Edit friendly DDPM noise space

Question

I was reading this paper, "An Edit Friendly DDPM Noise Space: Inversion and Manipulations". In page no. 4, they have mentioned that in DDPM, noise maps of consecutive steps are highly correlated while their edit friendly noise maps of consecutive steps are independent.



How are the noise maps in DDPM highly correlated? I couldn't figure out the difference in their noise maps and in those of native DDPM. I understand that xt and xt-1 can have high or low variance depending on the value of alpha, but how will epsilon of consecutive steps have high or low variance when they are being randomly sampled from unit normal distribution.
Also, in abstract, they have mentioned complete opposite thing, "As opposed to the native DDPM noise space, the edit-friendly noise maps do not have a standard normal distribution and are not statistically independent across timesteps."

Am I missing something or is there some mistake in my understanding?

Answer 1

Equation 2 refers to creating $x_t$ from $x_0$. In the proposed method, each $x_t$ was generated from $x_0$ by sampling a random noise, i.e.:

$\\x_1 = \sqrt{\bar{\alpha_1}}\cdot x_0 + \sqrt(1-\bar{\alpha_1})\cdot \epsilon_1$

$x_2 = \sqrt(\bar{\alpha_2})\cdot x_0 + \sqrt(1-\bar{\alpha_2})\cdot \epsilon_2$

s.t. $\epsilon_1, \epsilon_2 \sim N(0,I)$, which means that they are independent.

However, when creating $x_t$ from $x_{t-1}$, by rearranging the equation to be how $x_t$ was created from $x_0$, you will get that these noises are dependent, i.e.:

$\\x_1 = \sqrt(1-\beta_1)\cdot x_0 + \sqrt(\beta_1)\cdot \epsilon_1$

$x_2 = \sqrt(1-\beta_2)\cdot x_1 + \sqrt(\beta_2)\cdot \epsilon_2$

s.t. $\epsilon_1, \epsilon_2 \sim N(0,I)$

Rearranging:

$\\x_1 = \sqrt(1-\beta_1)\cdot x_0 + \sqrt(\beta_1)\cdot \epsilon_1$

$\\x_2 = \sqrt(1-\beta_2)\cdot \sqrt(1-\beta_1)\cdot x_0 + \sqrt(1-\beta_1)*\sqrt(\beta_1)\cdot \epsilon_1+\sqrt(\beta_2)\cdot \epsilon_2 = A\cdot x_0 + B*\epsilon_3$

which shows that $\epsilon_1$ and $\epsilon_3$ are dependent.

Hope it is clearer now.

Answer 2

"Why is the noise added in eq.(2) in native DDPM statistically non-independent compared to eq. (6)?" My take on this is that in (6) you have to iteratively calculate x_t+1 until you get x_T and at every step sample an independent gaussian. In (2) you can sample a single independent gaussian and get x_T immediately. Thus, the noise is dependent across timesteps. The downside of (6) this is that the forward process is slow now. You can check alg. 1 in the ddpm paper where they sample epsilon and do the forward process for any T to arrive at xt. Meaning only one noise sample.

"Also, in abstract, they have mentioned complete opposite thing (...)" Here, I think they mean the noise variances z. In the native ddpm you sample z at every step during denoising (independent).