Part4: Appendix
本文是前几篇文章中相关公式的详细推导部分，主要对论文中一些被省略的推导进行补充说明，对“扩散模型”感兴趣请查看前几篇文章。
- Part1: Overview of Diffusion Process
- Part2: DDPM as Example of Variational Inference
- Part3: Dive into DDPM
- Part4: Appendix
高斯分布
概率密度函数
若\(x \sim \mathcal{N}(\mu, \sigma^2)\)，则：
\[f(x ; \mu, \sigma)=\frac{1}{\sigma \sqrt{2 \pi}} \exp \left(-\frac{(x-\mu)^2}{2 \sigma^2}\right)
\]
两个高斯的KL散度
\[D_{\mathrm{KL}}\left(\mathcal{N(\mu_1, \sigma_1^2) \mid\mid N(\mu_2, \sigma_2^2)}\right) = \ln \frac{\sigma_2}{\sigma_1} + \frac{\sigma_1^2 + (\mu_1 - \mu_2)^2}{2\sigma_2^2} - \frac{1}{2}
\]
性质1
如果存在一个随机变量\(x \sim \mathcal{N}(\mu, \sigma^2)\)服从高斯分布，那么存在实数\(a, b\)，满足：
\[ax + b \sim \mathcal{N}(a\mu + b, (a\sigma)^2)
\]
因此，对于任意高斯分布\(\mathbf{x} \sim \mathcal{N}(\mu, \sigma^2)\)，可以将其表示为服从标准正态分布的随机变量\(\epsilon\)的变换，即:
\[\mathbf{x} = \epsilon * \sigma + \mu, \epsilon \sim \mathcal{N}(0, \mathbf{I})
\]
性质2
假定两个随机变量都服从高斯分布且相互独立，记作\(x \sim \mathcal{N}(\mu_x, \sigma_x^2),\ \ y \sim \mathcal{N}(\mu_y, \sigma_y^2)\)，则两个随机变量的和或差仍服从高斯分布，即：
\[\begin{aligned}
& U=x+y \sim N\left(\mu_x+\mu_y, \sigma_x^2+\sigma_y^2\right) \\
& V=x-y \sim N\left(\mu_x-\mu_y, \sigma_x^2+\sigma_y^2\right)
\end{aligned}\]
推导一
在\(\text{Diffusion Forward process}\)中，任意时刻\(t\)的状态\(\mathbf{x}_t\)如何基于\(\mathbf{x}_0\)表示？
解：
已知前向过程中，状态间的转换服从高斯分布，有：
\[q\left(\mathbf{x}_t \mid \mathbf{x}_{t-1}\right) = \mathcal{N}\left(\sqrt{1-\beta_t} \mathbf{x}_{t-1}, \beta_t \mathbf{I}\right)\tag{1}
\]
对\(\beta_{t}\)进行变换，定义：
\[\begin{aligned}
\alpha_t & =1-\beta_t \\
\bar{\alpha}_t & =\prod_{i=1}^t \alpha_i
\end{aligned}\]
对\((1)\)式展开如下：
\[\begin{aligned}
q\left(\mathbf{x}_t \mid \mathbf{x}_{t-1}\right) & =\mathcal{N}\left(\sqrt{1-\beta_t} \mathbf{x}_{t-1}, \beta_t \mathbf{I}\right) \\
\mathbf{x}_t & =\sqrt{1-\beta_t} \mathbf{x}_{t-1}+\sqrt{\beta_t} \epsilon, \quad \epsilon \sim \mathcal{N}(0, \mathbf{I}) \\
& =\sqrt{\alpha_t} \mathbf{x}_{t-1}+\sqrt{1-\alpha_t} \epsilon
\end{aligned}
\tag{2}
\]
已知\(\mathbf{x}_t = \sqrt{\alpha_t}\mathbf{x}_{t-1} + \sqrt{1 - \alpha_t} \epsilon\)，同理可得\(\mathbf{x}_{t-1} = \sqrt{\alpha_{t-1}}\mathbf{x}_{t-2} + \sqrt{1 - \alpha_{t-1}} \bar{\epsilon}\)，对\((2)\)改写，有：
\[\begin{aligned}
& \sqrt{\alpha_t}\mathbf{x}_{t-1} + \sqrt{1 - \alpha_t} \epsilon \\
=& \sqrt{\alpha_t}\left(\sqrt{\alpha_{t-1}}\mathbf{x}_{t-2} + \sqrt{1 - \alpha_{t-1}} \bar{\epsilon} \right) + \sqrt{1 - \alpha_t} \epsilon \\
=& \sqrt{\alpha_t \alpha_{t-1}} \mathbf{x}_{t-2} + \sqrt{\alpha_t \left(1 - \alpha_{t-1}\right)} \bar{\epsilon} + \sqrt{1 - \alpha_t} \epsilon
\end{aligned}
\tag{3}
\]
为了与\(\epsilon\)进行区分，使用\(\bar{\epsilon}\)表示另外一个服从标准高斯分布\(\mathcal{N}(0, \mathbf{I})\)的变量。
根据高斯分布的性质1，任意的高斯分布可由标准高斯分布转换得到，故：
\[\begin{aligned}
\epsilon \sim \mathcal{N}(0, \mathbf{I})
\quad &\Rightarrow \quad
\sqrt{1 - \alpha_t} \epsilon \sim \mathcal{N}(0, \left(1-\alpha_t\right)\mathbf{I}) \
\\
\bar{\epsilon} \sim \mathcal{N}(0, \mathbf{I})
\quad &\Rightarrow \quad
\sqrt{\alpha_t \left(1 - \alpha_{t-1}\right)} \epsilon \sim \mathcal{N}(0, \alpha_t\left(1-\alpha_{t-1}\right)\mathbf{I})
\end{aligned} \tag{a}\]
由于\(\sqrt{1 - \alpha_t} \epsilon\)与\(\sqrt{\alpha_t \left(1 - \alpha_{t-1}\right)} \bar{\epsilon}\)独立且都服从高斯分布，记\(U = \sqrt{1 - \alpha_t} \epsilon + \sqrt{\alpha_t \left(1 - \alpha_{t-1}\right)} \bar{\epsilon}\)，由性质2可知\(U\)也服从高斯分布，有：
\[\begin{aligned}
\sqrt{1 - \alpha_t} \epsilon + \sqrt{\alpha_t \left(1 - \alpha_{t-1}\right)} \bar{\epsilon} &\sim \mathcal{N}(0, \left(1-\alpha_t\right)\mathbf{I} +\alpha_t\left(1-\alpha_{t-1}\right)\mathbf{I}) \\ \Rightarrow U & \sim \mathcal{N}(0, \left(1-\alpha_t\alpha_{t-1}\right)\mathbf{I})
\end{aligned} \tag{b}\]
基于高斯分布的性质1，将\(U\)使用标准高斯分布表示：
\[\begin{aligned}
U &\sim \mathcal{N}(0, \left(1-\alpha_t\alpha_{t-1}\right)\mathbf{I})
\Rightarrow
U = \sqrt{1 - \alpha_t\alpha_{t-1}} \epsilon
\end{aligned} \tag{c}\]
将\((c)\)代入\((3)\)，可得：
\[\begin{aligned}
& \sqrt{\alpha_t}\mathbf{x}_{t-1} + \sqrt{1 - \alpha_t} \epsilon \\
=& \sqrt{\alpha_t}\left(\sqrt{\alpha_{t-1}}\mathbf{x}_{t-2} + \sqrt{1 - \alpha_{t-1}} \bar{\epsilon} \right) + \sqrt{1 - \alpha_t} \epsilon \\
=& \sqrt{\alpha_t \alpha_{t-1}} \mathbf{x}_{t-2} + \sqrt{\alpha_t \left(1 - \alpha_{t-1}\right)} \bar{\epsilon} + \sqrt{1 - \alpha_t} \epsilon \\
=& \sqrt{\alpha_t \alpha_{t-1}}\mathbf{x}_{t-2} + \sqrt{1 - \alpha_t \alpha_{t-1}} \epsilon
\end{aligned}\]
由数学归纳法，易知：
\[\begin{aligned}
q\left(\mathbf{x}_t \mid \mathbf{x}_{t-1}\right) & =\mathcal{N}\left(\sqrt{1-\beta_t} \mathbf{x}_{t-1}, \beta_t \mathbf{I}\right) \\
\mathbf{x}_t & =\sqrt{1-\beta_t} \mathbf{x}_{t-1}+\sqrt{\beta_t} \epsilon, \quad \epsilon \sim \mathcal{N}(0, \mathbf{I}) \\
& =\sqrt{\alpha_t} \mathbf{x}_{t-1}+\sqrt{1-\alpha_t} \epsilon \\
& =\sqrt{\alpha_t \alpha_{t-1}} \mathbf{x}_{t-2}+\sqrt{1-\alpha_t \alpha_{t-1}} \epsilon \\
& =\ldots \\
& =\sqrt{\bar{\alpha}_t} \mathbf{x}_0+\sqrt{1-\bar{\alpha}_t} \epsilon
\end{aligned}
\]
因此，\(q\left(\mathbf{x}_t \mid \mathbf{x}_{0}\right) = \mathcal{N}\left(\sqrt{\bar{\alpha}_t} \mathbf{x}_{0}, \sqrt{1 - \bar{\alpha}_t} \mathbf{I}\right)\)
推导二
在\(diffusion\)中，定义\(q\)服从高斯分布，故对\(q\left(\mathbf{x}_{t-1} \mid \mathbf{x}_t, \mathbf{x}_0\right)\)定义如下：
\[\begin{aligned}
q\left(\mathbf{x}_{t-1} \mid \mathbf{x}_t, \mathbf{x}_0\right) & =\mathcal{N}\left(\mathbf{x}_{t-1} ;
\tilde{\boldsymbol{\mu}}_t\left(\mathbf{x}_t, \mathbf{x}_0\right), \tilde{\beta}_t \mathbf{I}\right)
\end{aligned}
\]
那其中\(\tilde{\boldsymbol{\mu}}_t\left(\mathbf{x}_t, \mathbf{x}_0\right)\)与\(\tilde{\beta_t}\)如何得到？
此处先给出结论，下方是更详细的推导。
\[\begin{aligned}
\tilde{\boldsymbol{\mu}}_t\left(\mathbf{x}_t, \mathbf{x}_0\right) &:= \frac{\sqrt{\bar{\alpha}_{t-1}} \beta_t}{1-\bar{\alpha}_t} \mathbf{x}_0+\frac{\sqrt{\alpha_t}\left(1-\bar{\alpha}_{t-1}\right)}{1-\bar{\alpha}_t} \mathbf{x}_t, \\ \tilde{\beta}_t &:= \frac{1-\bar{\alpha}_{t-1}}{1-\bar{\alpha}_t} \beta_t
\end{aligned}
\]
解：
回顾贝叶斯公式，对\(q\left(\mathbf{x}_{t-1} \mid \mathbf{x}_t, \mathbf{x}_0\right)\)改写，有：
\[q\left(\mathbf{x}_{t-1} \mid \mathbf{x}_t, \mathbf{x}_0\right)
=q\left(\mathbf{x}_t \mid \mathbf{x}_{t-1}, \mathbf{x}_0\right) \frac{q\left(\mathbf{x}_{t-1} \mid \mathbf{x}_0\right)}{q\left(\mathbf{x}_t \mid \mathbf{x}_0\right)}\tag{1}\]
由于Diffusion基于马尔可夫链建模，由马尔可夫性易知每个状态只依赖于前一个状态，故
\[q\left(\mathbf{x}_t \mid \mathbf{x}_{t-1}, \mathbf{x}_0\right) = q\left(\mathbf{x}_t \mid \mathbf{x}_{t-1}\right)
\]
\((1)\)式写作\((2)\)式：
\[\begin{aligned}
q\left(\mathbf{x}_{t-1} \mid \mathbf{x}_t, \mathbf{x}_0\right)
&=q\left(\mathbf{x}_t \mid \mathbf{x}_{t-1}, \mathbf{x}_0\right) \frac{q\left(\mathbf{x}_{t-1} \mid \mathbf{x}_0\right)}{q\left(\mathbf{x}_t \mid \mathbf{x}_0\right)} \\
&=q\left(\mathbf{x}_t \mid \mathbf{x}_{t-1}\right) \frac{q\left(\mathbf{x}_{t-1} \mid \mathbf{x}_0\right)}{q\left(\mathbf{x}_t \mid \mathbf{x}_0\right)}
\end{aligned}
\tag{2}\]
基于推导一的结论，易知：
\[\begin{aligned}
q\left(\mathbf{x}_t \mid \mathbf{x}_{0}\right) &= \mathcal{N}\left(\sqrt{\bar{\alpha}_t} \mathbf{x}_{0}, \sqrt{1 - \bar{\alpha}_t} \mathbf{I}\right) \\
q\left(\mathbf{x}_{t-1} \mid \mathbf{x}_{0}\right) &= \mathcal{N}\left(\sqrt{\bar{\alpha}_{t-1}} \mathbf{x}_{0}, \sqrt{1 - \bar{\alpha}_{t-1}} \mathbf{I}\right)
\end{aligned}\]
由高斯分布的概率密度函数，对\((2)\)展开，有：
\[\begin{aligned}
q\left(\mathbf{x}_{t-1} \mid \mathbf{x}_t, \mathbf{x}_0\right) & = q\left(\mathbf{x}_t \mid \mathbf{x}_{t-1}\right) \frac{q\left(\mathbf{x}_{t-1} \mid \mathbf{x}_0\right)}{q\left(\mathbf{x}_t \mid \mathbf{x}_0\right)} \\ & \propto \exp \left(-\frac{1}{2}\left(\frac{\left(\mathbf{x}_t-\sqrt{\alpha_t} \mathbf{x}_{t-1}\right)^2}{\beta_t}+\frac{\left(\mathbf{x}_{t-1}-\sqrt{\bar{\alpha}_{t-1}} \mathbf{x}_0\right)^2}{1-\bar{\alpha}_{t-1}}-\frac{\left(\mathbf{x}_t-\sqrt{\bar{\alpha}_t} \mathbf{x}_0\right)^2}{1-\bar{\alpha}_t}\right)\right) \end{aligned}\tag{3}\]
不论是\(\beta_t\)或是\(\bar{\alpha}_t\)皆非随机变量，故可省略。最终目标是使用随机变量\(\mathbf{x}_0\)与\(\mathbf{x}_{t}\)表示\(\mathbf{x}_{t-1}\)。对\((3)\)式继续展开，有\((4)\)：
\[\begin{aligned} &q\left(\mathbf{x}_{t-1} \mid \mathbf{x}_t, \mathbf{x}_0\right) = q\left(\mathbf{x}_t \mid \mathbf{x}_{t-1}\right) \frac{q\left(\mathbf{x}_{t-1} \mid \mathbf{x}_0\right)}{q\left(\mathbf{x}_t \mid \mathbf{x}_0\right)} \\ & \propto \exp \left(-\frac{1}{2}\left(\frac{\left(\mathbf{x}_t-\sqrt{\alpha_t} \mathbf{x}_{t-1}\right)^2}{\beta_t}+\frac{\left(\mathbf{x}_{t-1}-\sqrt{\bar{\alpha}_{t-1}} \mathbf{x}_0\right)^2}{1-\bar{\alpha}_{t-1}}-\frac{\left(\mathbf{x}_t-\sqrt{\bar{\alpha}_t} \mathbf{x}_0\right)^2}{1-\bar{\alpha}_t}\right)\right) \\
&=\exp \left(-\frac{1}{2}\left(\frac{\mathbf{x}_t^2-2 \sqrt{\alpha_t} \mathbf{x}_t \mathbf{x}_{t-1}+\alpha_t \mathbf{x}_{t-1}^2}{\beta_t}+\frac{\mathbf{x}_{t-1}^2-2 \sqrt{\bar{\alpha}_{t-1}} \mathbf{x}_0 \mathbf{x}_{t-1}+\bar{\alpha}_{t-1} \mathbf{x}_0^2}{1-\bar{\alpha}_{t-1}}-\frac{\left(\mathbf{x}_t-\sqrt{\bar{\alpha}_t} \mathbf{x}_0\right)^2}{1-\bar{\alpha}_t}\right)\right) \\ &=\exp \left(-\frac{1}{2}\left(\left(\frac{\alpha_t}{\beta_t}+\frac{1}{1-\bar{\alpha}_{t-1}}\right) \mathbf{x}_{t-1}^2-\left(\frac{2 \sqrt{\alpha_t}}{\beta_t} \mathbf{x}_t+\frac{2 \sqrt{\bar{\alpha}_{t-1}}}{1-\bar{\alpha}_{t-1}} \mathbf{x}_0\right) \mathbf{x}_{t-1}+C\left(\mathbf{x}_t, \mathbf{x}_0\right)\right)\right) \end{aligned} \tag{4}\]
其中，倒数第二个等号右边是对上一步的平方展开；最后一个等号右边是以\(\mathbf{x}_{t-1}\)为变量，\(\mathbf{x}_0\)与\(\mathbf{x}_{t}\)为参数，构造完全平方公式，以形成高斯分布概率密度函数中的指数部分，形如\(-\frac{(\mathbf{x}_{t-1}-\tilde{\mu_t})^2}{2 \tilde{\beta_t}}\)。因此，不难得出：
\[\begin{aligned}
\tilde{\boldsymbol{\mu}}_t &= \frac{1}{\frac{\alpha_t}{\beta_t}+\frac{1}{1-\bar{\alpha}_{t-1}}} * \left(\frac{\sqrt{\alpha_t}}{\beta_t} \mathbf{x}_t+\frac{\sqrt{\bar{\alpha}_{t-1}}}{1-\bar{\alpha}_{t-1}} \mathbf{x}_0\right) \\
&= \frac{\left(1 - \bar{\alpha}_{t-1}\right) \beta_{t}}{\alpha_t\left(1 - \bar{\alpha}_{t-1}\right) + \beta_t} * \left(\frac{\sqrt{\alpha_t}}{\beta_t} \mathbf{x}_t+\frac{\sqrt{\bar{\alpha}_{t-1}}}{1-\bar{\alpha}_{t-1}} \mathbf{x}_0\right) \\
& = \frac{\left(1 - \bar{\alpha}_{t-1}\right)\sqrt{\alpha_t}}{\alpha_t\left(1 - \bar{\alpha}_{t-1}\right) + \beta_t} \mathbf{x}_t + \frac{\sqrt{\bar{\alpha}_{t-1}} \beta_{t}}{\alpha_t\left(1 - \bar{\alpha}_{t-1}\right) + \beta_t} \mathbf{x}_0 \\
\end{aligned}\tag{5}\]
\(\alpha_t = 1 - \beta_t\)，故：
\[\begin{aligned}
\alpha_t\left(1 - \bar{\alpha}_{t-1}\right) + \beta_t &= \alpha_t - \alpha_t \bar{\alpha}_{t-1} + \beta_t \\
&= 1 - \beta_t - \alpha_t \bar{\alpha}_{t-1} + \beta_t \\
&= 1 - \alpha_t \bar{\alpha}_{t-1} \\
&= 1 - \bar{\alpha}_{t}
\end{aligned}\tag{6}\]
将\((6)\)式代入\((5)\)，有：
\[\tilde{\boldsymbol{\mu}}_t\left(\mathbf{x}_t, \mathbf{x}_0\right) :=\frac{\sqrt{\bar{\alpha}_{t-1}} \beta_t}{1-\bar{\alpha}_t} \mathbf{x}_0+\frac{\sqrt{\alpha_t}\left(1-\bar{\alpha}_{t-1}\right)}{1-\bar{\alpha}_t} \mathbf{x}_t
\]
对于\(\tilde{\beta}_t\)，有：
\[\begin{aligned}
\tilde{\beta}_t &= \frac{1}{\frac{\alpha_t}{\beta_t}+\frac{1}{1-\bar{\alpha}_{t-1}}} \\
&= \frac{\left(1 - \bar{\alpha}_{t-1}\right) \beta_{t}}{\alpha_t\left(1 - \bar{\alpha}_{t-1}\right) + \beta_t} \\
&= \frac{1-\bar{\alpha}_{t-1}}{1-\bar{\alpha}_t} \beta_t
\end{aligned}\]
以上内容即\(\text{DDPM}\)中一些被省略的数学推导。
