I am currently trying to solve an exercise that involves estimating the position $\chi_t$ and and velocity $\dot\chi_t$ of a truck at time $t$. The truck moves on rails and is buffeted around by a random force $\ddot\chi_t$ with probability density function $P(\ddot\chi_t) = \mathcal{N}(\mu, \sigma^2)$. Time is discrete.
So I have already determined the following differential equation that describes the position and velocity of the truck relative to the last time step:
\begin{equation} x_t = \begin{bmatrix} \chi_t \\ \dot\chi_{t} \end{bmatrix} = \begin{bmatrix}1 & \Delta t \\ 0 & 1\end{bmatrix} \begin{bmatrix}\chi_{t-1} \\ \dot\chi_{t-1}\end{bmatrix} + \begin{bmatrix}\frac{\Delta t^2}{2} \\ \Delta t\end{bmatrix} \ddot\chi_{t-1} \end{equation}
Now, I would like to compute the probability $P(x_t | x_{t-1})$. According to the solution, it is:
\begin{equation} P(x_t|x_{t-1}) = \mathcal{N}(A_t x_{t-1}, R_t) \\ = \text{det}(2\pi R_t)^{-\frac{1}{2}} \text{ exp}\{-\frac{1}{2}(x_t - A_t x_{t-1})^T R_t^{-1} (x_t - A_t x_{t-1})\} \\ \text{ with } A_t = \begin{bmatrix}1 & \Delta t \\ 0 & 1\end{bmatrix} \text{ and } R_t = \begin{bmatrix} \frac{\Delta t^4}{4} & \frac{\Delta t^2}{2} \\ \frac{\Delta t^4}{4} & 1\end{bmatrix} \end{equation}
Could you please explain how to get from the first formula to the second one? I am specifically interested in how to calculate $R_t$.
Thank you for every answer!
note that, given to conditional probability:
$$ X_t = Ax_{t-1} + N $$
i think if we know the $x_{t-1}$ , then $$f_{{X_t}|{X_{t-1}}}(x_t|x_{t-1})$$ will be two dimensional Gaussian distributed. I am very busy currently, if it does not help you comment me.