## Appendix A Solving Parameters in \(\lambda-PDF\)--Expanded Method  
First, solve the parameters of the polynomial in the first expansion. For the first extended probability density function obtained in Equation (15), solve its \(l\)-th moment of origin:  
\[E=\int_{b_{0}-b_{1}}^{b_{0}+b_{1}}x^{l}f_{x}^{\lambda}(x)dx=\int_{-1}^{1}(b_{0} +b_{1}\xi)^{l}\kappa_{\lambda}\Big{(}1-\xi^{2}\Big{)}^{\lambda-0.5}d\xi, \tag{17}\]  
Then, its first and second origin moments are:  
\[E_{1}=\int_{b_{0}-b_{1}}^{b_{0}+b_{1}}x^{l}f_{x}^{\lambda}(x)dx=b_{0}, \tag{18}\]  
\[E_{2}=\int_{D_{0}-b_{1}}^{b_{0}+b_{1}}x^{l}f_{x}^{\lambda}(x)dx=b_{0}^{2}+b_{1 }^{2}\int_{-1}^{1}(b_{0}+b_{1}\xi)^{2}f_{\xi}^{\lambda}(\xi)d\xi, \tag{19}\]  
In the value interval of the random variable, the integral of the probability density function is 1:  
\[\int_{-1}^{1}k_{\lambda}\Big{(}1-\xi^{2}\Big{)}^{\lambda-0.5}d\xi=1, \tag{20}\]  
Therefore:  
\[\int_{-1}^{1}\Big{(}1-\xi^{2}\Big{)}^{\lambda-0.5}d\xi=\frac{1}{\kappa_{ \lambda}}=\frac{\Gamma(0.5)\Gamma(\lambda+0.5)}{\Gamma(\lambda+1)}|, \tag{21}\]  
Then, the \(l\)-th origin moment of the random variable \(\xi\) can be obtained as:  
\[C_{l}= \begin{cases}\Big{(}\frac{l-1}{2}\Big{)}!\times\Big{(}\frac{1}{\lambda+ \frac{l+1}{2}}\Big{)}!\times\kappa_{\lambda}\times\frac{\Gamma(0.5)\Gamma \big{(}\lambda+0.5+\frac{l}{2}\big{)}}{\Gamma(\lambda+1+\frac{l}{2})}&,l\text{ is even number}\\ 0,&l\text{ is old number}\end{cases} , \tag{22}\]  
Then, the second-, fourth-, sixth- and eighth-order origin moments of random variable \(\xi\) can be obtained:  
\[\left\{ \begin{array}{l}C_{2}=\frac{1}{2}\times\frac{1}{\lambda+1/2}\times \kappa_{\lambda}\times\frac{\Gamma(1/2)\Gamma(\lambda+3/2)}{\Gamma(\lambda+2)} \\ C_{4}=\frac{3}{4}\times\frac{1}{\lambda+1/2}\times\frac{1}{\lambda+3/2}\times \kappa_{\lambda}\times\frac{\Gamma(1/2)\Gamma(\lambda+5/2)}{\Gamma(\lambda+3)} \\ C_{6}=\frac{18}{8}\times\frac{1}{\lambda+1/2}\times\frac{1}{\lambda+3/2} \times\frac{1}{\lambda+5/2}\times\kappa_{\lambda}\times\frac{\Gamma(1/2) \Gamma(\lambda+7/2)}{\Gamma(\lambda+4)}\\ C_{8}=\frac{105}{8}\times\frac{1}{\lambda+1/2}\times\frac{1}{\lambda+3/2} \times\frac{1}{\lambda+5/2}\times\frac{1}{\lambda+7/2}\times\kappa_{\lambda} \times\frac{\Gamma(1/2)\Gamma(\lambda+9/2)}{\Gamma(\lambda+5)}\end{array}  \right., \tag{23}\]  
Then, the second-, third- and fourth-order origin moments of the extended random variable \(x\) are:  
\[\left\{ \begin{array}{l}E_{2}=b_{0}^{2}+b_{1}^{2}C_{2}\\ E_{3}=b_{0}^{3}+3b_{0}b_{1}^{2}C_{2}\\ E_{4}=b_{0}^{4}+6b_{0}^{2}b_{1}^{2}C_{2}+b_{1}^{4}C_{4}\end{array} \right., \tag{24}\]For a random variable, \(x\), the relationship between the central moment and the origin moment is:  
\[\left\{ \begin{array}{l}D_{2}=E_{2}-E_{1}^{2}\\ D_{3}=E_{3}-3E_{2}E_{1}+2E_{1}^{3}\\ D_{4}=E_{4}-4E_{3}E_{1}+6E_{2}E_{1}^{2}-3E_{1}^{4}\end{array} \right., \tag{10}\]  
At this point, when the mean, \(\mu\), variance, \(\sigma^{2}\), skewness, \(C_{s}\) and kurtosis, \(C_{k}\), of the random variable \(x\) are obtained, the following equations can be used to solve them simultaneously, so as to solve the parameters \(b_{0}\), \(b_{1}\) and \(\lambda\):  
\[\left\{ \begin{array}{l}\mu=E_{1}=b_{0}\\ \sigma^{2}=D_{2}\\ C_{s}=\frac{D_{3}}{\sigma^{3}}\\ C_{k}=\frac{D_{4}}{\sigma^{4}}\end{array} \right., \tag{11}\]  
In the same way, the first \(l\) origin moment of the second expanded random variable \(x\) is:  
\[E_{l}=\int_{-1}^{1}\left(b_{0}+b_{1}\xi+b_{2}\xi^{2}\right)^{l}\kappa_{\lambda }\times\left(1-\xi^{2}\right)^{\lambda-0.5}d\xi, \tag{12}\]  
Thus, the first four origin moments and the second-, third- and fourth-order central momentsare obtained:  
\[\left\{