Why does diffusion add Gaussian noise instead of any random corruption?

Diffusion adds Gaussian noise because Gaussian errors combine exactly. Each forward step is a linear shrink plus independent Gaussian error, so the noisy picture at any later step has a closed-form Gaussian distribution with a known mean and variance, and the reverse step becomes a Gaussian conditioning problem. That is the property any other corruption loses.

Write $\alpha_t=1-\beta_t$. If the values at step $t-1$ have variance 1, the next step has variance $(1-\beta_t)\times1+\beta_t\times1=1$. Take $\beta_t=0.02$, so $\alpha_t=0.98$. The variance after one step is $0.98\times1+0.02\times1=1.00$. The shrink is doing exactly enough to cancel the fresh noise. Drop it and use $x_t=x_{t-1}+\sqrt{0.02}\,\varepsilon$: after 100 steps the variance is $1+100\times0.02=3.00$. The path would then end somewhere that depends on how long it was run, and a sampler would not know which spread to start from. The Gaussian error is what lets this variance arithmetic be exact: multiplying by a constant scales variance by the square of that constant, and adding independent noise adds its variance.

Now apply the rule twice with the same $\alpha=0.98$. The coefficient on $x_0$ is $\sqrt{0.98\times0.98}=0.98$. The first step contributes noise variance $0.98\times0.02=0.0196$ to the second step, the second contributes $0.02$, and the two independent Gaussian errors add to one Gaussian error with variance $0.0196+0.02=0.0396$. That is exactly $1-0.98^2$, so two steps have the same form as one step with a combined $\alpha$. For a schedule, the product $\bar\alpha_t=\prod_{s=1}^{t}\alpha_s$ does the same job, and the whole chain collapses to one formula.

The natural guess is that any corruption would do: add uniform noise, or flip a random pixel, and train a network to undo it. That loses the closed form. Add independent uniform noise instead. After one step the values have a flat-topped distribution; after two steps the sum of two uniforms is triangular; after many steps it only approaches a bell shape. The variance still adds, but the distribution at step $t$ is not given by the one-line Gaussian formula, so producing a training pair at an arbitrary $t$ means running the chain rather than sampling $x_t$ directly from $x_0$. Flipping pixels is worse: it is not additive at all, so there is no variance recursion and no fixed point for the spread.

Because $x_t$ is $x_{t-1}$ multiplied by a number plus independent Gaussian error, the pair $(x_{t-1},x_t)$ is jointly Gaussian once $x_0$ is fixed. Conditioning one Gaussian on another gives a Gaussian. So $q(x_{t-1}\mid x_t,x_0)$ has a closed-form mean and a variance built from $\beta_t$ and the accumulated $\alpha$. A network only has to predict the missing $x_0$ or the noise $\varepsilon$; the rest of the reverse step is arithmetic fixed by the schedule. With uniform noise or pixel flips the reverse conditional has no such summary: the network would have to learn the whole shape of an arbitrary conditional distribution, not one mean and one known variance.

The exact composition survives any schedule, but the reverse transition $q(x_{t-1}\mid x_t)$ is a mixture over possible $x_0$ values and is not itself exactly Gaussian. The Gaussian formula is exact for $q(x_{t-1}\mid x_t,x_0)$, and sampling uses a network to approximate the mixture. The approximation is best when each $\beta_t$ is small, because each reverse step then stays close to a single Gaussian. If $\beta_t$ is large, one reverse step has to undo a lot and the local Gaussian picture is poor. If the noise is correlated across pixels, the algebra still works with covariance matrices, but the isotropic formula and the simple variance fixed point do not. If the noise has infinite variance, the fixed point argument fails altogether.