Having this properties:
$E[x(j)w^T(k)]=0$; $j,k\geq 0, j\leq k$.
$E[z(j)w^T(k)]=0$; $j,k\geq 0, j\leq k$.
Demonstrate it with the following assumptions:
- $x(k+1)=\Phi (k+1,k)x(k)+\Gamma (k+1,k)w(k)~$, $k\geq 0$; $x(0)=x_0~$ (is State equation)
- $z(k)=H(k)x(k)+v(k), k\geq0$ (observation equation)
- Initial state $x_0$, vector n-dimensional gaussian with cov. matrix $E[x_0x^T_o]=P_0$
The process $\{w(k);k\geq 0\}$ is a white noise succession, centred, with cov. matrix $E[w(k)w^T(k)]=Q(k), k\geq0$
- The process $\{v(k);k\geq 0\}$ is a white noise succession, centred, with cov. matrix $E[v(k)v^T(k)]=R(k), k\geq0$
The initial state and additive noises are mutually independent.
I think it has to be done with Doob theorem but in my case is not working. Thanks in advance,
$x_j$ is a linear combination of $x_0$ and $w_0,...,w_{j-1}$.
Per last assumptions, these are all independent of of the vectors $w_k$, $k\ge j$.
Therefore, the claim.