## 2 Problem Statement and Mathematical Models
### \(m_{\lambda}\) Measure and Expected Value Operator
_if_ \(\gamma\geq 0\)_, then_ \((\gamma\xi)_{\sup}(\alpha)=\gamma\xi_{\sup}(\alpha)\) _and_ \((\gamma\xi)_{\inf}(\alpha)=\gamma\xi_{\inf}(\alpha)\)_;_
4. _if_ \(\gamma<0\)_, then_ \((\gamma\xi)_{\sup}(\alpha)=\gamma\xi_{\inf}(\alpha)\) _and_ \((\gamma\xi)_{\inf}(\alpha)=\gamma\xi_{\sup}(\alpha)\)_._  
A fuzzy variable \(\xi\) with membership function \(\mu_{\xi}(x)\) is called a normalized fuzzy variable if it has the following characteristics: (1) there exists a real number \(x^{*}\) with \(\mu_{\xi}(x^{*})=1\); (2) \(\mu_{\xi}(x)\) is  
Figure 1: Expected value curves with respect to parameter \(\lambda\). nondecreasing for \(x\leq x^{*}\) and nonincreasing for \(x\geq x^{*}\). Typically, the interval fuzzy variables, triangular fuzzy variables, trapezoidal fuzzy variable and fuzzy variables with unimodal membership functions are all normalized fuzzy variables. Theorem 3.: _Let \(\xi\) be a normalized fuzzy variable with finite expected value. Then, we have:_  
\[E_{\lambda}[\xi]=\int_{0}^{1}\left(\lambda\xi_{\mathrm{sup}}(\alpha)+(1-\lambda )\xi_{\mathrm{inf}}(\alpha)\right)\mathrm{d}\alpha,\]  
_in which \(\xi_{\mathrm{sup}}(\alpha)\) and \(\xi_{\mathrm{inf}}(\alpha)\), respectively, are \(\alpha\)-optimistic and \(\alpha\)-pessimistic values to fuzzy variable \(\xi\)._  
Proof of Theorem 3.: Since \(\xi\) is a normalized fuzzy variable, there exists a real number \(x^{*}\) with \(\mu_{\xi}(x^{*})=1\), and \(\mu_{\xi}(x)\) is nondecreasing for all \(x<x^{*}\) and nonincreasing for all \(x\geq x^{*}\). In the following, we only consider the case of \(x^{*}>0\), and the other situation can be proven similarly. In this case, we have:  
\[E_{\lambda}[\xi] = \int_{0}^{+\infty}m_{\lambda}\{\xi\geq r\}\mathrm{d}r-\int_{- \infty}^{0}m_{1-\lambda}\{\xi\leq r\}\mathrm{d}r\] \[= \int_{0}^{x^{*}}m_{\lambda}\{\xi\geq r\}\mathrm{d}r+\int_{x^{*}} ^{+\infty}m_{\lambda}\{\xi\geq r\}\mathrm{d}r-\left[\int_{-\infty}^{x^{*}}m_{1 -\lambda}\{\xi\leq r\}\mathrm{d}r-\int_{0}^{x^{*}}m_{1-\lambda}\{\xi\leq r\} \mathrm{d}r\right]\] \[E_{\lambda_{1}}[\xi] = \int_{0}^{+\infty}m_{\lambda_{1}}\{\xi\geq r\}\mathrm{d}r-\int_{- \infty}^{0}m_{1-\lambda_{1}}\{\xi\leq r\}\mathrm{d}r\]0 \[E_{\lambda_{1}}[\xi] = \int_{0}^{+\infty}m_{\lambda_{1}}\{\xi\geq r\}\mathrm{d}r-\int_{- \infty}^{0}m_{1-\lambda_{1}}\{\xi\leq r\}\mathrm{d}r\]1 \[E_{\lambda_{1}}[\xi] = \int_{0}^{+\infty}m_{\lambda_{1}}\{\xi\geq r\}\mathrm{d}r-\int_{- \infty}^{0}m_{1-\lambda_{1}}\{\xi\leq r\}\mathrm{d}r\]2 \[E_{\lambda_{1}}[\xi] = \int_{0}^{+\infty}m_{\lambda_{1}}\{\xi\geq r\}\mathrm{d}r-\int_{- \infty}^{0}m_{1-\lambda_{1}}\{\xi\leq r\}\mathrm{d}r\]3  
The proof is completed. The linearity of the expected value operator of random variables is an important property in the real-world applications. Likewise, for the \(\lambda\)-expected value operator proposed in this study, we also have a similar linearity for the fuzzy variables. Theorem 4.: _Let \(\xi\) and \(\eta\) be normalized fuzzy variables with finite expected values. If \(a\) and \(b\) are two real numbers, we then have:_  
\[E_{\lambda_{1}}[\xi] = \int_{0}^{+\infty}m_{\lambda_{1}}\{\xi\geq r\}\mathrm{d}r-\int_{- \infty}^{0}m_{1-\lambda_{1}}\{\xi\leq r\}\mathrm{d}r\]4  
This theorem can be easily proven by using Theorems 2 and 3. In addition, based on the expected value operator, we can define the concept of variance to measure the diversity degree of the uncertain information, given below. Definition 3.: Let \(\xi\) be a fuzzy variable with \(\lambda\)-expected value \(e\). Then, \(\lambda\)-variance of \(\xi\) is defined as:  
\[E_{\lambda_{1}}[\xi] = \int_{0}^{+\infty}m_{\lambda_{1}}\{\xi\geq r\}\mathrm{d}r-\int_{- \infty}^{0}m_{1-\lambda_{1}}\{\xi\leq r\}\mathrm{d}r\]5  
Example 4.: _Let \(\xi=[a,b]\) be an interval fuzzy variable. Then, the \(\lambda\)-variance of \(\xi\) is:_  
\[E_{\lambda_{1}}[\xi] = \int_{0}^{+\infty}m_{\lambda_{1}}\{\xi\geq r\}\mathrm{d}r-\int_{- \infty}^{0}m_{1-\lambda_{1}}\{\xi\leq r\}\mathrm{d}r\]6  
**Proof. ** According to Example 3, we have \(E_{\lambda}[\xi]=(1-\lambda)a+\lambda b\).