Recently, I learn basic Kalman Filter knowlege via https://www.kalmanfilter.net/covextrap.html . And I am confused with the meaning of process noise and process noise covariance matrix.
For example, there is a constant velocity model, p denotes as displacement, and $\mathbf{x} = \begin{bmatrix}
p\\
\dot{p}
\end{bmatrix}$
Then,
$$\mathbf{\dot{x}} = \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}\mathbf{x}$$
Denote $ \mathbf{A} = \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix},\ \mathbf{F} = e^{\Delta tA}$ . Then $\mathbf{x}(t+\Delta t) = \mathbf{F}\mathbf{x}(t)$. Then $\mathbf{x}_{n+1} = \mathbf{F}\mathbf{x}_n$
I think the article suggest me to write $\mathbf{\hat{x}}_{n+1,n} = \mathbf{F}\mathbf{\hat{x}}_{n,n}+\mathbf{w}_n$. My explain is there exists some error when we do estimation, like initial measurement of the target equaling true value is impossible...etc. So, we must plus $\mathbf{w}_n$ to correct our estimate algorithm. My question is, $\mathbf{w}_n$ is an unknown for me. Can anyone give a mathmatical definition of $\mathbf{w}_n$ for me and explain why？I guess the ideal definition for $\mathbf{w}_n$ is $\mathbf{x}_{n+1}$ (the true value at time stamp $n+1$) subtracts $\mathbf{F}\mathbf{\hat{x}}_{n,n}$. However, it seems this is wrong.
I read and think that $\mathbf{E}[\mathbf{w}_n\mathbf{w}_n^\intercal]$ is the process noise covariance matrix. So, I want to understand what $\mathbf{w}_n$ exactly is first.
My another question is even I accept the process noise covariance matrix in the above example is $$ \mathbf{Q} = \begin{bmatrix} V(p) & COV(p,\dot{p}) \\ COV(\dot{p},p) & V(\dot{p})\end{bmatrix}$$ I can't understand the matrix elements explained in this article. For example, we have already known this is a constant velocity system. Then $V(\dot{p})$ should be zero. On the other hand, I guess the author means there is some small fluctuation in the system such that $V(\dot{p})$ is not zero, but in this case, how to compute $\mathbf{Q}$？
The noise term $w_k$ comes from the continuous-time process noise $w(t)$ which acts on the dynamics as
$$\dot{x}(t)=Ax(t)+w(t)$$
which, after discretization, yields
$$x_{k+1}=\exp(A\Delta T)x_k+\int_0^{\Delta T}\exp(A(\Delta T-s))w(t_k+s)ds.$$
So, we get that $F:=\exp(A\Delta T)$ and that $w_k:=\int_0^{\Delta T}\exp(A(\Delta T-s))w(t_k+s)ds$. There are ways to approximate this latter term into a form that is simpler to deal with; e.g. for covariance calculations. One of them is to consider that $w(t_k+s)$ is a constant random variable over the interval $[0,\Delta T]$ which has zero mean and covariance equal to $I$. As a result, $w_k$ will have zero mean and a covariance equal to
$$\left(\int_0^{\Delta T}\exp(As)ds\right)\left(\int_0^{\Delta T}\exp(As)ds\right)^T.$$
Note also that when $A$ is invertible, then $\int_0^{\Delta T}\exp(As)ds=\dfrac{A^{-1}(\exp(A\Delta T)-I)}{\Delta T}$.
Regarding your second question, your process will be noisy and has a result, the velocity will not be constant anymore but will fluctuate, giving then rise to a mon-zero variance.