probability density $P_v(y)$, depending on ${{\mathbf L}}$, such that $$\lim_{N\to\infty}\frac{1}{N}{\#}\!\left\{n\in\{1,\ldots,N\} : \frac{1}{v^2}
Z'(k_n,{{\mathbf L}})<R\right\}=\int_{-\infty}^{R} P_v(y){{\mathrm d}}y,$$ with $P_v(y)=0$ for $y<0$.

\[thm:4\] For each $v$ let the bond lengths $L_j$, $j=1,\ldots,v$ lie in the range $[{\bar{L}},{\bar{L}}+\Delta L]$ and be linearly independent over ${\mathbb Q}$. If $v\Delta L\to0$ as $v\to\infty$ then for any $R\in{\mathbb R}$, $$\int_{-\infty}^R P_v(y){{\mathrm d}}y\to\int_{-\infty}^R P(y){{\mathrm d}}y$$ as $v\to\infty$. The limiting density is given by the continuous function $$P(y)=\left\{
\begin{array}{lr}
\displaystyle \frac{\sqrt{{\bar{L}}}}{4\pi y^{3/2}}\int_{-\infty}^{\infty}
\exp\left(-\frac{\xi^2}{4}-\frac{{\bar{L}}m(\xi)^2}{4y}\right)m(\xi)
{{\mathrm d}}\xi,
&\mbox{$y>0$}\\
0, & \mbox{$y\leq0$},
\end{array}
\right.$$ where $$m(\xi):=\frac{2}{\sqrt{\pi}}{{\mathrm e}}^{-\xi^2/4}+\xi{\mathop{\rm erf}}(\xi/2).$$

By comparison, the value distribution for the logarithm of the derivative of the characteristic polynomial of a matrix drawn from the CUE of random matrix theory, evaluated at an eigenvalue in the limit as matrix size tends to infinity, is Gaussian [@hug:rmt].

The following results refer to the value distribution of $A_i(n,{{\mathbf L}};v)$.

\[thm:5\] Assume the conditions of theorem \[thm:3\] are satisfied. Then there exists a probability density $Q_v(\eta)$ such that $$\lim_{N\to\infty}\frac{1}{N}{\#}\!\left\{ n\in\{1,\ldots,N\}: v^2A_i
(n,{{\mathbf L}};v)<R\right\}=\int_0^{R} Q_v(\eta){{\mathrm d}}\eta$$ where the density $Q_v(\eta)$ is independent of the choice of bond $i$ but depends on ${{\mathbf L}}$.

\[thm:6\] Assume the conditions of theorem \[thm:4\] are satisfied. Then for each $R>0$ $$\int_{0}^R Q_v(\eta){{\mathrm d}}\eta\to\int_0^R Q(\eta){{\mathrm d}}\eta.$$ 