## 3 The Derivation and Design of the \(k\)-\(\mu\) Estimators  
The moment estimation method has always been one of the most commonly used methods for extracting channel parameters, owing to its advantages of closed-form simplicity, low algorithm complexity, and small estimation error. It has been widely used in the research of fading channel model parameter estimators, such as the \(K\)-factor estimator based on the Rice distribution [28], the M-parameter estimator based on the Nakagami distribution [29], and the \(k\)-parameter estimator based on the \(k\)-\(\mu\) distribution [25], among others. Moreover, the various-order moment values of the signal envelope can be easily obtained through experimental data. As the sample size approaches infinity, the theoretical \(n\)-th moment value and the expected value of the \(n\)-th power of the measured data have unbiased asymptotic properties. Therefore, in this paper, we choose the moment estimation method to estimate the \(k\) and \(\mu\) parameters of the \(k\)-\(\mu\) distribution. From the moment functions of the \(k\)-\(\mu\) distribution in Equation (5), it is evident that each order of moment functions contains gamma functions and Kummer confluent hypergeometric functions. The expansion of the right-hand side of the moment function equation is a complex and irreducible combination polynomial, making it difficult to solve for \(k\) and \(\mu\) using inverse function methods or substitution-based elimination techniques. At this point, we observe that even-order moments share the same functional structure. By utilizing two even-order moment functions as the numerator and denominator, we can eliminate the gamma function terms and exp(-\(k\mu\)) terms in Equation (5), thus obtaining the simplified second-, fourth-, and sixth-order moments of the kappa-mu distribution (for the detailed derivation process, please refer to Appendix A). \[E(r^{2})=\Omega \tag{8}\]  
\[E(r^{4})=\Omega^{2}\cdot\left(1+\frac{1+2k}{\mu(1+k)^{2}}\right) \tag{9}\]  
\[E(r^{6})=\Omega^{3}\cdot\left(\frac{\mu(\mu+1)(\mu+2)+3k\mu(\mu+1)(\mu+2)+3(k \mu)^{2}(\mu+2)+(k\mu)^{3}}{{(1+k)}^{3}\mu^{3}}\right) \tag{10}\]  
Equations (8)-(10) form a channel parameter estimation model with respect to variables \(k\) and \(\mu\). In this model, \(\Omega\) represents the total signal power, while \(E(r^{2}),E(r^{4}),E(r^{6})\) can be estimated from either empirical measurements or simulation data. Transforming Equation (9) into a function with respect to \(\mu\), we obtain:  
\[\mu=\frac{E^{2}(r^{2})}{Var[r^{2}]}\frac{{(1+k)}^{2}}{{(1+2k)}}=\eta\frac{{(1+ k)}^{2}}{{(1+2k)}} \tag{11}\]  
where \(\eta=E^{2}(r^{2})/Var[r^{2}]\), \(Var[\cdot]\) denotes the variance operator. By substituting Equation (11) into the sixth-order moment function given by Equation (10), we can simplify it to obtain Equation (12) as:  
\[\frac{E(r^{6})}{E^{3}(r^{2})}=\frac{{(2+6k)(1+k)}}{{\eta^{2}(1+2k)}^{2}}+ \frac{3}{\eta}+1 \tag{12}\]  
Expanding Equation (12) allows us to obtain that  
\[[4(\frac{E(r^{6})}{E^{3}(r^{2})}\eta^{2}-3\eta-\eta^{2})-6]\cdot k^{2}+[4( \frac{E(r^{6})}{E^{3}(r^{2})}\eta^{2}-3\eta-\eta^{2})-8]\cdot k+(\frac{E(r^{6} )}{E^{3}(r^{2})}\eta^{2}-3\eta-\eta^{2})-2=0, \tag{13}\]Considering the initial conditions that \(k\) and \(\mu\) must both be greater than zero, we can eliminate an invalid solution concerning \(k\). Consequently, our first closed-form estimate for \(k\) and \(\mu\) derived in this paper is as follows, denoted by a delta subscript (\(\Delta\)):  
\[\left\{ \begin{array}{l}\hat{k}_{\Delta}=\frac{\Delta+\sqrt{\Delta}}{2-2\Delta} \\ \hat{\mu}_{\Delta}=\eta\frac{(1+2\hat{k}_{\Delta})}{(1+\hat{k}_{\Delta})^{2}} \end{array} \right.,\left\{ \begin{array}{l}\eta=\frac{E^{2}(r^{2})}{Var(r^{2})} \\ \Delta=4-2(\frac{E(r^{6})}{[E(r^{2})]}\eta^{2}-3\eta-\eta^{2})\end{array}