I have read the following derivation in a book about correlation theory (Correlation theory of stationary and related random functions) and I need help understanding how the correlation function is derived.

The paper states that a random process can be established through convolving a function (typically smoothing kernel) $k(s)$ with another random process $x(s)$. Lets assume that $x(s)$ here is a Gaussian white noise process

$$y(s)=\int_{-\infty}^{\infty} \! k(u-s) x(u).du$$ Based on this the covariance is written as a function of $d=s-s'$ as follows $$Cov\left\{y(s),y(s')\right\}=E\left\{y(s)y(s') \right\}=$$ $$E\left\{\int_{-\infty}^{\infty} \! k(u-s) x(u)\,du\int_{-\infty}^{\infty} \! k(u'-s') x(u')\,du') \right\}= \int_{-\infty}^{\infty} \! k(u-d) k(u).du$$

The only conditions needed is that $x(s)$ is a continuous Gaussisan white noise process and that $k(s)$ is absolutely integrable.

I have been trying to understand how this equation is derived. Some points I was able to understand are

1) Integration here is possible since the continuous Gaussisan white noise is defined through a Dirac function $\delta(s)$ which is integrable

2) The stationarity assumption allows the covariance to be written in term of $d=s-s'$

I would really appreciate a detailed derivation to help me understand this approach for establishing random fields.


1 Answer 1


Using $$E\left\{x(u)x(u')\right\}=\delta(u-u') \qquad (1)$$ (white noise property), we obtain

\begin{equation} \begin{split} E\left\{\int_{-\infty}^{\infty} \! k(u-s) x(u)du\int_{-\infty}^{\infty} \! k(u'-s') x(u')du') \right\}&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}du\,du'k(u-s)k(u'-s') E\left\{x(u) x(u')\right\} \\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}du\,du'k(u-s)k(u'-s') \delta(u-u') \\ &=\int_{-\infty}^{\infty}du\,k(u-s)k(u-s') \\ &=\int_{-\infty}^{\infty}du\,k(u-s+s')k(u) \\ &=\int_{-\infty}^{\infty}du\,k(u-d)k(u). \\ \end{split} \end{equation}

In the first line, we used the linearity of $E$ to pull the integrals out. In the second line, we applied (1). In the third line, we evaluated the integral over $u'$, where the delta distribution tells us to replace $u'$ by $u$. In the fourth line we used the substitution $u\leftarrow u-s'$. Finally, in the last line we used the definition $d=s-s'$.

One comment: There is no such thing as a "continuous white-noise process". Your $x(s)$ is everywhere discontinuous. That's the reason that we prefer using stochastic calculus for these sorts of questions, where $x(s)ds$ would be written as the increment of a Wiener process $dW_s$. The formalism using $x(s)$ works to some extent, but it has severe limitations that become apparent once you start to dig deeper.


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