# How to derive the variance of the forward step of Variational Diffusion Models in terms of the log signal-to-noise ratio $\lambda_t$?

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.

• More papers are using the same expression.
• I think the definition of $$\sigma_{t|s}$$ Eq. (22) in the original VDM paper is straightforward with respect to DDPM's definitions.
• The expression seems to be quite similar to Eq.(32) of VDM.

{\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 $$\alpha_t^2 + \sigma_t^2 = 1 \\ \alpha_{t|s}^2 + \sigma_{t|s}^2 = 1 \\ e^{\lambda_t} = {\alpha_t^2 \over \sigma_t^2}$$ .
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 $$e^{\lambda_t} = {\alpha_t^2 \over \sigma_t^2}$$ .