In this slide on diffusion models, it reads

$q(x_t)=\int q(x_0)q(x_t |x_0)dx_0=\int q(x_0)q(x_t|x_0)dx_0$ The diffusion kernel is Gaussian convolution.

I but don't understand why it is Gaussian convolution.

Convolution, by its definition, would be of following form: $$f_Z(z) = \int_{-\infty}^{\infty}f_X(t)f_Y(z-t)dt $$

In the original paper: Deep Unsupervised Learning using Nonequilibrium Thermodynamics, the author states in 2.1:

The data distribution is gradually converted into a well behaved (analytically tractable) distribution $\pi(y) $ by repeated application of a Markov diffusion kernel $T_{\pi}(y|y';\beta)$ for $\pi(y)$.

$$ \pi(y) = \int dy' T_\pi (y|y';\beta)\pi(y')$$

... Table App.1 gives the diffusion kernels for both Gaussian...

In Table App.1, we know the kernel is: $$\mathcal{N}(x^{(t)};x^{(t-1)}\sqrt{(1-\beta_t)},I\beta_t)$$

Referring back to convolution definition with variable $z$ and $t$, the kernel $T_\pi (y|y')$ is of form: $${\displaystyle {\frac {1}{\sigma }}\varphi \left({\frac {y- y'\sqrt{(1-\beta_t)}}{\sigma }}\right)}$$

This gives the coefficient $\sqrt{(1-\beta_t)}$ to $y'$. But, In the definition, in contrast of $f_Y(z-t)$, $t$ has no coefficient (except for a $-1$).

Is this still a Gaussian convolution?


1 Answer 1


You can convert this to the standard convolution form by substitution; set $u = \sqrt{1-\beta_t}y'$ then you get $$\int \frac{1}{\sigma \sqrt{1-\beta_t}} \phi \left( \frac{y - u}{\sigma} \right) \pi'(u) du$$ where $$ \pi'(u) = \pi\left(\frac{u}{\sqrt{1-\beta_t}}\right) = \pi(y')$$


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