# An Effective Adaptive Combination Strategy for Distributed Learning Network
## 3 Adaptive Combination Scheme
### Fixed-Point Iteration Solution  
First, we introduce a transform matrix \(\mathbf{P}_{k}\), defined as \(\mathbf{P}_{k}\triangleq[l\text{th column of }\mathbf{I}_{N}]_{l\in\mathcal{N}_{k}}\). Then, \(\mathbf{a}_{k}\) in (6) can be expressed by \(\mathbf{a}_{k}=\mathbf{P}_{k}\mathbf{b}_{k}\), with \(\mathbf{b}_{k}\in\mathbb{R}^{n_{k}}\). Therefore, the minimization problem (6) can be transformed into  
\[\min_{\mathbf{b}_{k}} \;J(\mathbf{b}_{k})\triangleq\mathbf{b}_{k}^{\mathrm{T}}\Re(\mathbf{Q}_\mathbf{{\Psi}_{k}})\mathbf{b}_{k} \tag{7}\] \[\text{ s.t. }\mathbf{1}_{n_{k}}^{T}\mathbf{b}_{k}=1;b_{k,i}\geq 0,\text{ for }i=1, \ldots,n_{k},\]  
where \(\mathbf{Q}_{\mathbf{\Psi}_{k}}\triangleq\mathbf{P}_{k}^{\mathrm{T}}\mathbf{Q}_{\mathbf{\Psi}}\mathbf{ P}_{k}=E\{(\mathbf{\Psi}_{k}(t)-E\{\mathbf{\Psi}_{k}(t)\})^{\mathrm{H}}(\mathbf{\Psi}_{k}(t)-E\{ \mathbf{\Psi}_{k}(t)\})\}\) with \(\mathbf{\Psi}_{k}(t)\triangleq\mathbf{\Psi}(t)\mathbf{P}_{k}\), and \(b_{k,i}\) is the \(i\)th element of vector \(\mathbf{b}_{k}\). By introducing the Lagrange multipliers \(\mathbf{\alpha}\) and \(\mathbf{\omega}\), we obtain the cost function  
\[\mathcal{L}(\mathbf{b}_{k},\mathbf{\omega},\mathbf{\alpha})\triangleq J(\mathbf{b}_{k})-\mathbf{ \omega}^{\mathrm{T}}\mathbf{b}_{k}+\alpha(\mathbf{1}_{n_{k}}^{T}\mathbf{b}_{k}-1). \tag{8}\]  
Taking the gradient of (8) with respect to \(\mathbf{b}_{k}\) yields  
\[\
abla_{\mathbf{b}_{k}}\mathcal{L}(\mathbf{b}_{k},\mathbf{\omega},\mathbf{\alpha}) =\
abla_{\mathbf{b}_{k}}J(\mathbf{b}_{k})-\mathbf{\omega}+\mathbf{1}_{n_{k}}\mathbf{\alpha}\] \[=\Re(\mathbf{Q}_{\mathbf{\Psi}_{k}})\mathbf{b}_{k}-\mathbf{\omega}+\mathbf{1}_{n_{k}} \mathbf{\alpha}. \tag{9}\]According to the Karush-Kuhn-Tucker (KKT) condition [21], the optimal value of the tuple \((\mathbf{b}_{k},\mathbf{\omega},\alpha)\) should obey  
\[\left\{ \begin{array}{l}\Re(\mathbf{Q}_{\mathbf{\tau}_{k}})\mathbf{b}_{k}-\mathbf{\omega}+\mathbf{1 }_{n_{k}}\alpha=0\\ \left[\mathbf{\omega}\right]_{i}b_{k,i}=0\end{array} \right.. \tag{10}\]  
or equivalently,  
\[[\Re(\mathbf{Q}_{\mathbf{\tau}_{k}})\mathbf{b}_{k}+\mathbf{1}_{n_{k}}\alpha]_{i}b_{k,i}=0. \tag{11}\]  
We introduce a positive definite diagonal matrix \(\mathbf{D}_{k}(t-1)\), whose \(i\)th diagonal element is an arbitrary positive function with respect to \(\mathbf{b}_{k}(t-1)\), denoted as \(f_{i}(\mathbf{b}_{k}(t-1))\). Obviously, \(\mathbf{D}_{k}(t-1)\big{(}-\
abla_{\mathbf{b}_{k}}\mathcal{L}(\mathbf{b}_{k},\mathbf{\omega}, \alpha)\big{)}\) is still the descent direction of the cost function (8) [21]. Therefore, we have the adaptive solution of problem (7) through the fixed-point iteration method  
\[ \begin{array}{l}b_{k,i}(t)=b_{k,i}(t-1)-\eta_{k,i}(t)f_{i}(\mathbf{b}_{k}(t-1))b _{k,i}(t-1)\\ \left[\Re(\mathbf{Q}_{\mathbf{\tau}_{k}})\mathbf{b}_{k}(t-1)+\mathbf{1}_{n_{k}}\alpha\right]_ {i},\end{array}  \tag{12}\]  
where \(b_{k,i}(t)\) is the estimate of \(b_{k,i}\) at time instant \(t\) and \(\eta_{k,i}\) is the learning factor of \(b_{k,i}(t)\). To simplify the problem, we choose the same learning factor \(\eta_{k,i}\) for all \(i\) at any time \(t\). Hence, we can rewrite (12) as a vector,  
\[\mathbf{b}_{k}(t)= \mathbf{b}_{k}(t-1)-\eta_{k}(t)\mathbf{\Gamma}_{k}(t-1)[\Re(\mathbf{Q}_\mathbf{{\tau}_{k}})\mathbf{b}_{k}(t-1)+\mathbf{1}_{n_{k}}\alpha]. \tag{13}\]  
where \(\mathbf{\Gamma}_{k}(t-1)\triangleq\text{diag}\{\mathbf{D}_{k}(t-1)\mathbf{b}_{k}(t-1)\}\).