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?
