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：

$$f(x;\mu ,\sigma )=\frac{1}{\sigma \sqrt{2\pi }}\exp (-\frac{(x-\mu {)}^2}{2{\sigma }^2})$$

两个高斯的KL散度

$$D_{\mathrm{KL}}\left(\mathcal{N}\mathcal{(}{\mu }_{\mathcal{1}}\mathcal{,}{\sigma }_{\mathcal{1}}^{\mathcal{2}}\mathcal{)}\mid \mid \mathcal{N}\mathcal{(}{\mu }_{\mathcal{2}}\mathcal{,}{\sigma }_{\mathcal{2}}^{\mathcal{2}}\mathcal{)}\right)=\ln \frac{{\sigma }_2}{{\sigma }_1}+\frac{{\sigma }_1^2+({\mu }_1-{\mu }_2{)}^2}{2{\sigma }_2^2}-\frac{1}{2}$$

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$ϵ$的变换，即:

$$\mathbf{x}=ϵ\ast \sigma +\mu ,ϵ\sim \mathcal{N}(0,\mathbf{I})$$

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：

$$\begin{array}{rl}& U=x+y\sim N({\mu }_x+{\mu }_y,{\sigma }_x^2+{\sigma }_y^2)\\ & V=x-y\sim N({\mu }_x-{\mu }_y,{\sigma }_x^2+{\sigma }_y^2)\end{array}$$

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：

$$\begin{array}{}\text{(1)}& q({\mathbf{x}}_t\mid {\mathbf{x}}_{t-1})=\mathcal{N}(\sqrt{1-{\beta }_t}{\mathbf{x}}_{t-1},{\beta }_t\mathbf{I})\end{array}$$

对${\beta }_t$进行变换，定义：

$$\begin{array}{rl}{\alpha }_t& =1-{\beta }_t\\ {\overline{\alpha }}_t& =\prod _{i=1}^t{\alpha }_i\end{array}$$

对$(1)$式展开如下：

$$\begin{array}{}\text{(2)}& \begin{array}{rl}q({\mathbf{x}}_t\mid {\mathbf{x}}_{t-1})& =\mathcal{N}(\sqrt{1-{\beta }_t}{\mathbf{x}}_{t-1},{\beta }_t\mathbf{I})\\ {\mathbf{x}}_t& =\sqrt{1-{\beta }_t}{\mathbf{x}}_{t-1}+\sqrt{{\beta }_t}ϵ,ϵ\sim \mathcal{N}(0,\mathbf{I})\\ & =\sqrt{{\alpha }_t}{\mathbf{x}}_{t-1}+\sqrt{1-{\alpha }_t}ϵ\end{array}\end{array}$$

已知${\mathbf{x}}_t=\sqrt{{\alpha }_t}{\mathbf{x}}_{t-1}+\sqrt{1-{\alpha }_t}ϵ$，同理可得${\mathbf{x}}_{t-1}=\sqrt{{\alpha }_{t-1}}{\mathbf{x}}_{t-2}+\sqrt{1-{\alpha }_{t-1}}\overline{ϵ}$，对$(2)$改写，有：

$$\begin{array}{}\text{(3)}& \begin{array}{rl}& \sqrt{{\alpha }_t}{\mathbf{x}}_{t-1}+\sqrt{1-{\alpha }_t}ϵ\\ =& \sqrt{{\alpha }_t}(\sqrt{{\alpha }_{t-1}}{\mathbf{x}}_{t-2}+\sqrt{1-{\alpha }_{t-1}}\overline{ϵ})+\sqrt{1-{\alpha }_t}ϵ\\ =& \sqrt{{\alpha }_t{\alpha }_{t-1}}{\mathbf{x}}_{t-2}+\sqrt{{\alpha }_t(1-{\alpha }_{t-1})}\overline{ϵ}+\sqrt{1-{\alpha }_t}ϵ\end{array}\end{array}$$

为了与$ϵ$进行区分，使用$\overline{ϵ}$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：

$$\begin{array}{}\text{(a)}& \begin{array}{rl}ϵ\sim \mathcal{N}(0,\mathbf{I})& \Rightarrow \sqrt{1-{\alpha }_t}ϵ\sim \mathcal{N}(0,(1-{\alpha }_t)\mathbf{I})\\ \overline{ϵ}\sim \mathcal{N}(0,\mathbf{I})& \Rightarrow \sqrt{{\alpha }_t(1-{\alpha }_{t-1})}ϵ\sim \mathcal{N}(0,{\alpha }_t(1-{\alpha }_{t-1})\mathbf{I})\end{array}\end{array}$$

由于$\sqrt{1-{\alpha }_t}ϵ$与$\sqrt{{\alpha }_t(1-{\alpha }_{t-1})}\overline{ϵ}$独立且都服从高斯分布，记$U=\sqrt{1-{\alpha }_t}ϵ+\sqrt{{\alpha }_t(1-{\alpha }_{t-1})}\overline{ϵ}$，由性质2可知$U$也服从高斯分布，有：

$$\begin{array}{}\text{(b)}& \begin{array}{rl}\sqrt{1-{\alpha }_t}ϵ+\sqrt{{\alpha }_t(1-{\alpha }_{t-1})}\overline{ϵ}& \sim \mathcal{N}(0,(1-{\alpha }_t)\mathbf{I}+{\alpha }_t(1-{\alpha }_{t-1})\mathbf{I})\\ \Rightarrow U& \sim \mathcal{N}(0,(1-{\alpha }_t{\alpha }_{t-1})\mathbf{I})\end{array}\end{array}$$

基于高斯分布的性质1，将$U$使用标准高斯分布表示：

$$\begin{array}{}\text{(c)}& \begin{array}{rl}U& \sim \mathcal{N}(0,(1-{\alpha }_t{\alpha }_{t-1})\mathbf{I})\Rightarrow U=\sqrt{1-{\alpha }_t{\alpha }_{t-1}}ϵ\end{array}\end{array}$$

将$(c)$代入$(3)$，可得：

$$\begin{array}{rl}& \sqrt{{\alpha }_t}{\mathbf{x}}_{t-1}+\sqrt{1-{\alpha }_t}ϵ\\ =& \sqrt{{\alpha }_t}(\sqrt{{\alpha }_{t-1}}{\mathbf{x}}_{t-2}+\sqrt{1-{\alpha }_{t-1}}\overline{ϵ})+\sqrt{1-{\alpha }_t}ϵ\\ =& \sqrt{{\alpha }_t{\alpha }_{t-1}}{\mathbf{x}}_{t-2}+\sqrt{{\alpha }_t(1-{\alpha }_{t-1})}\overline{ϵ}+\sqrt{1-{\alpha }_t}ϵ\\ =& \sqrt{{\alpha }_t{\alpha }_{t-1}}{\mathbf{x}}_{t-2}+\sqrt{1-{\alpha }_t{\alpha }_{t-1}}ϵ\end{array}$$

由数学归纳法，易知：

$$\begin{array}{rl}q({\mathbf{x}}_t\mid {\mathbf{x}}_{t-1})& =\mathcal{N}(\sqrt{1-{\beta }_t}{\mathbf{x}}_{t-1},{\beta }_t\mathbf{I})\\ {\mathbf{x}}_t& =\sqrt{1-{\beta }_t}{\mathbf{x}}_{t-1}+\sqrt{{\beta }_t}ϵ,ϵ\sim \mathcal{N}(0,\mathbf{I})\\ & =\sqrt{{\alpha }_t}{\mathbf{x}}_{t-1}+\sqrt{1-{\alpha }_t}ϵ\\ & =\sqrt{{\alpha }_t{\alpha }_{t-1}}{\mathbf{x}}_{t-2}+\sqrt{1-{\alpha }_t{\alpha }_{t-1}}ϵ\\ & =\dots \\ & =\sqrt{{\overline{\alpha }}_t}{\mathbf{x}}_0+\sqrt{1-{\overline{\alpha }}_t}ϵ\end{array}$$

因此，$q({\mathbf{x}}_t\mid {\mathbf{x}}_0)=\mathcal{N}(\sqrt{{\overline{\alpha }}_t}{\mathbf{x}}_0,\sqrt{1-{\overline{\alpha }}_t}\mathbf{I})$

推导二

在$diffusion$中，定义$q$服从高斯分布，故对$q({\mathbf{x}}_{t-1}\mid {\mathbf{x}}_t,{\mathbf{x}}_0)$定义如下：

$$\begin{array}{rl}q({\mathbf{x}}_{t-1}\mid {\mathbf{x}}_t,{\mathbf{x}}_0)& =\mathcal{N}({\mathbf{x}}_{t-1};{\tilde{\mathit{\mu }}}_t({\mathbf{x}}_t,{\mathbf{x}}_0),{\tilde{\beta }}_t\mathbf{I})\end{array}$$

那其中${\tilde{\mathit{\mu }}}_t({\mathbf{x}}_t,{\mathbf{x}}_0)$与$\widetilde{{\beta }_t}$如何得到？

此处先给出结论，下方是更详细的推导。

$$\begin{array}{rl}{\tilde{\mathit{\mu }}}_t({\mathbf{x}}_t,{\mathbf{x}}_0)& :=\frac{\sqrt{{\overline{\alpha }}_{t-1}}{\beta }_t}{1-{\overline{\alpha }}_t}{\mathbf{x}}_0+\frac{\sqrt{{\alpha }_t}(1-{\overline{\alpha }}_{t-1})}{1-{\overline{\alpha }}_t}{\mathbf{x}}_t,\\ {\tilde{\beta }}_t& :=\frac{1-{\overline{\alpha }}_{t-1}}{1-{\overline{\alpha }}_t}{\beta }_t\end{array}$$

解：
回顾贝叶斯公式，对$q({\mathbf{x}}_{t-1}\mid {\mathbf{x}}_t,{\mathbf{x}}_0)$改写，有：

$$\begin{array}{}\text{(1)}& q({\mathbf{x}}_{t-1}\mid {\mathbf{x}}_t,{\mathbf{x}}_0)=q({\mathbf{x}}_t\mid {\mathbf{x}}_{t-1},{\mathbf{x}}_0)\frac{q({\mathbf{x}}_{t-1}\mid {\mathbf{x}}_0)}{q({\mathbf{x}}_t\mid {\mathbf{x}}_0)}\end{array}$$

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

$$q({\mathbf{x}}_t\mid {\mathbf{x}}_{t-1},{\mathbf{x}}_0)=q({\mathbf{x}}_t\mid {\mathbf{x}}_{t-1})$$

$(1)$式写作$(2)$式：

$$\begin{array}{}\text{(2)}& \begin{array}{rl}q({\mathbf{x}}_{t-1}\mid {\mathbf{x}}_t,{\mathbf{x}}_0)& =q({\mathbf{x}}_t\mid {\mathbf{x}}_{t-1},{\mathbf{x}}_0)\frac{q({\mathbf{x}}_{t-1}\mid {\mathbf{x}}_0)}{q({\mathbf{x}}_t\mid {\mathbf{x}}_0)}\\ & =q({\mathbf{x}}_t\mid {\mathbf{x}}_{t-1})\frac{q({\mathbf{x}}_{t-1}\mid {\mathbf{x}}_0)}{q({\mathbf{x}}_t\mid {\mathbf{x}}_0)}\end{array}\end{array}$$

基于推导一的结论，易知：

$$\begin{array}{rl}q({\mathbf{x}}_t\mid {\mathbf{x}}_0)& =\mathcal{N}(\sqrt{{\overline{\alpha }}_t}{\mathbf{x}}_0,\sqrt{1-{\overline{\alpha }}_t}\mathbf{I})\\ q({\mathbf{x}}_{t-1}\mid {\mathbf{x}}_0)& =\mathcal{N}(\sqrt{{\overline{\alpha }}_{t-1}}{\mathbf{x}}_0,\sqrt{1-{\overline{\alpha }}_{t-1}}\mathbf{I})\end{array}$$

由高斯分布的概率密度函数，对$(2)$展开，有：

$$\begin{array}{}\text{(3)}& \begin{array}{rl}q({\mathbf{x}}_{t-1}\mid {\mathbf{x}}_t,{\mathbf{x}}_0)& =q({\mathbf{x}}_t\mid {\mathbf{x}}_{t-1})\frac{q({\mathbf{x}}_{t-1}\mid {\mathbf{x}}_0)}{q({\mathbf{x}}_t\mid {\mathbf{x}}_0)}\\ & \propto \exp (-\frac{1}{2}(\frac{{({\mathbf{x}}_t-\sqrt{{\alpha }_t}{\mathbf{x}}_{t-1})}^2}{{\beta }_t}+\frac{{({\mathbf{x}}_{t-1}-\sqrt{{\overline{\alpha }}_{t-1}}{\mathbf{x}}_0)}^2}{1-{\overline{\alpha }}_{t-1}}-\frac{{({\mathbf{x}}_t-\sqrt{{\overline{\alpha }}_t}{\mathbf{x}}_0)}^2}{1-{\overline{\alpha }}_t}))\end{array}\end{array}$$

不论是${\beta }_t$或是${\overline{\alpha }}_t$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$(4)$：

$$\begin{array}{}\text{(4)}& \begin{array}{rl}& q({\mathbf{x}}_{t-1}\mid {\mathbf{x}}_t,{\mathbf{x}}_0)=q({\mathbf{x}}_t\mid {\mathbf{x}}_{t-1})\frac{q({\mathbf{x}}_{t-1}\mid {\mathbf{x}}_0)}{q({\mathbf{x}}_t\mid {\mathbf{x}}_0)}\\ & \propto \exp (-\frac{1}{2}(\frac{{({\mathbf{x}}_t-\sqrt{{\alpha }_t}{\mathbf{x}}_{t-1})}^2}{{\beta }_t}+\frac{{({\mathbf{x}}_{t-1}-\sqrt{{\overline{\alpha }}_{t-1}}{\mathbf{x}}_0)}^2}{1-{\overline{\alpha }}_{t-1}}-\frac{{({\mathbf{x}}_t-\sqrt{{\overline{\alpha }}_t}{\mathbf{x}}_0)}^2}{1-{\overline{\alpha }}_t}))\\ & =\exp (-\frac{1}{2}(\frac{{\mathbf{x}}_t^2-2\sqrt{{\alpha }_t}{\mathbf{x}}_t{\mathbf{x}}_{t-1}+{\alpha }_t{\mathbf{x}}_{t-1}^2}{{\beta }_t}+\frac{{\mathbf{x}}_{t-1}^2-2\sqrt{{\overline{\alpha }}_{t-1}}{\mathbf{x}}_0{\mathbf{x}}_{t-1}+{\overline{\alpha }}_{t-1}{\mathbf{x}}_0^2}{1-{\overline{\alpha }}_{t-1}}-\frac{{({\mathbf{x}}_t-\sqrt{{\overline{\alpha }}_t}{\mathbf{x}}_0)}^2}{1-{\overline{\alpha }}_t}))\\ & =\exp (-\frac{1}{2}((\frac{{\alpha }_t}{{\beta }_t}+\frac{1}{1-{\overline{\alpha }}_{t-1}}){\mathbf{x}}_{t-1}^2-(\frac{2\sqrt{{\alpha }_t}}{{\beta }_t}{\mathbf{x}}_t+\frac{2\sqrt{{\overline{\alpha }}_{t-1}}}{1-{\overline{\alpha }}_{t-1}}{\mathbf{x}}_0){\mathbf{x}}_{t-1}+C({\mathbf{x}}_t,{\mathbf{x}}_0)))\end{array}\end{array}$$

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$-\frac{({\mathbf{x}}_{t-1}-\widetilde{{\mu }_t}{)}^2}{2\widetilde{{\beta }_t}}$。因此，不难得出：

$$\begin{array}{}\text{(5)}& \begin{array}{rl}{\tilde{\mathit{\mu }}}_t& =\frac{1}{\frac{{\alpha }_t}{{\beta }_t}+\frac{1}{1-{\overline{\alpha }}_{t-1}}}\ast (\frac{\sqrt{{\alpha }_t}}{{\beta }_t}{\mathbf{x}}_t+\frac{\sqrt{{\overline{\alpha }}_{t-1}}}{1-{\overline{\alpha }}_{t-1}}{\mathbf{x}}_0)\\ & =\frac{(1-{\overline{\alpha }}_{t-1}){\beta }_t}{{\alpha }_t(1-{\overline{\alpha }}_{t-1})+{\beta }_t}\ast (\frac{\sqrt{{\alpha }_t}}{{\beta }_t}{\mathbf{x}}_t+\frac{\sqrt{{\overline{\alpha }}_{t-1}}}{1-{\overline{\alpha }}_{t-1}}{\mathbf{x}}_0)\\ & =\frac{(1-{\overline{\alpha }}_{t-1})\sqrt{{\alpha }_t}}{{\alpha }_t(1-{\overline{\alpha }}_{t-1})+{\beta }_t}{\mathbf{x}}_t+\frac{\sqrt{{\overline{\alpha }}_{t-1}}{\beta }_t}{{\alpha }_t(1-{\overline{\alpha }}_{t-1})+{\beta }_t}{\mathbf{x}}_0\end{array}\end{array}$$

${\alpha }_t=1-{\beta }_t$，故：

$$\begin{array}{}\text{(6)}& \begin{array}{rl}{\alpha }_t(1-{\overline{\alpha }}_{t-1})+{\beta }_t& ={\alpha }_t-{\alpha }_t{\overline{\alpha }}_{t-1}+{\beta }_t\\ & =1-{\beta }_t-{\alpha }_t{\overline{\alpha }}_{t-1}+{\beta }_t\\ & =1-{\alpha }_t{\overline{\alpha }}_{t-1}\\ & =1-{\overline{\alpha }}_t\end{array}\end{array}$$

将$(6)$式代入$(5)$，有：

$${\tilde{\mathit{\mu }}}_t({\mathbf{x}}_t,{\mathbf{x}}_0):=\frac{\sqrt{{\overline{\alpha }}_{t-1}}{\beta }_t}{1-{\overline{\alpha }}_t}{\mathbf{x}}_0+\frac{\sqrt{{\alpha }_t}(1-{\overline{\alpha }}_{t-1})}{1-{\overline{\alpha }}_t}{\mathbf{x}}_t$$

对于${\tilde{\beta }}_t$，有：

$$\begin{array}{rl}{\tilde{\beta }}_t& =\frac{1}{\frac{{\alpha }_t}{{\beta }_t}+\frac{1}{1-{\overline{\alpha }}_{t-1}}}\\ & =\frac{(1-{\overline{\alpha }}_{t-1}){\beta }_t}{{\alpha }_t(1-{\overline{\alpha }}_{t-1})+{\beta }_t}\\ & =\frac{1-{\overline{\alpha }}_{t-1}}{1-{\overline{\alpha }}_t}{\beta }_t\end{array}$$

以上内容即$\text{DDPM}$中一些被省略的数学推导。

Papers

1. Deep unsupervised learning using nonequilibrium thermodynamics, 2015.
2. Denoising diffusion probabilistic models, 2020.
3. Improved denoising diffusion probabilistic models, 2021.