6
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.