# Construction Method of Probabilistic Boolean Networks Based on Imperfect Information
## 4 Example  
In this example, we consider the PBN with three states. Suppose that the index set for each \(i\) is given by \(\mathcal{N}_{1}=\{3\}\), \(\mathcal{N}_{2}=\{1,3\}\), and \(\mathcal{N}_{3}=\{1,2\}\), and the sample mean for each \(i\) is given by \(s_{1}^{k+1}=s_{1}^{k}=0.8\), \(s_{2}^{k+1}=s_{2}^{k}=0.4\), and \(s_{3}^{k+1}=s_{3}^{k}=0.7\) (i.e., \(p=1\)). Suppose also that one of the candidates of Boolean functions is given by  
\[\left\{\begin{array}{l}x_{1}(k+1)=\
eg x_{3}(k),\\ x_{2}(k+1)=x_{1}(k)\wedge\
eg x_{3}(k),\\ x_{3}(k+1)=x_{1}(k)\wedge\
eg x_{2}(k).\end{array}\right.\]  
We demonstrate the proposed method in the case of \(x_{2}\). The matrix \(\tilde{B}^{(2)}\) can be obtained as  
\[\tilde{B}^{(2)}=\begin{bmatrix}1-a_{1}^{(2)}&1-a_{2}^{(2)}&1-a_{3}^{(2)}&1-a_ {4}^{(2)}\\ a_{1}^{(2)}&a_{2}^{(2)}&a_{3}^{(2)}&a_{4}^{(2)}\end{bmatrix},\]  
where \(a_{1}^{(2)}\), \(a_{2}^{(2)}\), \(a_{3}^{(2)}\), and \(a_{4}^{(2)}\) are decision variables. From the given Boolean function \(x_{2}(k+1)=x_{1}(k)\wedge\
eg x_{3}(k)\), the matrix \(\tilde{B}_{d}^{(2)}\) can be obtained as  
\[\tilde{B}_{d}^{(2)}=\begin{bmatrix}1&1&0&1\\ 0&0&1&0\end{bmatrix},\]  
i.e., \(b_{1}^{(2)}=b_{2}^{(2)}=b_{4}^{(2)}=0\) and \(b_{4}^{(2)}=1\). Then, the objective function \(J^{(2)}\) in the LP problem for \(x_{2}\) is given by  
\[J^{(2)}=(1-a_{1}^{(2)})+(1-a_{2}^{(2)})+a_{3}^{(2)}+(1-a_{4}^{(2)}).\]  
The constraint condition (5) is given by  
\[\begin{bmatrix}1-0.4\\ 0.4\end{bmatrix}=\begin{bmatrix}1-a_{1}^{(2)}&1-a_{2}^{(2)}&1-a_{3}^{(2)}&1-a_ {4}^{(2)}\\ a_{1}^{(2)}&a_{2}^{(2)}&a_{3}^{(2)}&a_{4}^{(2)}\end{bmatrix}\begin{bmatrix}(1- 0.8)(1-0.7)\\ (1-0.8)0.7\\ 0.8(1-0.7)\\ 0.8\cdot 0.7\end{bmatrix}.\]By solving the LP problem in which \(J^{(2)}\) is maximized under the above constraint condition, we can obtain  
\[\bar{B}^{(2)}=\begin{bmatrix}1&1&0&0.7143\\ 0&0&1&0.2857\end{bmatrix}.\]  
This matrix can be decomposed as follows:  
\[\bar{B}^{(2)}=0.7143\begin{bmatrix}1&1&0&1\\ 0&0&1&0\end{bmatrix}+0.2857\begin{bmatrix}1&1&0&0\\ 0&0&1&1\end{bmatrix}.\]  
Hence, we can obtain two Boolean functions and those selection probabilities as follows:  
\[f^{(2)}=\begin{cases}f_{1}^{(2)}=x_{1}(k)\wedge\
eg x_{3}(k),&c_{1}^{(2)}=0.71 43,\\ f_{2}^{(2)}=\
eg x_{3}(k),&c_{2}^{(2)}=0.2857,\end{cases}\]  
where \(q(2)=2\). In a similar way, we can obtain two Boolean functions and those selection probabilities for \(x_{1}\) and \(x_{3}\) as follows:  
\[f^{(1)}=\begin{cases}f_{1}^{(1)}=\
eg x_{3}(k),&c_{1}^{(1)}=0.2857,\\ f_{2}^{(1)}=1,&c_{2}^{(1)}=0.7143,\\ f^{(3)}=\begin{cases}f_{1}^{(3)}=x_{1}(k)\wedge\
eg x_{2}(k),&c_{1}^{(3)}=0. 5417,\\ f_{2}^{(3)}=\
eg(x_{1}(k)\oplus x_{2}(k)),&c_{2}^{(3)}=0.4583.\end{cases}\]  
Thus, we can obtain the PBN.  
In the case of Problem 3 (i.e., the case where there is no objective function in Problem 4), we can obtain the PBN as follows:  
\[f^{(1)}=\begin{cases}f_{1}^{(1)}=\
eg x_{3}(k),&c_{1}^{(1)}=0.1379,\\ f_{2}^{(1)}=1,&c_{2}^{(1)}=0.8621,\\ f^{(2)}=\begin{cases}f_{1}^{(2)}=x_{1}(k)\wedge\
eg x_{3}(k),&c_{1}^{(2)}=0. 4391,\\ f_{2}^{(2)}=\
eg x_{1}(k)\lor x_{3}(k),&c_{2}^{(2)}=0.3580,\\ f_{3}^{(2)}=\
eg x_{1}(k),&c_{3}^{(2)}=0.1065,\\ f_{4}^{(2)}=1,&c_{4}^{(2)}=0.0761,\\ f_{5}^{(2)}=\
eg x_{1}(k)\wedge\
eg x_{3}(k),&c_{5}^{(2)}=0.0203,\\ \end{cases}\]  
\[f^{(3)}=\begin{cases}f_{1}^{(3)}=x_{1}(k)\wedge\
eg x_{2}(k),&c_{1}^{(3)}=0. 4548,\\ f_{2}^{(3)}=\
eg(x_{1}(k)\oplus x_{2}(k)),&c_{2}^{(3)}=0.3167,\\ f_{3}^{(3)}=x_{2}(k),&c_{3}^{(3)}=0.1154,\\ f_{4}^{(3)}=\
eg x_{1}(k)\lor s_{2}(k),&c_{4}^{(3)}=0.1108,\\ f_{5}^{(3)}=\
eg x_{1}(k),&c_{5}^{(3)}=0.0023.\end{cases}\]  
From these results, we see that the probability that the Boolean function given in advance is chosen becomes higher by introducing the objective function. Thus, if a dominant Boolean function is known, then the proposed method is effective.