How is the variance for a diffusion kernel derived for a diffusion model? So I'm watching this video tutorial from CVPR this year on diffusion models, and I am confused by the variance term in the distribution on the left on the video. I understand that in the forward process, we can track intermediate distributions
$$q(\mathbf{x}_t|\mathbf{x}_{t-1}) = N(\mathbf{x}_t;\sqrt{1-\beta_t}\mathbf{x}_{t-1},\beta_t\mathbf{I})$$
And that the joint distributions of all the intermediate steps conditioned on the input is given by
$$q(\mathbf{x}_{1:T}|\mathbf{x}_{0}) = \prod_{t=1}^Tq(\mathbf{x}_t|\mathbf{x}_{t-1}).$$
If we define $\bar{\alpha} = \prod_{s = 1}^t(1 - \beta_t),$ then we are supposed to derive the diffusion kernel
$$q(\mathbf{x}_t|\mathbf{x}_{0}) = N(\mathbf{x}_t;\sqrt{\bar{\alpha}}\mathbf{x}_0,(1 - \bar{\alpha}_t)\mathbf{I})$$
I can definitely see why the mean is what it is, but I'm having a hard time seeing where we get the variance from. How is the variance of $(1 - \bar{\alpha}_t)\mathbf{I}$ derived?

I struggled for hours on this question too until I finally found useful resource here: https://lilianweng.github.io/posts/2021-07-11-diffusion-models/
The key is to use re-parametrization trick and combine the sum of the noise as a new gaussian distribution.
Hope this helps!
