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Regarding Eq. (1) in Progressive Distillation for Fast Sampling of Diffusion Models, $$ q(\mathbf{z}_t|\mathbf{z}_s) = \mathcal{N}(\mathbf{z}_t; (\alpha_t/\alpha_s)\mathbf{z}_s, \sigma _{t|s}^2 \mathbf{I}) \tag{1} $$ , it says $\sigma_{t|s}^2 = (1 - e^{\lambda_t - \lambda_s})\sigma_t^2$.

I believe $\sigma_{t|s}$ is from DDPM's $\beta$, then it's equivalent to $1 - \alpha_t^2 / \alpha_s^2$ here. Note that $\lambda_t = \log (\alpha_t^2 / \sigma_t^2)$. As far as I understand, $\sigma_t^2 + \alpha_t^2 = 1$. Maybe I'm taking something wrong.

Anyway, how could $\sigma_{t|s}^2 = (1 - e^{\lambda_t - \lambda_s})\sigma_t^2$ be derived from $1 - \alpha_t^2 / \alpha_s^2$?

Some more notes.

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2 Answers 2

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One of my colleagues gave me a solution. (Thanks Ken!)

$$ {\begin{aligned} \sigma_{t|s}^2 &= 1 - \alpha_{t|s}^2 \\ &= 1 - {\alpha_t^2 \over \alpha_s^2} \\ &= {\alpha_s^2 - \alpha_t^2 \over \alpha_s^2} \\ &= {(1 - \sigma_s^2)-(1-\sigma_t^2) \over \alpha_s^2} \\ &= {\sigma_t^2 - \sigma_s^2 \over \alpha_s^2} \\ &= {\sigma_t^2 (\alpha_s^2 + \sigma_s^2) - \sigma_s^2 (\alpha_t^2 + \sigma_t^2) \over \alpha_s^2} \\ &= {\sigma_t^2 \alpha_s^2 - \sigma_s^2 \alpha_t^2 \over \alpha_s^2} \\ &= \left(1 -{\sigma_s^2 \alpha_t^2 \over \alpha_s^2 \sigma_t^2}\right) \sigma_t^2 \\ &= \left(1 -{1 \over e^{\lambda_s}}\cdot e^{\lambda_t}\right) \sigma_t^2 \\ &= \left(1 - e^{\lambda_t - \lambda_s}\right) \sigma_t^2 \\ \end{aligned}} $$ given $$ \alpha_t^2 + \sigma_t^2 = 1 \\ \alpha_{t|s}^2 + \sigma_{t|s}^2 = 1 \\ e^{\lambda_t} = {\alpha_t^2 \over \sigma_t^2} $$ .

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There is another derivation: $$ {\begin{aligned} \sigma_{t|s}^2 &:= \sigma_t^2 - {\alpha_{t}^2 \over \alpha_s^2} \sigma_s^2 \\ &= \left( 1 - {\alpha_t^2 \sigma_s^2\over \sigma_t^2\alpha_s^2} \right) \sigma_t^2\\ &= \left(1 -e^{\lambda_t}\cdot {1 \over e^{\lambda_s}}\right) \sigma_t^2 \\ &= \left(1 - e^{\lambda_t - \lambda_s}\right) \sigma_t^2 \\ \end{aligned}} $$

given $$ e^{\lambda_t} = {\alpha_t^2 \over \sigma_t^2} $$ .

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