Asymptotic Properties of the KF

So we can say that:

NOTE: The $ii)$ condition can be relaxed and just technical, it grantees that there is a unique solution to the ARE (Algebraic Riccati Equation), you want to avoid non-reachable eigenvalues values on the unit circumference. While the $i)$ condition is unavoidable, and it says that the non-observable eigenvalues have to be stable $|{\lambda }_i|<1$, if you cannot measure something it has to go to zero.

1. $lim$ of the error $\tilde{x}$ : Both limits go to $0$, no matter the initial value $\tilde{x}(0|-1)$:

Solution of the ARE (Algebraic Riccati Equation), in a specific linear case:

• $A\to a$
• $C\to 1$
• $Q\to 0$
• $R\to {\sigma }_v^2$

Why this is true, will be shown later.

3. $li{m}_{t\to \infty }\bar{A}(t)$ Where, $\bar{A}(t)$ was defined in the “KF as a dynamic system”: Taking the results from the previous point (2.) of $P_{\infty }$

4. Let’s take a simple version of the DRE (Discrete Riccati Equation): We study the curve: So we obtain something like this: Let’s take a random $p(0)$: To compute $p(1)$ we simply do: So we take project $p(1)$ on the $x$ axis, using the $45°$ line $p(t+1)=p(t)$ and calculate $p(2)$: As we iterate, as we can see we will have that $p(t\to \infty )=0$ So we have shown graphically that that is true $\forall p(0)>0$

Let’s see what happens for $|a|>1$, the starting graph looks something like: The point where the two curves intersect is: So now doing like we did before we will obtain the limit: This is how we can find the two different solution of the DRE for $|a|<1$ and $|a|>1$.

NOTE: $p(0)=0$ is the only exception, for $p(0)=0$, $p(\infty )=0$ and not ${\sigma }_v^2\left(a^2-1\right)$

Let’s now see the case with the disturbance processes $w(t)$ and $v(t)$ : The theorem holds, so the ARE has an unique solution, as we can see solving the equation: Where:

• The positive solution will be our man, $p_{\infty }$
• The negative solution, I don’t care, the variance is always positive. Even if $p(0)=0$ the solution will always be $p(\infty )$.