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In papers and other material regarding diffusion models the forward diffusion process is defined by adding a small amount of Gaussian noise to an image $x_0$ for $T$ time steps. In each time step the noise has a variance of $\beta_t$. This process produces a sequence of noisy samples: $x_1, x_2, x_3... x_T$ such that: $q(x_t|x_{t-1}) = N(x_t; \sqrt{1-\beta_t} x_{t-1}, \beta_tI)$

I don't understand why this is $q(x_t|x_{t-1})$ distribution. When adding a constant $c$ to a normal random variable with mean $\mu$ and variance $\sigma^2$ we get a new random variable with the same variance and a mean of $c+\mu$. Therefore, I expect $q(x_t|x_{t-1})$ to be: $q(x_t|x_{t-1}) = x_{t-1} + \epsilon_t = N(x_t; x_{t-1}, \beta_t I)$ where $\epsilon_t=N(\epsilon_t; 0, \beta_t I)$

Any help will be appreciated.

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4 Answers 4

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The relationship between $x_t$ and $x_{t-1}$ is as follows: $$ x_t = \sqrt{1-\beta_t}x_{t-1}+\sqrt{\beta_t}\epsilon_t,\quad \epsilon_t\sim\mathcal{N}(0,I). $$ Not only is a small amount of noise added, the original image is also scaled down slightly.

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The image data $\mathbf{x}_{t-1}$ is not a constant $\mathbf{c}$. It's itself a distribution. Different permutations of pixels have different probabilities.

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  • $\begingroup$ Can you please expand this answer? How is the sqrt(1-beta_t) derived? $\endgroup$
    – PascalIv
    Commented Jan 4, 2023 at 14:08
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The purpose of $q(x_t|x_{t-1})$ is: given that the random variable $x_{t-1}$ is sampled to be a specified value, what is the probability density function for $x_t$? So we're using the sampled value of $x_{t-1}$ to calculate the mean of the probability density function, which is $\sqrt{1-\beta_t} x_{t-1}$. We're not adding $x_{t-1}$ and $\epsilon_t$. Instead, we are steering the mean towards zero a little bit each time step (because $\sqrt{1-\beta_t}<1$), while adding a little noise.

To implement this, we execute the sampling of $x_t$ by taking the mean (the constant $\sqrt{1-\beta_t} x_{t-1}$) and adding a sample from the standard normal distribution scaled by $\sqrt{\beta_t}$.

Of course, the probability chain rule then provides us the desired (nonconditional) pdf for $x_T$ at the final time.

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Maybe this video will help with the understanding.

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  • $\begingroup$ Hi @sumit dugar and welcome to AI Stack Exchange! If possible, please include some details of the video in your answer. Thank you for posting, and we hope to see you again soon on this site! $\endgroup$
    – DeepQZero
    Commented Jan 2 at 15:17

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