}}}-\int q_{j}{\rm ln}q_{j}d{\rm\mathbf{z_{j}}}+{\rm const} \\ | ||
| &=\int q_{j}{\rm ln}\tilde{p}({\rm \mathbf{x},\mathbf{z_{j}}})d{\rm\mathbf{z_{j}}}-\int q_{j}{\rm ln}q_{j}d{\rm\mathbf{z_{j}}}+{\rm const} | ||
| \end{align} | ||
| $$ | ||
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| [推导] | ||
| $$ | ||
| \mathcal{L}(q)=\int \prod_{i}q_{i}\bigg\{ {\rm ln}p({\rm \mathbf{x},\mathbf{z}})-\sum_{i}{\rm ln}q_{i}\bigg\}d{\rm\mathbf{z}}=\int\prod_{i}q_{i}{\rm ln}p({\rm \mathbf{x},\mathbf{z}})d{\rm\mathbf{z}}-\int\prod_{i}q_{i}\sum_{i}{\rm ln}q_{i}d{\rm\mathbf{z}} | ||
| $$ | ||
| 公式可以看做两个积分相减，我们先来看左边积分$\int\prod_{i}q_{i}{\rm ln}p({\rm \mathbf{x},\mathbf{z}})d{\rm\mathbf{z}}$的推导。 | ||
| $$ | ||
| \begin{align} | ||
| \int\prod_{i}q_{i}{\rm ln}p({\rm \mathbf{x},\mathbf{z}})d{\rm\mathbf{z}} &= \int q_{j}\prod_{i\ne j}q_{i}{\rm ln}p({\rm \mathbf{x},\mathbf{z}})d{\rm\mathbf{z}} \\ | ||
| &= \int q_{j}\bigg\{\int{\rm ln}p({\rm \mathbf{x},\mathbf{z}})\prod_{i\ne j}q_{i}d{\rm\mathbf{z_{i}}}\bigg\}d{\rm\mathbf{z_{j}}}\qquad (先对{\rm\mathbf{z_{j}}}求积分，再对{\rm\mathbf{z_{i}}}求积分) | ||
| \end{align} | ||
| $$ | ||
| 这个就是教材中的$14.36$左边的积分部分。 | ||
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| 我们现在看下右边积分的推导$\int\prod_{i}q_{i}\sum_{i}{\rm ln}q_{i}d{\rm\mathbf{z}}$的推导。 | ||
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| 在此之前我们看下$\int\prod_{i}q_{i}{\rm ln}q_{k}d{\rm\mathbf{z}}$的计算 | ||
| $$ | ||
| \begin{align} | ||
| \int\prod_{i}q_{i}{\rm ln}q_{k}d{\rm\mathbf{z}}&= \int q_{i^{\prime}}\prod_{i\ne i^{\prime}}q_{i}{\rm ln}q_{k}d{\rm\mathbf{z}}\qquad (选取一个变量q_{i^{\prime}}, i^{\prime}\ne k) \\ | ||
| &=\int q_{i^{\prime}}\bigg\{\int\prod_{i\ne i^{\prime}}q_{i}{\rm ln}q_{k}d{\rm\mathbf{z_{i}}}\bigg\}d{\rm\mathbf{z_{i^{\prime}}}} | ||
| \end{align} | ||
| $$ | ||
| $\bigg\{\int\prod_{i\ne i^{\prime}}q_{i}{\rm ln}q_{k}d{\rm\mathbf{z_{i}}}\bigg\}$部分与变量$q_{i^{\prime}}$无关，所以可以拿到积分外面。又因为$\int q_{i^{\prime}}d{\rm\mathbf{z_{i^{\prime}}}}=1$，所以 | ||
| $$ | ||
| \begin{align} | ||
| \int\prod_{i}q_{i}{\rm ln}q_{k}d{\rm\mathbf{z}}&=\int\prod_{i\ne i^{\prime}}q_{i}{\rm ln}q_{k}d{\rm\mathbf{z_{i}}} \\ | ||
| &= \int q_{k}{\rm ln}q_{k}d{\rm\mathbf{z_k}}\qquad (所有k以外的变量都可以通过上面的方式消除) | ||
| \end{align} | ||
| $$ | ||
| 有了这个结论，我们再来看公式 | ||
| $$ | ||
| \begin{align} | ||
| \int\prod_{i}q_{i}\sum_{i}{\rm ln}q_{i}d{\rm\mathbf{z}}&= \int\prod_{i}q_{i}{\rm ln}q_{j}d{\rm\mathbf{z}} + \sum_{k\ne j}\int\prod_{i}q_{i}{\rm ln}q_{k}d{\rm\mathbf{z}} \\ | ||
| &= \int q_{j}{\rm ln}q_{j}d{\rm\mathbf{z_j}} + \sum_{z\ne j}\int q_{k}{\rm ln}q_{k}d{\rm\mathbf{z_k}}\qquad (根据上面结论) \\ | ||
| &= \int q_{j}{\rm ln}q_{j}d{\rm\mathbf{z_j}} + {\rm const} \qquad (这里我们关心的是q_{j}，其他变量可以视为{\rm const}) | ||
| \end{align} | ||
| $$ | ||
| 这个就是$14.36$右边的积分部分。 |
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| Original file line number | Diff line number | Diff line change |
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| @@ -0,0 +1,17 @@ | ||
| ## 8.3 | ||
| $$ | ||
| \begin{aligned} P(H(\boldsymbol{x}) \neq f(\boldsymbol{x})) &=\sum_{k=0}^{\lfloor T / 2\rfloor} \left( \begin{array}{c}{T} \\ {k}\end{array}\right)(1-\epsilon)^{k} \epsilon^{T-k} \\ & \leqslant \exp \left(-\frac{1}{2} T(1-2 \epsilon)^{2}\right) \end{aligned} | ||
| $$ | ||
| [推导]:由基分类器相互独立，设X为T个基分类器分类正确的次数，因此$\mathrm{X} \sim \mathrm{B}(\mathrm{T}, 1-\mathrm{\epsilon})$ | ||
| $$ | ||
| \begin{aligned} P(H(x) \neq f(x))=& P(X \leq\lfloor T / 2\rfloor) \\ & \leqslant P(X \leq T / 2) | ||
| \\ & =P\left[X-(1-\varepsilon) T \leqslant \frac{T}{2}-(1-\varepsilon) T\right] | ||
| \\ & =P\left[X- | ||
| (1-\varepsilon) T \leqslant -\frac{T}{2}\left(1-2\varepsilon\right)]\right] | ||
| \end{aligned} | ||
| $$ | ||
| 根据Hoeffding不等式$P(X-(1-\epsilon)T\leqslant -kT) \leq \exp (-2k^2T)$ | ||
| 令$k=\frac {(1-2\epsilon)}{2}$得 | ||
| $$ | ||
| \begin{aligned} P(H(\boldsymbol{x}) \neq f(\boldsymbol{x})) &=\sum_{k=0}^{\lfloor T / 2\rfloor} \left( \begin{array}{c}{T} \\ {k}\end{array}\right)(1-\epsilon)^{k} \epsilon^{T-k} \\ & \leqslant \exp \left(-\frac{1}{2} T(1-2 \epsilon)^{2}\right) \end{aligned} | ||
| $$ |
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