Consider the 1st order filter $\frac{1}{1+\tau s}$ with $s$ the Laplace operator.
Say $B$ is a white noise process with zero mean and constant variance $Q_B$.
Now let's define $X(t)$ by
- for $t=0$, $X(0)=X_0$
- for $t \gt 0$, $\frac{X}{B}=\frac{1}{1+\tau s}$
I'm trying to evaluate the variance of $\dot X(t)$. Ideally for any $t$, but at least when ${t\to \infty}$.
My reasoning was that I could simply write $$\dot X=\frac{B-X}{\tau}$$
Then :
- if $B(t)$ and $X(t)$ are independant
- using the outside knowledge that $\lim_{t\to \infty} var(X(t))=\frac{Q_B}{2\tau}$
I simply deduced
$$\lim_{t\to \infty}var\left(\dot X(t)\right)=\frac{Q_B}{\tau^2}\left(1+\frac{1}{2\tau}\right)$$
My questions are :
is this right ? In particular, is it really true that $B(t)$ and $X(t)$ are independant ? Simply explicitly calculating $cov\left(B(t),X(t)\right)$ (by replacing $X(t)$ with its integral solution) wants me to say yes, but I tend to get scared with white noise processes as a lot of things you're usually allowed to do are not valid anymore.
How can I most cleanly evaluate $var(\dot X(t))$ for other, non-$\infty$ values of $t$ ? I'm still quite shaky on Ito calculus etc.
Additionally,
I have tried studying $$Y(t,L)=\frac{X(t+L)-X(t)}{L}={\left(e^{-L/\tau}-1\right)X(t)+e^{-L/\tau}\int_{t}^{t+L} e^{(u-t)/\tau}B(u)du \over L}$$
The variance of which is $$var\left(Y(t,L)\right)={\left(e^{-L/\tau}-1\right)^2var(X(t))+\left(1-e^{-2L/\tau}\right)\frac{Q_B\tau}{2} \over L^2}$$
Then at least for $t\to \infty$, I replace $var(X(t))$ by $\frac{Q_B}{2\tau}$ and get $$var\left(Y_{\infty}(L)\right)=\left[\left(e^{-L/\tau}-1\right)^2\tau^{-1}+\left(1-e^{-2L/\tau}\right)\tau\right]\frac{Q_B}{2L^2}$$
"Problem" : when $L \to 0$, we get $$var\left(Y_{\infty}(L)\right)\sim\frac{Q_B}{L}$$ which is not at all what was obtained while studying $\dot X(t)$.
Am I wrong to expect $var(Y(t,L))$ to converge to $var(\dot X(t))$ when $L\to0$ ? Or have I made mistakes ?
Ok, I think I'm now more clear on a number of things and can actually start answering (bits of) the question myself.
The first result, that is $$\lim_{t\to \infty}var\left(\dot X(t)\right)=\frac{Q_B}{\tau^2}\left(1+\frac{1}{2\tau}\right)$$ is in fact, wrong, because $var(B(t))=Q_B\delta_t$ ($\delta_t$ being a Dirac) and not simply $Q_B$.
Additionally, $B(t)$ and $X(t)$ are not independant : since $$X(t)=e^{-t/\tau}\left(X_0+\int_0^t e^{u/\tau}B(u)du\right)$$ we have $$cov(B(t),X(t))=cov\left(B(t),e^{-t/\tau}\int_0^te^{+u/\tau}B(u)du\right)=e^{-2t/\tau}\int_0^tcov\left(B(t),e^{+u/\tau}B(u)\right)du=e^{-2t/\tau}\int_0^te^{+2u/\tau}Q_B\delta_tdu=Q_B$$
In the end, this first calculus would return $$\lim_{t\to \infty}var\left(\dot X(t)\right)=\frac{Q_B}{\tau^2}\left(\delta_t+\frac{1}{2\tau}-1\right)$$ which is infinite/undefined due to the Dirac.
This is to be expected since the power of white noise is infinite/undefined, and $\dot X(t)$ is in fact simply scaled white noise plus a finite variable ($X(t)$).
Trying to evaluate $var(\dot X(t))$ for finite $t$ wouldn't make any more sense (it would be infinite/undefined as well).
Now, let's see what we can do about $Y(t)$.
Since my understanding of the problem has grown to a point that I've been able to correct some important mistakes + I realize my original question was probably poorly worded (I'd better use $dW_t$ instead of $B(t)dt$, I think I'll just create another cleaner, hopefully more rigorous question. I hope this is ok.