DDIM公式草稿3
已知q(xt∣x0)q(xt−1∣x0)不使用q(xt∣xt−1),求q(xt−1∣xt,x0)⎩⎪⎨⎪⎧xt−1=mtxt+ntx0+σtε1xt=αtx0+1−αtε2xt−1=αt−1x0+1−αt−1ε3xt−1解得：xt−1xt−1if=mt(αtx0+1−αtε2)+ntx0+σtε1=(mtαt+nt)x0+mt1−αtε2+σtε1{mtαt+nt=αt−1mt2(1−αt)+σt2=1−αt−1mt=1−αt1−αt−1−σt2nt=αt−1−1−αtαt(1−αt−1−σt2)=1−αt1−αt−1−σt2xt+(αt−1−1−αtαt(1−αt−1−σt2))x0+σtε1=αt−1x0+1−αt−1−σt2(1−αt1xt−1−αtαtx0)+σtε1=αt−1x0+1−αt−1−σt21−αtxt−αtx0+σtε1x0=x0∣t^1−αtxt−αtx0=εθ(xt,t)σt=σDDPMDDIM→DDPM
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