This is a question related to statistics, but my major concern relates to the setup and evaluation of integrals. So I decided this question was better suited for Mathematics Exchange than CV.
I know the following...
$ E[S(x)]=\mu , \\ Var[S(x)]= \sigma^{2} , \\ \rho(h)= \mathrm{e}^{\left(-\tfrac{h}{\phi} \right)} , \\ R(x)=(2\theta)^{-1} \int\limits_{x-\theta}^{x+\theta} S(u) \, du , \\ \implies Cov \left[ S(u), S(v) \right] = \gamma(h) =\sigma^{2} \mathrm{exp} \left\{-\frac{|h|}{\phi} \right\} = \sigma^{2} \mathrm{exp} \left\{-\frac{| u-v|}{\phi} \right\}.$
I also know,
$E[R(x)]= (2\theta)^{-1} \mu \ (2\theta) = \mu.$
What I want to know is $Cov[R(x), R(y)]$.
Here is what I have so far...
$$ \begin{align} Cov \left[R(x), R(y) \right] &=Cov \left[(2\theta)^{-1} \int\limits_{x-\theta}^{x+\theta} S(u) \, du, \ (2\theta)^{-1} \int\limits_{y-\theta}^{y+\theta} S(v) \, dv \right] \\ &= (2\theta)^{-2} \ Cov \left[ \ \int\limits_{x-\theta}^{x+\theta} S(u) \, du, \ \int\limits_{y-\theta}^{y+\theta} S(v) \, dv \right] \\ &= (2\theta)^{-2} \int\limits_{x-\theta}^{x+\theta} \int\limits_{y-\theta}^{y+\theta} \ Cov \left[ S(u), S(v) \right] \ du \ dv \\ &= (2\theta)^{-2} \int\limits_{x-\theta}^{x+\theta} \int\limits_{y-\theta}^{y+\theta} \sigma^{2} \mathrm{exp} \left\{-\tfrac{| u-v|}{\phi} \right\} \ du \ dv \\ &= (2\theta)^{-2} \ \sigma^{2} \int\limits_{x-\theta}^{x+\theta} \int\limits_{y-\theta}^{y+\theta} \mathrm{exp} \left\{-\tfrac{| u-v|}{\phi} \right\} \ du \ dv \end{align} $$
Here, I am having trouble understanding how to properly set up this integral. I have been told this is what it should look like... $$ (2\theta)^{-2} \ \sigma^{2} \int\limits_{x-\theta}^{x+\theta} \int\limits_{v}^{y+\theta} \mathrm{exp} \left\{-\frac{(u-v)}{\phi} \right\} \ du \ dv + (2\theta)^{-2} \ \sigma^{2} \int\limits_{y-\theta}^{y+\theta} \int\limits_{u}^{x+\theta} \mathrm{exp} \left\{-\frac{(v-u)}{\phi} \right\} \ dv \ du $$
I have a few questions. What is the intuition regarding splitting the two integrals into two double integrals? How do you deiced on the bounds and order of the integration? Why is the order of $u$ and $v$ in the exponent changed? And how would I go about integrating the following...
$$\int\limits_{x-\theta}^{x+\theta} \int\limits_{v}^{y+\theta} \mathrm{exp} \left\{-\frac{(u-v)}{\phi} \right\} \ du \ dv$$
Thanks for the help!