Purpose of scaling mean by $\sqrt{1 - \beta_t}$ in forward diffusion process In the forward diffusion process described by Ho, et al. the probability distribution for the next step is:
$$q(\mathbf{x}_t|\mathbf{x}_{t-1}) = N(\mathbf{x}_t;\sqrt{1-\beta_t}\mathbf{x}_{t-1},\beta_t\mathbf{I})$$
What is the purpose of scaling the mean of the distribution by $\sqrt{1 - \beta_t}$? I think this has something to do with maintaining the same variance across steps, but don't understand why keeping variance fixed would be helpful.
Ho, J., Jain, A., Abbeel, P. (2020). Denoising Diffusion Probabilistic Models. Proceedings of the 34th Conference on Neural Information Processing Systems. NeurIPS 2020, Vancouver, CA.

The scaling factor is needed just to avoid making the variance "explode".
Intuition. In a forward diffusion process you start with some vector $\mathbf{x}$ and add a bit of noise at each step $t$. If you do not scale $\mathbf{x}$ before adding the noise, all the noise will add up to increase the overall variance. This is undesired, because the range of magnitude of the vectors you get at the end will depend on the number $T$ of steps you perform.
If you start with just a number $x_0 = 0$, after $T$ steps you will have a value $x_T \in [-T, T]$.
Proof. To solve that, at step $t$, we need to "scale down a bit" $\mathbf{x_{t-1}}$ before adding the noise.
For simplicity, consider $x_0 \sim \mathcal{N}(\mu, 1)$, so we can perform exact computations (note that the input is normalized to unit variance). We would like to keep the variance at $1$ at every step $t$.
So, for the first step $t=1$, we scale $x_0$ by a factor $a$, and then add $\epsilon_1 \sim \mathcal{N}(0, \beta_1)$:
$$ x_1 = a\, x_0 + \sqrt{\beta_1} \epsilon_1 $$
The variance of $x_1$ is given by:
$$ \mathrm{Var}(x_1) = a^2 + \beta_1 $$
If we force $ \mathrm{Var}(x_1) = 1 $ we can solve for the required scaling factor $a$ to keep the variance unchanged:
$$ a^2 + \beta_1 = 1 \Rightarrow a = \sqrt{1 - \beta_1} $$
This of course works at every step $t$, not just the first one. So we get:
$$ \mathbf{x_t} = \sqrt{1-\beta_t} \mathbf{x_{t-1}} + \sqrt{\beta_t} \mathbf{\epsilon_{t-1}} $$
which corresponds to the probability distribution you started from:
$$ q(\mathbf{x_t} | \mathbf{x_{t-1}} ) = \mathcal{N}(\mathbf{x_t}; \sqrt{1-\beta_t} \mathbf{x_{t-1}}, \beta_t \mathbb{I} ) $$
