# Why the forward process of a Diffusion Model is Gaussian Convolution?

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: $${\frac {1}{\sigma }}\varphi \left({\frac {y- y'\sqrt{(1-\beta_t)}}{\sigma }}\right)}$$

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?

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 $$\pi'(u) = \pi\left(\frac{u}{\sqrt{1-\beta_t}}\right) = \pi(y')$$