Diffusion-DDPM-扩散模型-公式速查

Diffusion-DDPM-扩散模型-公式速查

---

扩散过程

操作	分布形式	计算形式	备注
由 ${\mathbf{x}}_0$ 求 ${\mathbf{x}}_t$	$q({\mathbf{x}}_t|{\mathbf{x}}_0)=\mathcal{N}({\mathbf{x}}_t;\sqrt{{\bar{\alpha }}_t}{\mathbf{x}}_0,(1-{\bar{\alpha }}_t)\mathbf{I})$	$$\begin{gathered} {\mathbf{x}}_t=\sqrt{{\bar{\alpha }}_t}{\mathbf{x}}_0+\sqrt{1-{\bar{\alpha }}_t}\mathbf{ϵ} \\ {\mathbf{x}}_0=\frac{1}{\sqrt{{\bar{\alpha }}_t}}{\mathbf{x}}_t-\frac{\sqrt{1-{\bar{\alpha }}_t}}{\sqrt{{\bar{\alpha }}_t}}\mathbf{ϵ} \end{gathered}$$	$$\begin{gathered} \mathbf{ϵ}\sim \mathcal{N}(0,\mathbf{I}) \\ {\alpha }_t=1-{\beta }_t \\ {\bar{\alpha }}_t={\Pi }_{s=1}^t{\alpha }_s={\alpha }_1{\alpha }_2...{\alpha }_t \end{gathered}$$
由 ${\mathbf{x}}_{t-1}$ 求 ${\mathbf{x}}_t$	$q({\mathbf{x}}_t|{\mathbf{x}}_{t-1})=\mathcal{N}({\mathbf{x}}_t;\sqrt{1-{\beta }_t}{\mathbf{x}}_{t-1},{\beta }_t\mathbf{I})$	${\mathbf{x}}_t=\sqrt{1-{\beta }_t}{\mathbf{x}}_{t-1}+\sqrt{{\beta }_t}\mathbf{ϵ}$	$\mathbf{ϵ}\sim \mathcal{N}(0,\mathbf{I})$

---

逆扩散过程

操作	分布形式	计算形式	备注
${\mathbf{x}}_T$ 的采样	$p({\mathbf{x}}_T)=\mathcal{N}({\mathbf{x}}_T;0,\mathbf{I})$		${\mathbf{x}}_T\sim \mathcal{N}(0,\mathbf{I})$
由 ${\mathbf{x}}_t$ 得到 ${\mathbf{x}}_{t-1}$	$p_{\theta }({\mathbf{x}}_{t-1}|{\mathbf{x}}_t)=\mathcal{N}({\mathbf{x}}_{t-1};{\mathbf{\mu }}_{\theta }({\mathbf{x}}_t,t),{\mathbf{\Sigma }}_{\theta }({\mathbf{x}}_t,t))$	${\mathbf{x}}_{t-1}={\mathbf{\mu }}_{\theta }({\mathbf{x}}_t,t)+\sqrt{{\mathbf{\Sigma }}_{\theta }({\mathbf{x}}_t,t)}{\mathbf{ϵ}}_t$	$$\begin{gathered} {\mathbf{ϵ}}_t\sim \mathcal{N}(0,\mathbf{I}) \\ {\mathbf{\Sigma }}_{\theta }({\mathbf{x}}_t,t)={\sigma }_t^2\mathbf{I} \\ {\sigma }_t^2={\beta }_{t}or{\tilde{\beta }}_t \end{gathered}$$
给定 ${\mathbf{x}}_0$ 和 ${\mathbf{x}}_t$，求 ${\mathbf{x}}_{t-1}$	$q({\mathbf{x}}_{t-1}|{\mathbf{x}}_t,{\mathbf{x}}_0)=\mathcal{N}({\mathbf{x}}_{t-1};{\bar{\mathbf{\mu }}}_t({\mathbf{x}}_t,{\mathbf{x}}_0),{\tilde{\beta }}_t\mathbf{I})$	${\mathbf{x}}_{t-1}={\bar{\mathbf{\mu }}}_t({\mathbf{x}}_t,{\mathbf{x}}_0)+\sqrt{{\tilde{\beta }}_t}{\mathbf{ϵ}}_t$	$$\begin{gathered} {\bar{\mathbf{\mu }}}_t({\mathbf{x}}_t,{\mathbf{x}}_0)=\frac{\sqrt{{\bar{\alpha }}_{t-1}}{\beta }_t}{1-{\bar{\alpha }}_t}{\mathbf{x}}_0+\frac{\sqrt{{\alpha }_t}(1-{\bar{\alpha }}_{t-1})}{1-{\bar{\alpha }}_t}{\mathbf{x}}_t \\ {\tilde{\beta }}_t=\frac{1-{\bar{\alpha }}_{t-1}}{1-{\bar{\alpha }}_t}{\beta }_t \end{gathered}$$