If we consider $x\_t$ an Ornstein-Uhlenbeck process (with $W\_t$ the Wiener process), does anyone know what would be the variance of the convolution of $x\_t$ with a given filter $A$ i.e. $V(x\_t \star A)$ ?


I've been trying to derive the solution starting from the analytical solution such as


\begin{align}
x\_t = {} & \mu\_x \int\_0^\infty A(p-t) (1-e^{-p/T}) \, dp \\ & + \left(\frac{2\sigma} T \right)^{1/2} \int\_0^\infty \left(A(p-t) \int\_0^p e^{-(p-s)/T} \, dW\_s \right) dp
\end{align}


and using properties of stochastic integrals and Itô isometry but I can't get the end of it...


So far, my "most successful" unsuccessful developments are the following. Since the integration of a stochastic process is a stochastic process such that


$$\int\_0^p e^{-(p-s)/T} \, dW\_s = X\_p = \mathcal{N} \left(0,\int\_0^p e^{-2(p-s)/T} \, ds \right) = \mathcal{N}\left(0,\frac{T}{2}(1-e^{-2p/T}) \right)$$


and that one can consider (with $W\_p$ a Wiener process)


$$X\_p = \left(\frac{\frac{T}{2}(1-e^{-2p/T})} p \right)^{1/2} \, W\_p $$


then


\begin{align}
x\_t = {} & \mu\_x \int\_0^\infty A(p-t) (1-e^{-p/T}) \, dp \\
& + \left( \frac{2\sigma}{T} \right)^{1/2} \int\_0^\infty A(p-t) \left(\frac{\frac{T}{2}(1-e^{-2p/T})} p \right)^{1/2} W\_p \, dp
\end{align}


and given that


$$ \int\_0^a g'(p) W\_p \, dp \sim \mathcal{N}\left(0, \int\_0^a \left(g(s)-g(p)\right)^2 \, ds\right)$$


one therefore gets with $a=\infty$


$$\sigma\_x^2 = \sigma^2 \int\_0^\infty \left( \int\_p^\infty A(u-t) \left(\frac{1-e^{-2u/T}}{u}\right)^{1/2} \, du \right)^2 \, dp$$


But I can't go anywhere from the last equation. Even taking $A$ as a dirac, it's not easy...any help would be greatly appreciated !


Thanks!



Variance of convolution between filter $A$ and Ornstein-Uhlenbeck process $x_t$
Welcome to MO! This is not quite a research level question, but here is the answer anyway.


Represent the Orenstein-Uhlenbeck process as white noise passing through a low-pass (this is really the representation in the equation for $x\_t$ you wrote). Call the transfer function of the low-pass $H$. Then your process is white noise passing through the filter $AH$. Now the variance you ask about is simply the square of the
$L^2$ norm of $AH$, i.e. $\int |AH(\omega)|^2 d\omega$.


