Variance of Time-Integrated Ornstein-Uhlenbeck Process

Question

I'm attempting to filter white noise from a deterministic, finite-power signal using a low-pass filter. This filter can be described using an exponentially-decaying response function:

$$ h(t) = \gamma \exp (-\gamma t) $$

Evaluating the Wiener integral for the filtered noise results in a zero-mean Ornstein-Uhlenbeck process, whose variance is:

$$\sigma^2_{OU} =\frac{\gamma}{2}\left( 1-\exp(-2\gamma t)\right)$$

The task now is to find the variance of the time-integrated process. My question is whether the following procedure, based on Wiener integrals, is correct, and if not, what is the correct procedure?

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First, we use the fact that Gaussian random variables are determined by their means and variances to express the zero-mean OU process as the product of a deterministic function and a Wiener process:

$$ \sigma^2_W = t \quad \therefore \quad \sigma^2_{F\times W} = F^2(t)\times t $$ $$ F(t) = \frac{\gamma}{2} \sqrt{\frac{1-\exp(-2\gamma t)}{t}} \quad \rightarrow \quad \sigma^2_{F\times W} = \sigma^2_{OU}$$

We perform integration by parts:

$$ \int_0^t W(\tau)F(\tau)d\tau = \left. \left[W(\tau)\int_{0}^{\tau}F(\tau ')d\tau ' \right] \right \vert_0^t - \int_0^t \left( \int_{0}^{\tau}F(\tau ')d\tau ' \right)dW(\tau)$$

Labelling $\int_{0}^{\tau}F(\tau ')d\tau '$ as $\tilde{F}(\tau)$, we write the variance of the integrated process as the sum of the variances of the two terms above. The variance of the left term is $ \tilde{F}^2(t)\times t$, since this is just the Wiener process, multiplied by a deterministic function. The term on the right is a Wiener integral, whose variance is $\int_0^t \tilde{F}^2(t) dt$. I'll leave $\tilde{F}(t)$ unspecified here, since the integral is difficult.

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So, have I made some error in deriving the variance?

Answer 1

An Ornstein-Uhlenbeck process is not the product of a Wiener process with a deterministic function. A Gaussian process $X(t)$ is not determined by its mean and variance for each $t$, but rather by its mean for each $t$ and by the covariance of $X(s)$ and $X(t)$ for each $s,t$.

Answer 2

(In what follows, I'll use autocorrelations instead of autocovariances, since the processes being discussed are zero-mean.)

When a random process is subjected to a linear, time-invariant system, the autocorrelation of the resulting process can be found using a convolution formula:

$$ R_{\textrm{out}}(t) = R_{\textrm{in}}(t) \ast r_{\textrm{sys}}(t) $$

Here, $r_{\textrm{sys}}(t) = h_{\textrm{sys}}(t) \ast h_{\textrm{sys}}(-t)$, where $h_{\textrm{sys}}(t)$ is the transfer function of the system, in the time domain.

Integration can be represented as such a system, as an integral is just a convolution with a unit box function:

$$\int_0^t X(\tau)d\tau = X(\tau) \ast B(\tau/t-1/2)$$

Here, the box function $B(x)$ is defined to be $0$ wherever $\vert x \vert > 1/2$, and $1$ elsewhere.

The variance of a process $X$, $\mathbb{E}(X^2(t))$ is $R_X(0)=\mathbb{E}(X(t)X(t+\tau))$ at $\tau=0$.

The Ornstein-Uhlenbeck process is exponentially correlated, $R_{OU}(t) = \exp(-\gamma \vert t \vert)$, convolution with $ \int_{-\infty}^{\infty}B(t'/t-1/2)B((t'+\tau)/t-1/2)dt'$ results in a variance of

$$ t + \dfrac{\exp(- \gamma t)-1}{ \gamma} $$

However, I don't know whether this is correct. Comments?