Gaussian 1D

$p(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2} \right)$

Gaussian MD

$p(\mathbf{x}) = \frac{1}{\sqrt{(2\pi)^k |\boldsymbol{\Sigma}|}} \exp\left( -\frac{1}{2} (\mathbf{x}-\boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} (\mathbf{x}-\boldsymbol{\mu}) \right)$

$q(x_0,x_{t-1}, x_t) = q(x_t|x_{t-1},x_0)q(x_{t-1}|x_0)q(x_0)~~~markov~property~:q(x_t|x_{t-1},x_0)=q(x_t|x_{t-1})$

$q(x_0,x_{t-1}, x_t) = q(x_{t-1}|x_{t},x_0)q(x_{t}|x_0)q(x_0)$

$q(x_t|x_{t-1},x_0)q(x_{t-1}|x_0)q(x_0)=q(x_{t-1}|x_{t},x_0)q(x_{t}|x_0)q(x_0)$

$q(x_{t-1}|x_t,x_0)=\frac{q(x_t|x_{t-1})q(x_{t-1}|x_0)}{q(x_t|x_0)}$

Let’s find out the distrbution what $q(x_{t-1}|x_t,x_0)$ is following

$q(x_t|x_{t-1})$

which is follow $\mathcal{N}(x_t;\sqrt{\alpha_t}x_{t-1},\beta_tI)$

Exponential term

$\exp[-\frac{(x_t-\sqrt{\alpha_t}x_{t-1})^2}{2\beta_t}]$

$q(x_{t-1}|x_0)$

which is follow $\mathcal{N}(x_{t-1};\sqrt{\bar\alpha_{t-1}}x_{0},(1-\sqrt{\bar\alpha_{t-1}})I)$

Exponential term

$\exp[-\frac{(x_{t-1}-\sqrt{\bar\alpha_{t-1}}x_{0})^2}{2(1-\bar\alpha_{t-1})}]$

$q(x_t|x_0)$