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)

$$X_p = \left(\frac{\frac{T}{2}(1-e^{-2p/T})} p \right)^{1/2} \, W_p $$


\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 !



1 Answer 1


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. $\int |AH(\omega)|^2 d\omega$.

  • $\begingroup$ Thank you for your anwser ! And sorry, I couldn't get any anwser on mathstackexchange so I thought my question was maybe a complicated one...although it was not. Indeed, going spectral makes the reasoning much easier...Although I don't find it to be straight forward to derive the filter H...I'll think of it. Thanks ! $\endgroup$
    – Romain
    May 8, 2018 at 11:30
  • $\begingroup$ So H can be calculated using the link between the autocorrelation of $x_t$ and the spectral density. Here an even more general derivation math.stackexchange.com/questions/1647793/… $\endgroup$
    – Romain
    May 8, 2018 at 13:54

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