Noise Without Running the Chain

The forward process has nothing to learn. At step $i$ it keeps a fraction of what it was handed and mixes in fresh noise, following a fixed schedule $\beta_1, \beta_2, \ldots, \beta_T$. Step $i$ keeps $\alpha_i = 1 - \beta_i$, so the signal remaining after $t$ steps is the running product

$\bar{\alpha}_t = \alpha_1 \alpha_2 \cdots \alpha_t$

Training would be hopeless if every example meant running the chain $t$ times. It doesn't: one calculation lands you at any noise level you like.

$x_t = \sqrt{\bar{\alpha}_t}\; x_0 \; + \; \sqrt{1 - \bar{\alpha}_t}\; \varepsilon$

Here $\varepsilon$ is the noise you drew — and therefore the answer the denoiser will later be scored against.

Task: write noisy_at_step(x0, betas, noise, t), returning $x_t$ as a plain list, rounded to 4 decimal places.

• betas[0] is $\beta_1$, the amount of noise added by the first step.
• x0 and noise are lists of the same length; everything is elementwise.
• t runs from 0 up to len(betas). At t = 0 nothing has happened yet.

Notice that the two coefficients are square roots rather than the fractions themselves. That is what holds the mixture at a constant scale: $(\sqrt{\bar{\alpha}_t})^2 + (\sqrt{1 - \bar{\alpha}_t})^2 = 1$, so a barely touched image and an almost fully destroyed one arrive at the network with the same typical magnitude. One network handles every noise level partly because of this.