# Two Universality Properties Associated with the Monkey Model of Zipf's Law
## 2 Proof that \(-\beta\) Tends Towards \(-1\) from the Asymptotics of Log-Spacings
To understand how this behavior can emerge, we define spacings through a random division of the unit interval and then state the Shao-Hahn limit law. Let \(X_{1},X_{2},\ldots,X_{K-1}\) be a sample of \(K-1\) i.i.d. random variables drawn from a distribution on \([0,1]\) with a bounded density function \(h(x)\leq M\). [PERSON] and [PERSON] present the conditions for this limit in a more general way that reduces to this simpler statement when a density function exists. Write the order statistics of the sample as \(X_{1:K-1}\geq X_{2:K-1}\cdots\geq X_{K-1:K-1}\). The \(K\)_spacings_\(D_{i}\) are defined as the differences between the successive order statistics: \(D_{1}=1-X_{1:K-1}\), \(D_{i}=X_{i-1:K-1}-X_{i:K-1}\) for \(2\leq i\leq K-1\) and \(D_{K}=X_{K-1:K-1}\). We'll refer to this as a _generalized broken stick process_. By [PERSON] and [PERSON] [3] Corollary 3.6, we have  
\[\frac{1}{K}\sum_{i=1}^{K}\ln(KD_{i})\xrightarrow{a.s}-\int_{0}^{1}h(x)\ln h( x)dx-\lambda, \tag{4}\]  
as \(K\to\infty\) and where _a.s._ signifies _almost sure convergence_, \(\lambda=0.577\ldots\) is the Euler constant and \(-\int_{0}^{1}h(x)\ln h(x)dx\) is the differential entropy of \(h(x)\). Clearly,  
\[\frac{1}{K}\sum_{i=1}^{K}\ln(KD_{i})=\ln K+\frac{1}{K}\sum_{i=1}^{K}\ln D_{i}, \tag{5}\]  
so dividing through by \(\ln K\) gives  
\[\frac{\ln K}{\ln K}+\frac{1}{K}\frac{\sum_{i=1}^{K}\ln D_{i}}{\ln K}\] \[\xrightarrow{a.s}\frac{-\int_{0}^{1}h(x)\ln h(x)dx}{\ln K} - \frac{\lambda}{\ln K}, \tag{6}\]as \(K\to\infty\). The right side of the limit in (6) goes to \(0\) because \(-\int_{0}^{1}h(x)\ln h(x)dx/\ln K\) goes to \(0\) by the boundedness of the density \(h(x)\). Expressing logarithms with a radix \(=K\) then leads to the limit  
\[\frac{\sum_{i=1}^{K}\log_{K}D_{i}}{K}\xrightarrow{a.s}-1\;\;\text{as}\;K\to\infty. \tag{7}\]  
Our _universality property_ for \(\beta\) will now follow almost immediately from this. We use sample spacings to populate the \(K\) letter probabilities for the monkey keyboard. Since \(s\) is the probability of the space character, define \(q_{i}=(1-s)D_{i}\) (\(1\leq i\leq K\)) so that \(\sum_{i=1}^{K}q_{i}=1-s\). Let \(\overline{m}_{K}=\sum_{i=1}^{K}\log_{K}q_{i}/K\). Then from the limit (7), we have  
\[\overline{m}_{K} = \sum_{i=1}^{K}\frac{\log_{K}q_{i}}{K}\!=\sum_{i=1}^{K}\frac{\log _{K}((1-s)D_{i})}{K} \tag{8}\] \[= \log_{K}(1-s)+\frac{\sum_{i=1}^{K}\log_{K}D_{i}}{K}\xrightarrow{ a.s}-1\]  
as \(K\to\infty\). Observe that \(\log_{K}(1-s)=\ln(1-s)/\ln K\to 0\). Since \(\overline{m}_{K}\xrightarrow{a.s}-1\) from (8) and since \(-\beta<-1\), showing that \(\overline{m}_{K}\leq-\beta\) will prove the argument that \(-\beta\approx-1\) for finite but sufficiently large \(K\). To see that this is the case, note:  
\[q_{1}^{1/\beta}+q_{2}^{1/\beta}+\cdots+q_{K}^{1/\beta}=1, \tag{2}\]0 \[q_{1}^{1/\beta}+q_{2}^{1/\beta}+\cdots+q_{K}^{1/\beta}=1, \tag{2}\]1 (10) \[q_{1}^{1/\beta}+q_{2}^{1/\beta}+\cdots+q_{K}^{1/\beta}=1, \tag{2}\]2  
where the inequality in Equation (10) follows from the geometric-arithmetic mean inequality. Therefore, from  
\[q_{1}^{1/\beta}+q_{2}^{1/\beta}+\cdots+q_{K}^{1/\beta}=1, \tag{2}\]3  
it has now been established that \(\overline{m}_{K}\leq-\beta\). In the special case of [PERSON]'s model using all equal letter probabilities, \(\overline{m}_{K}=-\beta\). We want to complement this analytic result for \(-\beta\) using a natural and insightful approximation Prof. [PERSON] kindly brought to our attention after looking at a preprint of this article. Consider the function \(p(x)=\sum_{i=1}^{K}q_{i}^{1/x}\) for \(x>0\). Then \(p^{\prime}(x)=-\sum_{i=1}^{K}q^{1/x}\ln q_{i}/x^{2}\) and a first order approximation from a [PERSON] expansion about \(x=1\) gives \(p(x)\approx p(1)+(x-1)p^{\prime}(1)=1-s+(x-1)\)\((-\sum_{i=1}^{K}q_{i}\ln q_{i})\). Since \(\beta\) satisfies \(p(\beta)=1\), it follows that \(1=p(\beta)\approx 1-s+(\beta-1)(-\sum_{i=1}^{K}q_{i}\ln q_{i})\), giving \(\beta\approx 1+s/(-\sum_{i=1}^{K}q_{i}\ln q_{i})\).