Given $x_t=(1-t)x_0+ty+t\epsilon_t\eta_t$, where $\eta_t\sim N(0,I)$, how can the analytic form of $p(x_{t-\delta}|x_t,x_0)$ be derived ?
Since what we have is $p(x_{t-\delta}|x_0,y)$ and $p(x_{t-\delta}|x_0,y)$, I tried to rewrite $p(x_{t-\delta}|x_t,x_0)$ as, \begin{align*} p(x_{t-\delta}|x_t,x_0) = \int_{y} q(x_{t-\delta}|x_t,x_0,y)*q(y|x_0,x_t) \,dy \end{align*} I don't know how to continue from here.
While I don't know how to solve the problem using the basics of probability, a so called reparameterization trick may help. Since $x_t=(1-t)x_0+ty+\epsilon_t\eta_t$, we have \begin{align*} y=\frac{1}{t}x_t-\frac{1-t}{t}x_0-\epsilon_t*\eta_t \end{align*} . Does this means $y\sim N(\frac{1}{t}x_t-\frac{1-t}{t}x_0,\epsilon_tI)$ ?
Since $x_{t-\delta}=(1-t-\delta)x_0+(t-\delta)y+(t-\delta)\epsilon_{t-\delta}\eta_{t-\delta}$, we may substitute the expression for y and get \begin{align*} x_{t-\delta}=(1-t-\delta)x_0+\frac{t-\delta}{t}x_t-\frac{t-\delta}{t}(1-t)x_0-(t-\delta)\epsilon_t\eta_t+(t-\delta)\epsilon_{t-\delta}\eta_{t-\delta} =\frac{\delta}{t}x_0+\frac{t-\delta}{t}x_t+(t-\delta)(\epsilon_t\eta_t+\epsilon_{t-\delta}\eta_{t-\delta}) \end{align*}, therefore we may conclude $$ p(x_{t-\delta}|x_t,x_0)=N(x_{t-\delta};\frac{\delta}{t}x_0+\frac{t-\delta}{t}x_t,(t-\delta)\sqrt{\epsilon_t^2+\epsilon_{t-\delta}^2}I) $$ This is indeed the solution given in the paper (equation 10) https://openreview.net/pdf?id=VmyFF5lL3F. Is the solution derived by using the reparameterization trick make sense? If so how can we achieve the same solution using the basics of probability ?
Thanks for your help in advance !