【深度学习】Diffusion扩散模型原理解析2

由于篇幅受限，CSDN不能发布超过一定次数的文章，故在此给出上一篇链接：【深度学习】diffusion原理解析

3.2、目标函数求解

里面的最后一项，$q(x_T|x_0)$我们前面提到过，其近似服从标准正态，而对于$P(x_T)$，我们是假定为标准正态，这两项都可以求出来，所以没有任何可学习的参数

真正需要优化的是第一项和第二项。第一项就是重构损失；而第二项，是KL散度。里面的$P(x_{t-1}|x_t)$需要用神经网络去逼近。

论文提到，$q(x_{t-1}|x_t)$是正态分布，但由于$q(x_{t-1}|x_t)$是无法求出来的，所以选择用$P(x_{t-1}|x_t)$去逼近

$q(x_{t-1}|x_t,x_0)$服从正态分布(证明)，我们可以求出来。

直接把$q(x_{t-1}|x_t,x_0)$配成正态分布求解期望和方差比较麻烦，我们不如反过来推

假设多维高斯分布P(x)，我们有
$$\begin{array}{rl}P(x)=& \frac{1}{(2\pi {)}^{\frac{p}{2}}|\Sigma {|}^{\frac{1}{2}}}\exp \left\{-\frac{1}{2}(x-\mu {)}^T{\Sigma }^{-1}(x-\mu )\right\}\\ =& \frac{1}{(2\pi {)}^{\frac{p}{2}}|\Sigma {|}^{\frac{1}{2}}}\exp \left\{-\frac{1}{2}(x^T{\Sigma }^{-1}x-{\mu }^T{\Sigma }^{-1}x-x^T{\Sigma }^{-1}\mu +\mu {\Sigma }^{-1}\mu )\right\}\\ =& \frac{1}{(2\pi {)}^{\frac{p}{2}}|\Sigma {|}^{\frac{1}{2}}}\exp \left\{-\frac{1}{2}(x^T{\Sigma }^{-1}x-2{\mu }^T{\Sigma }^{-1}x+\mu {\Sigma }^{-1}\mu )\right\}\end{array}\text{\hspace{1em}(11)}$$
对于随机变量x，里面有关的只有$x^T{\Sigma }^{-1}x$和$2{\mu }^T{\Sigma }^{-1}x$。其中第一项有两个x，为二次项。第二项有一个x，为一次项。

同理，对于$q(x_{t-1}|x_t,x_0)$，我们只需要找出对应的一次项跟二次项，就能够得出期望跟协方差了

$$\begin{array}{rl}q(x_{t-1}|x_t,x_0)=& \frac{q(x_{t-1},x_t|x_0)}{q(x_t|x_0)}\\ =& \frac{q(x_t|x_{t-1},x_0)q(x_{t-1}|x_0)}{q(x_t|x_0)}\\ =& \frac{q(x_t|x_{t-1})q(x_{t-1}|x_0)}{q(x_t|x_0)}\\ =& \frac{N(\sqrt{{\alpha }_t}{x}_{t-1},(1-{\alpha }_t)I)N(\sqrt{{\bar{\alpha }}_{t-1}}{x}_0,(1-{\bar{\alpha }}_{t-1})I)}{N(\sqrt{{\bar{\alpha }}_t}{x}_0,(1-{\bar{\alpha }}_t)I)}\\ \propto & \exp -\left\{\frac{(x_t-\sqrt{{\alpha }_t}{x}_{t-1}{)}^T(x_t-\sqrt{{\alpha }_t}{x}_{t-1})}{2(1-{\alpha }_t)}+\frac{(x_{t-1}-\sqrt{{\bar{\alpha }}_{t-1}}{x}_0{)}^T(x_{t-1}-\sqrt{{\bar{\alpha }}_{t-1}}{x}_0)}{2(1-{\bar{\alpha }}_{t-1})}-\frac{(x_t-\sqrt{{\bar{\alpha }}_t}{x}_0{)}^T(x_t-\sqrt{{\bar{\alpha }}_t}{x}_0)}{2(1-{\bar{\alpha }}_t)}\right\}\\ =& \exp -\left\{\frac{x_t^T{x}_t-2\sqrt{{\alpha }_t}{x}_t^T{x}_{t-1}+{\alpha }_t{x}_{t-1}^T{x}_{t-1}}{2(1-{\alpha }_t)}+\frac{x_{t-1}^T{x}_{t-1}-2\sqrt{{\bar{\alpha }}_{t-1}}{x}_0^T{x}_{t-1}+{\alpha }_{t-1}{x}_0^T{x}_0}{2(1-{\bar{\alpha }}_{t-1})}-\frac{(x_t-\sqrt{{\bar{\alpha }}_t}{x}_0{)}^T(x_t-\sqrt{{\bar{\alpha }}_t}{x}_0)}{2(1-{\bar{\alpha }}_t)}\right\}\\ =& \exp \left\{-\frac{1}{2}\left(x_{t-1}^T\frac{1-{\bar{\alpha }}_t}{{\beta }_t(1-{\bar{\alpha }}_{t-1})}{x}_{t-1}-2\frac{\sqrt{{\alpha }_t}(1-{\bar{\alpha }}_{t-1})x_t^T+\sqrt{{\bar{\alpha }}_{t-1}}(1-{\alpha }_t)x_0^T}{1-{\bar{\alpha }}_t}\frac{1-{\bar{\alpha }}_t}{{\beta }_t(1-{\bar{\alpha }}_{t-1})}\right)x_{t-1}+C\right\}\end{array}由式（11）可得$$

$$q(x_{t-1}|x_t,x_0)\sim N(x_{t-1}|\frac{\sqrt{a_t}(1-{\bar{\alpha }}_{t-1})x_t+\sqrt{{\bar{\alpha }}_{t-1}}(1-{\alpha }_t)x_0}{1-{\bar{\alpha }}_t},\frac{1-{\bar{\alpha }}_{t-1}}{1-{\bar{\alpha }}_t}{\beta }_{t}I)$$

再简单变化一下，可得
$$q(x_{t-1}|x_t,x_0)\sim N(x_{t-1}|\frac{\sqrt{a_t}(1-{\bar{\alpha }}_{t-1})x_t}{1-{\bar{\alpha }}_t}+\frac{\sqrt{{\bar{\alpha }}_{t-1}}{\beta }_t{x}_0}{1-{\bar{\alpha }}_t},\frac{1-{\bar{\alpha }}_{t-1}}{1-{\bar{\alpha }}_t}{\beta }_{t}I)$$
那么接下来，就可以求解$KL(q(x_{t-1}|x_t,x_0)||P(x_{t-1}|x_t))$

记$q(x_{t-1}|x_t,x_0)\sim N(x_{t-1}|{\mu }_{\varphi }^{t-1},{\Sigma }_{\varphi }^{t-1})$，为了简便，我隐去t-1时刻，记作$q(x_{t-1}|x_t,x_0)\sim N(x_{t-1}|{\mu }_{\varphi },{\Sigma }_{\varphi })$

由于$q(x_{t-1}|x_t,x_0)$的协方差与$x_0,x_t$无关，是一个固定的值。

所以，设$P(x_{t-1}|x_t)\sim N(x_{t-1}|{\mu }_{\theta }^{t-1},{\Sigma }_{\theta }^{t-1})$，为了简便，我依然隐去时刻，表达为$P(x_{t-1}|x_t)\sim N(x_{t-1}|{\mu }_{\theta },{\Sigma }_{\theta })$

里面的${\Sigma }_{\theta }$直接等于$q(x_{t-1}|x_t,x_0)$的协方差，也就是${\Sigma }_{\varphi }={\Sigma }_{\theta }$

下面给出两个正态分布的KL散度公式

其中，n表示随机变量x的维度

推导请看参考③

直接代入公式可得：
$$\begin{array}{rl}KL(q(x_{t-1}|x_t,x_0)||P(x_{t-1}|x_0))=& \frac{1}{2}\left[({\mu }_{\varphi }-{\mu }_{\theta }{)}^T{\Sigma }_{\theta }^{-1}({\mu }_{\varphi }-{\mu }_{\theta })-\log \det ({\Sigma }_{\theta }^{-1}{\Sigma }_{\varphi })+Tr({\Sigma }_{\theta }^{-1}{\Sigma }_{\varphi })-n\right]\\ =& \frac{1}{2}\left[({\mu }_{\varphi }-{\mu }_{\theta }{)}^T{\Sigma }_{\theta }^{-1}({\mu }_{\varphi }-{\mu }_{\theta })-\log 1+n-n\right]\\ =& \frac{1}{2}\left[({\mu }_{\varphi }-{\mu }_{\theta }{)}^T{\Sigma }_{\theta }^{-1}({\mu }_{\varphi }-{\mu }_{\theta })\right]\\ =& \frac{1}{2{\sigma }_t^2}\left[||{\mu }_{\varphi }-{\mu }_{\theta }(x_t,t)|{|}^2\right]\end{array}\text{\hspace{1em}(12)}$$
${\sigma }_t^2$是方差${\Sigma }_{\theta }$的表达，由于是给定的，所以为了简单起见，写成这样。

但论文里面对他进行了比较，不论${\sigma }_t^2$直接取成${\Sigma }_{\theta }$，还是${\beta }_t、\bar{\beta }$，都得到了差不多的实验结果

所以，便得到了最终的损失函数

在论文中，还将该损失函数写成了其他形式，我们前面写到
$${\mu }_{\varphi }=\frac{\sqrt{a_t}(1-{\bar{\alpha }}_{t-1})x_t}{1-{\bar{\alpha }}_t}+\frac{\sqrt{{\bar{\alpha }}_{t-1}}{\beta }_t{x}_0}{1-{\bar{\alpha }}_t}\text{\hspace{1em}(13)}$$
那么对于$P(x_{t-1}|x_t)$而言，里面其实只有一个未知数，也就是$x_0$，所以，我们只需要让神经网络预测$x_0$就可以了，记神经网络预测的$x_0$为$f_{\theta }(x_t,t)$所以式（12）可进行如下变化：
$$\begin{array}{rl}\frac{1}{2{\sigma }_i^2}\left[||{\mu }_{\varphi }-{\mu }_{\theta }(x_t,t)|{|}^2\right]=& \frac{1}{2{\sigma }_t^2}\left[||\left(\frac{\sqrt{a_t}(1-{\bar{\alpha }}_{t-1})x_t}{1-{\bar{\alpha }}_t}+\frac{\sqrt{{\bar{\alpha }}_{t-1}}{\beta }_t{x}_0}{1-{\bar{\alpha }}_t}\right)-\left(\frac{\sqrt{a_t}(1-{\bar{\alpha }}_{t-1})x_t}{1-{\bar{\alpha }}_t}+\frac{\sqrt{{\bar{\alpha }}_{t-1}}{\beta }_t{f}_{\theta }(x_t,t)}{1-{\bar{\alpha }}_t}|{|}^2\right)\right]\\ =& \frac{{\bar{\alpha }}_{t-1}{\beta }_t^2}{2{\sigma }_t^2(1-{\bar{\alpha }}_t{)}^2}\left[||x_0-f_{\theta }(x_t,t)|{|}^2\right]\end{array}\text{\hspace{1em}(14)}$$
除此之外，还可以去预测噪声

由
$$x_t=\sqrt{{\bar{\alpha }}_t}{x}_0+\sqrt{1-{\bar{\alpha }}_t}{ϵ}_t\to x_0=\frac{x_t-\sqrt{1-{\bar{\alpha }}_t}{ϵ}_t}{\sqrt{{\bar{\alpha }}_t}}\text{\hspace{1em}(14)}$$
将$x_0$代入式（13）
$$\begin{array}{rl}{\mu }_{\varphi }=& \frac{\sqrt{a_t}(1-{\bar{\alpha }}_{t-1})x_t}{1-{\bar{\alpha }}_t}+\frac{\sqrt{{\bar{\alpha }}_{t-1}}{\beta }_t}{1-{\bar{\alpha }}_t}\frac{x_t-\sqrt{1-{\bar{\alpha }}_t}{ϵ}_t}{\sqrt{{\bar{\alpha }}_t}}\\ =& \frac{\sqrt{a_t}(1-{\bar{\alpha }}_{t-1})x_t}{1-{\bar{\alpha }}_t}+\frac{\sqrt{{\bar{\alpha }}_{t-1}}{\beta }_t{x}_t}{(1-{\bar{\alpha }}_t)\sqrt{{\bar{\alpha }}_t}}-\frac{\sqrt{{\bar{\alpha }}_{t-1}}{\beta }_t\sqrt{1-{\bar{\alpha }}_t}{ϵ}_t}{(1-{\bar{\alpha }}_t)\sqrt{{\bar{\alpha }}_t}}\\ =& \frac{\sqrt{a_t}(1-{\bar{\alpha }}_{t-1})x_t}{1-{\bar{\alpha }}_t}+\frac{{\beta }_t{x}_t}{(1-{\bar{\alpha }}_t){\sqrt{\alpha }}_t}-\frac{{\beta }_t\sqrt{1-{\bar{\alpha }}_t}{ϵ}_t}{(1-{\bar{\alpha }}_t){\sqrt{\alpha }}_t}\\ =& \frac{1}{\sqrt{{\alpha }_t}}\left[\frac{a_t(1-{\bar{\alpha }}_{t-1})x_t}{1-{\bar{\alpha }}_t}+\frac{{\beta }_t{x}_t}{1-{\bar{\alpha }}_t}-\frac{{\beta }_t\sqrt{1-{\bar{\alpha }}_t}{ϵ}_t}{1-{\bar{\alpha }}_t}\right]\\ =& \frac{1}{\sqrt{{\alpha }_t}}\left[\left(\frac{{\alpha }_t(1-{\bar{\alpha }}_{t-1})+{\beta }_t}{1-{\bar{\alpha }}_t}\right)x_t-\frac{{\beta }_t\sqrt{1-{\bar{\alpha }}_t}{ϵ}_t}{1-{\bar{\alpha }}_t}\right]\\ =& \frac{1}{\sqrt{{\alpha }_t}}\left[x_t-\frac{{\beta }_t}{\sqrt{1-{\bar{\alpha }}_t}}{ϵ}_t\right]\end{array}\text{\hspace{1em}(15)}$$
此时我们发现，对于$P(x_{t-1}|x_t)$中，只剩下${ϵ}_t$是未知数，所以，我们用神经网络去预测噪声
$$\begin{array}{rl}\frac{1}{2{\sigma }_i^2}\left[||{\mu }_{\varphi }-{\mu }_{\theta }(x_t,t)|{|}^2\right]=& \frac{1}{2{\sigma }_t^2}||\frac{1}{\sqrt{{\alpha }_t}}\left(x_t-\frac{{\beta }_t}{\sqrt{1-{\bar{\alpha }}_t}}{ϵ}_t\right)-\frac{1}{\sqrt{{\alpha }_t}}\left(x_t-\frac{{\beta }_t}{\sqrt{1-{\bar{\alpha }}_t}}{ϵ}_{\theta }(x_t,t)\right)|{|}^2\\ =& \frac{{\beta }_t^2}{2{\sigma }_t^2(1-{\bar{\alpha }}_t){\alpha }_t}||{ϵ}_t-{ϵ}_{\theta }(x_t,t)|{|}^2\end{array}$$
至此，我们中遇得到了KL散度的优化目标函数

接下来，我们来看重构损失
$$\begin{array}{rl}\max {\mathbb{E}}_{q(x_1|x_0)}[\log P(x_0|x_1)]\approx & \max \frac{1}{n}\sum _{i=1}^n\log P(x_0^i|x_1^i)\\ =& \max \frac{1}{n}\sum _{i=1}^n\log \frac{1}{{2\pi }^{D/2}|{\Sigma }_{\theta }{|}^{1/2}}\exp \left\{-\frac{1}{2}(x_0^i-{\mu }_{\theta }(x_1^i,1){)}^T{\Sigma }_{\theta }^{-1}(x_0^i-{\mu }_{\theta }(x_1^i,1))\right\}\\ =& \max \frac{1}{n}\sum _{i=1}^n\log \frac{1}{{2\pi }^{D/2}|{\Sigma }_{\theta }{|}^{1/2}}-\frac{1}{2}(x_0^i-{\mu }_{\theta }(x_1^i,1){)}^T{\Sigma }_{\theta }^{-1}(x_0^i-{\mu }_{\theta }(x_1^i,1))\\ \propto & \min \frac{1}{n}\sum _{i=1}^n(x_0^i-{\mu }_{\theta }(x_1^i,1){)}^T(x_0^i-{\mu }_{\theta }(x_1^i,1))\\ =& \min \frac{1}{n}\sum _{i=1}^n||x_0^i-{\mu }_{\theta }(x_1^i,1)|{|}^2\end{array}$$

将式（14）的$x_0$和式（15）代入
$$\begin{array}{rl}||x_0^i-{\mu }_{\theta }(x_1^i,1)|{|}^2=& ||\frac{x_1-\sqrt{1-{\bar{\alpha }}_1}{ϵ}_1}{\sqrt{{\bar{\alpha }}_1}}-\frac{1}{\sqrt{{\alpha }_1}}\left[x_1-\frac{{\beta }_1}{\sqrt{1-{\bar{\alpha }}_1}}{ϵ}_{\theta }(x_1,1)\right]|{|}^2\\ \propto & ||{ϵ}_{\theta }(x_1,1)-{ϵ}_1|{|}^2\end{array}$$
所以，最终的流程为：

4、结束

好了，以上就是本文所有内容了，如有问题，还望指出，阿里嘎多！

5、参考

①一文解释 Diffusion Model (一) DDPM 理论推导 - 知乎 (zhihu.com)

②Diffusion Model入门（8）——Denoising Diffusion Probabilistic Models（完结篇） - 知乎 (zhihu.com)

③两个多元正态分布的KL散度、巴氏距离和W距离 - 科学空间|Scientific Spaces (kexue.fm)

④什么是扩散模型？

⑤diffusion model 最近在图像生成领域大红大紫，如何看待它的风头开始超过 GAN ？ - 知乎 (zhihu.com)