Correlation Theory for stationary Random process 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.
 A: 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.
