# Mutation Rate Model Used in the DNA VIEW Program
## 4 Hardy-Weinberg Equilibrium  
In this section, we consider how the above mutation models satisfy conditions I and II (Hardy-Weinberg hypothesis). **Theorem 1**. : _Let us consider the sequence of positive numbers \(\{\alpha_{i},1\leq i\leq N\}\) and the array of non-negative numbers \(\{\beta_{i,j},1\leq i,j\leq N\}\) such that_  
_(i)_ \(\beta_{i,j}>0,\ \beta_{i,i}\geq 0,1\leq i\
eq j\leq N\)_,_  
_(ii)_ \(\beta_{i,j}=\beta_{j,j},1\leq i,j\leq N\)_,_  
_(iii)_ \(\alpha_{i}=\sum_{i=1}^{N}\beta_{i,j}=\sum_{j=1}^{N}\beta_{j,j},1\leq i\leq N\)_Let us define a matrix_  
\[A_{N}=\left[ \begin{array}{ccccc}-\alpha_{1}+\beta_{1,1}&\beta_{1,2}&\beta_{1,3}&...&\beta_{1,N}\\ \beta_{2,1}&-\alpha_{2}+\beta_{2,2}&\beta_{2,3}&...&\beta_{2,N}\\ \beta_{3,1}&\beta_{3,2}&-\alpha_{3}+\beta_{3,3}&...&\beta_{3,N}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \beta_{N,1}&\beta_{N,2}&\beta_{N,3}&...&-\alpha_{N}+\beta_{N,N}\end{array}  \right], \tag{13}\]  
_then \(rank(A_{N})=N-1\)._  
Proof of Theorem 1.: Let \(A_{k},1\leq k\leq N\) denote the \(k\)-th main minor of the matrix \(A_{N}\). Assume contrary that \(det(A_{k})=0\) for some \(1\leq k<N\) (cf. Exercise 11 (e) p. 48 [25]). It means that there exist numbers \(\lambda_{1},\lambda_{2},...,\lambda_{k-1}\) such that  
\[\sum_{i=1}^{k-1}|\lambda_{i}|>0, \tag{14}\]  
and  
\[v_{k}=\sum_{i=1}^{k-1}\lambda_{i}v_{i}, \tag{15}\]  
where \(v_{i}\) denotes the \(i\)-th column of the matrix \(A_{k}\). It can be rewritten (after renaming rows and columns in the minor) precisely as  
\[\left\{ \begin{array}{lcl}\beta_{k,1}&=&(-\alpha_{1}+\beta_{1,1})\lambda_{1} +\beta_{1,2}\lambda_{2}+...+\beta_{1,k-1}\lambda_{k-1},\\ \beta_{k,2}&=&\beta_{2,1}\lambda_{1}+(-\alpha_{2}+\beta_{2,2})\lambda_{2}+...+ \beta_{2,k-1}\lambda_{k-1},\\ \vdots&\vdots\\ -\alpha_{k}+\beta_{k,k}&=&\beta_{k,1}\lambda_{1}+\beta_{k,2}\lambda_{2}+...+ \beta_{k,k-1}\lambda_{k-1}.\end{array} \right. \tag{16}\]  
Among the numbers \(\lambda_{1},\lambda_{2},\ldots,\lambda_{k-1}\) there are positive and non-positive ones. By \(I_{1}\) we denote a set of positive numbers. This set is nonempty. Indeed, if \(I_{1}=\emptyset\), then summing up both sides of (16), we get  
\[-\alpha_{k}+\sum_{i=1}^{k}\beta_{k,i}=\sum_{j=1}^{k-1}(-\alpha_{j}+\sum_{i=1} ^{k}\beta_{i,j})\lambda_{j}, \tag{17}\]  
and on the left-hand side we have a negative number, whereas on the right hand side it is non-negative. Let \(t\) be an index of the maximal \(\lambda_{i}\) (if there is more than one such index, we take the smallest one),  
\[t=\min\{i:\lambda_{i}=\max_{j}\lambda_{j}\}. \tag{18}\]  
Then, taking the \(t\)-th equation in (16), we get  
\[\beta_{k,t}=(-\alpha_{t}+\beta_{t,t})\lambda_{t}+\sum_{i<k,\ i\
eq t}\beta_{t,i}\lambda_{i}\leq(-\alpha_{t}+\sum_{i=1}^{k-1}\beta_{t,i})\lambda_{t} \tag{19}\]  
and we have a positive number on the left-hand side of the above inequality and a non-positive number on its right-hand side. Thus, this contradiction leads us to \(\det(A_{k})\
eq 0\), for every \(1\leq k<N\). Now we prove that \(\det(A_{N})=0\). It follows from assumption (iii) which can be rewritten as  
\[\left\{ \begin{array}{lcl}-\beta_{N,1}&=&-\alpha_{1}+\beta_{1,1}+\beta_{1,2} +...+\beta_{1,N-1},\\ -\beta_{N,2}&=&\beta_{2,1}+-\alpha_{2}+\beta_{2,2}+...+\beta_{2,N-1},\\ \vdots&\vdots\\ \alpha_{N}-\beta_{N,N}&=&\beta_{N,1}+\beta_{N,2}+...+\beta_{N,N-1}.\end{array} \right. \tag{20}\]  
**Corollary 1**. : _(\(p\). 62 in [26] or Definitions 7.1.2 and 7.1.4 and Theorem 7.1.3 in [27]) Let \(A_{N}\) be a matrix satisfying the assumptions of Theorem 1. Let the matrix \(A_{N}^{\prime}\) be obtained from \(A_{N}\) by multiplying the rows of \(A_{N}\) by positive numbers possibly different for different rows. Then,_  
\[rank(A_{N}^{\prime})=N-1. \tag{21}\]  
_The same result can be obtained if we multiply columns instead of rows._  
**Corollary 2**. : _Mutation models 1-3 satisfy Condition I, but Condition II is only satisfied when \(\bigwedge_{1\leq i\leq N}v_{i}=\frac{1}{N}\).