DDIM公式草稿1

q(xtx0)=N(xt;αtx0,(1αt)I)xt=αtx0+1αtεεN(0,I)q(xt1xt)=q(xt,xt1)q(xt)=q(xtxt1)q(xt1)q(xt)xt1xtx0q(xt1xt)=q(xt1xt,x0)q(xt1xt,x0)=q(xtxt1)q(xt1x0)q(xtx0)=N(xt1;μ(xt,x0),σt2I)q(xtxt1)=N(xt;αtxt1,(1αt)I)μ(xt,x0)=αt1αt11αtxt+αt1βt1αtx0σt2=1αt11αtβtx0^=1αt(xt1αtεθ)DDPM:x0t^=1αt(xt1αt  εθ(xt,t) )xt1=μ(xt,x0t^)+σtεεN(0,I)q(x_t|x_0)=N(x_t;\sqrt{\overline{\alpha_{t}}}x_0,(1-\overline{\alpha_t})I) \\ \quad \\ x_t=\sqrt{\overline{\alpha_t}}x_{0}+\sqrt{1-\overline{\alpha_t}}\varepsilon \qquad \varepsilon \sim N(0,I) \\ \quad \\ q(x_{t-1}|x_t)=\frac{q(x_t,x_{t-1})}{q(x_t)}=\frac{q(x_t|x_{t-1})q(x_{t-1})}{q(x_t)} \\ \quad \\ 根据马尔可夫链,x_{t-1}只于x_t相关,与x_0独立\quad q(x_{t-1}|x_t) =q(x_{t-1}|x_t,x_0)\\ \quad \\ q(x_{t-1}|x_t,x_0)=\frac{q(x_t|x_{t-1})q(x_{t-1}|x_0)}{q(x_t|x_0)}=N(x_{t-1};\mu(x_t,x_0),\sigma^2_{t}I) \\ q(x_t|x_{t-1})=N(x_t;\sqrt{\alpha_t}x_{t-1},(1-\alpha_t)I) \\ \quad \\ \mu(x_t,x_0)=\sqrt{\alpha_t}\frac{1-\overline{\alpha_{t-1}}}{1-\overline{\alpha_t}}x_t+\frac{\sqrt{\overline{\alpha_{t-1}}}\beta_t}{1-\overline{\alpha_t}}x_0 \qquad \sigma^2_t=\frac{1-\overline{\alpha_{t-1}}}{1-\overline{\alpha_t}}\beta_t \\ \quad \\ \hat{x_0}=\frac{1}{\sqrt{\overline{\alpha_t}}}(x_t-\sqrt{1-\overline{\alpha_t}}\varepsilon_{\theta}) \\ \quad \\ \begin{aligned} DDPM: ①&\quad \hat{x_{0|t}}=\frac{1}{\sqrt{\overline{\alpha_t}}}(x_t-\sqrt{1-\overline{\alpha_t}} \ \ \varepsilon_{\theta}(x_t,t)\ )\\ ②&\quad x_{t-1}=\mu(x_t,\hat{x_{0|t}})+\sigma_t \varepsilon \qquad \varepsilon \sim N(0,I) \end{aligned}
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10-24 11:10
山西大学 Java
若梦难了:哥们,面试挂是很正常的。我大中厂终面挂,加起来快10次了,继续努力吧。
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