I think I have found the discrepancy.The short answer is that the distribution for case (2) is a doubly noncentral beta distribution. The bulk correlation is the same under each case.
In the solution below, I've used slightly different notation. In particular, $\boldsymbol{s }$ $\rightarrow$ $\boldsymbol{w}$. I've also used $\hat{A}$ to describe the maximum liklehood estimate for the amplitude of the sample vector $\boldsymbol{w}$ that maximally correlates with $\boldsymbol{x}$. This notation and formulation is targeted at detection theory applications.
The relative square error in approximating approximating a data stream $ \boldsymbol{x} $ using a waveform template $\boldsymbol{w }$ is the ratio of the least-squares error $\vert\vert \boldsymbol{e} \vert\vert^{2}$ to measure signal energy $ \vert \vert \boldsymbol{x} \vert \vert^{2}$. This error may be re-written as a ratio of quadratic forms that has a doubly noncentral $\beta$ distribution:
\begin{equation}
\begin{split}
\cfrac{\vert \vert \boldsymbol{e} \vert \vert^{2}}{\vert \vert \boldsymbol{x} \vert \vert^{2}} &= \cfrac{\vert \vert \boldsymbol{x} - \hat{A} \boldsymbol{w } \vert \vert^{2} }{ \vert \vert \boldsymbol{x} \vert \vert^{2}}
\\
&= \cfrac{\vert \vert \boldsymbol{x} - \cfrac{\langle \boldsymbol{x},\, \boldsymbol{w} \rangle}{ \vert \vert \boldsymbol{w } \vert \vert^{2}} \boldsymbol{w } \vert \vert^{2}} {\vert \vert \boldsymbol{x} \vert \vert^{2}}
\\
&= \cfrac{ \vert \vert P_{\boldsymbol{w}}^{\perp} \left( \boldsymbol{x} \right) \vert \vert ^{2}} { \vert \vert P_{\boldsymbol{w}}^{\perp}\left( \boldsymbol{x} \right) \vert \vert ^{2} + \vert \vert P_{\boldsymbol{w}} \left( \boldsymbol{x} \right) \vert \vert ^{2}}
\\
&\overset{d}{=} \cfrac{ \chi_{1}^{2}( \lambda^{\perp} )} { \chi_{1}^{2}( \lambda ) + \chi_{N_{E} - 1}^{2}( \lambda^{\perp} ) },
\end{split}
\end{equation}
where the noncentrality parameters are defined by $\lambda$ $=$ $\cfrac{\vert \vert P_{\boldsymbol{w}} \left( \boldsymbol{x} \right) \vert \vert ^{2}}{\sigma^{2}}$ and $\lambda^{\perp}$ $=$ $\cfrac{\vert \vert P_{\boldsymbol{w}}^{\perp} \left( \boldsymbol{x} \right) \vert \vert ^{2}}{\sigma^{2}}$, and where $\overset{d}{=}$ indicates distributional equality. This ratio is also related to the sample correlation coefficient $r$:
\begin{equation}
\begin{split}
\cfrac{\vert \vert \boldsymbol{x} - \cfrac{\langle \boldsymbol{x},\, \boldsymbol{w} \rangle}{ \vert \vert \boldsymbol{w } \vert \vert^{2}} \boldsymbol{w } \vert \vert^{2}} {\vert \vert \boldsymbol{x} \vert \vert^{2}}
&=
1- \cfrac{\langle \boldsymbol{x},\, \boldsymbol{w} \rangle^{2} }{ \vert \vert \boldsymbol{w } \vert \vert^{2} \vert \vert \boldsymbol{x } \vert \vert^{2}}
\\
&=
1 - r^{2}
\end{split}
\end{equation}
Therefore:
\begin{equation}
\begin{split}
r^{2} &\overset{d}{=} \cfrac{ \chi_{1}^{2}( \lambda )} { \chi_{1}^{2}( \lambda ) + \chi_{N_{E} - 1}^{2}( \lambda^{\perp} ) }
\\
&\sim \text{Beta} \left( \frac{1}{2}, \frac{1}{2}N_{E} ; \lambda, \lambda^{\perp} \right)
\end{split}
\end{equation}
Distinct hypotheses regarding the distribution for $\boldsymbol{x}$ simplify the form for this distribution. When the data stream contains only noise, the hypothesis $\mathcal{H}_{0}$ is satisfied and $r^{2}$ has a central Beta distribution, where $\lambda^{\perp}$ $=$ $\lambda$ $=$ $0$. In the presence of signal, a data stream $\boldsymbol{x}$ will generally have a non-zero projection $P_{\boldsymbol{w}}^{\perp}\left( \boldsymbol{x} \right)$ orthogonal to the noise-contaminated template data vector $\boldsymbol{w}$. In this case $\lambda$, $\lambda^{\perp}$ $\ne$ $0$, and $r^{2}$ has doubly noncentral Beta distribution. If the template signal has a very large SNR, then $ \boldsymbol{x}$ $\cong$ $A \boldsymbol{w}$ $+$ $\boldsymbol{n}$, $\lambda^{\perp}$ $=$ $0$, and $r^{2}$ is reasonably approximated by a noncentral Beta distribution. The noncentral Beta distribution therefore provides an absolute upper bound on the detection performance of a correlation detector.
$rank(\mathbf{R})$ equals to the number of independent random variables in $\mathbf{x}$. If $\mathbf{R}$ is full rank ($rank(\mathbf{R}) = n$), then it means that all components of $\mathbf{x}$ are linearly independent. If $rank(\mathbf{R}) = k \lt n$, that means there are only $k$ independent random variables in $\mathbf{x}$, the other $n-k$ random variables can be constructed by a linear combination of other components of $\mathbf{x}$.
Your equation (1b) doesn't lead to $rank(\mathbf{R}) = 1$. With certainly conditions (for example, $\mathbf{x}_i$ i.i.d normal), your equation (2) should approach $\mathbf{R}$, and $rank(\mathbf{R_{xx}})$ approaches $rank(\mathbf{R})$.
Best Answer
Assuming the numbers are random, the correlation between pairs at any distance should be independent of the distance. That is not surprising. When you convert the sum to the integral, you should say $$\frac{1}{N}\sum_{i=1}^{N}r_ir_{i+k} \approx \int_{2.5}^{7.5}x\ dx \int_{2.5}^{7.5}y\ dyP\left(x,y\right)xy $$ The way you did it the integral doesn't change if you change the range to $(100,105)$ but the product of two elements has to get much larger.