## 4 PEM Considering Boundary Constraints
### Power Transformation Method  
The power transformation method can transform an asymmetric probability distribution into a symmetric or approximately symmetric distribution. A symmetrical distribution function can reduce the possibility of estimated points falling outside the boundary constraints. The definition of a power transformation is  
\[y=x^{a}. \tag{41}\]  
The skewness of \(y\) is  
\[\lambda_{y,3}(a)=\frac{\int_{-\infty}^{+\infty}{(x^{a}-\mu_{x^{a}})}^{3}f(x^{a })dx}{\sigma_{x^{a}}^{3}}. \tag{42}\]  
The wind speed samples \(x\) are discrete, so the values of \(y\) are discrete:  
\[\overline{\lambda}_{y,3}(a)=\frac{\frac{1}{N}\sum\limits_{j=1}^{N}{(x_{j}^{a}- \frac{1}{N}\sum\limits_{k=1}^{N}{x_{k}^{a}})^{3}}}{[\sum\limits_{j=1}^{N}{(x_{ j}^{a}-\frac{1}{N}\sum\limits_{k=1}^{N}{X_{k}^{a}})^{2}}]} \tag{43}\]  
It can be inferred that when \(\overline{\lambda}_{y,3}\) is equal to 0, the power-transformed function is a symmetric function. Then, the estimated points can be directly obtained as follows:  
\[\overline{y_{i}}=\mu_{y}+{(-1)}^{i+1}\sigma_{y},i=1,2. \tag{44}\]  
Substituting \(\overline{y_{i}}\) into function \(f\) with the range constraints to obtain \(f(x_{i})\) yields  
\[f(\overline{x_{i}})=f(\sqrt[4]{y_{i}})=f(\sqrt[4]{\mu_{y}+{(-1)}^{i+1}\sigma_ {y}}),i=1,2. \tag{45}\]  
However, if \(x_{i}\) remains outside of the range, the power transformation method will be invalid.  
Usually, it is difficult to find \(\alpha\) such that \(\widetilde{\lambda}_{y,3}\) is 0, so the following iterative method is used to find an \(\alpha\) such thats within the limits:  
Step 1. Calculate the variable's original skewness \(\widetilde{\lambda}_{y,3}\). Let \(a_{j}=ka_{j-1}\), \(j\geq 0\), and \(a_{0}=1\); if \(\widetilde{\lambda}_{y,3}\) is greater than 0, set the coefficient value to \(k=\left(1+\widetilde{\lambda}_{x,3}\right)^{-1/9}\). If the skewness coefficient is less than 0, set it to \(k\)\(=(1-\widetilde{\lambda}_{x,3})^{1/9}\).  
Step 2. Set the maximum number of steps \(j\) to 100; calculate each step \(a_{j}\) to obtain \(y_{i,j}(i=1,2)\) and \(\widetilde{\lambda}_{j,x,3}\). Then, calculate \(y_{i,j}\) and determine whether \(x_{i,j}\) remains continuously within the limits. If it meets this criterion, the iteration stops.  
Step 3. If \(\alpha\) cannot be determined using the above steps, find the smallest \(\widetilde{\lambda}_{l,x,3}\). Let the number of iterations used to obtain \(\widetilde{\lambda}_{l,x,3}\) be \(l\); if \(\widetilde{\lambda}_{l,x,3}\) is smaller than 0.01, a suitable value for \(\alpha\) cannot be found such that is \(x_{i}\) is within the limits.  
Step 4. If \(\widetilde{\lambda}_{l,x,3}\) is greater than 0.01, set \(n=1\) and change \(k=\left(1+\widetilde{\lambda}_{x,3}\right)^{-1/9}\) to \(k=\left(1+\widetilde{\lambda}_{x,3}\right)^{-\frac{1}{99}}\), \(m=\frac{1}{3^{n}}\) in iteration \(l-1\). Calculate \(y_{i,j}\) and then determine whether \(x_{i,j}\) is within the allowed range from iteration \(l-1\) to iteration \(l+3^{n}\). If \(x_{i,j}\) is outside the limits, set \(n=n+1\) and repeat the above steps until \(x_{i,j}\) within the range; then, \(\alpha\) is determined. However, if \(\widetilde{\lambda}_{l,x,3}\) is less than 0.01, then \(\alpha\) cannot be found.