## 3 Training Signal Optimization
### Suboptimal Power Allocation in Closed-Form  
The numerical approach for optimal power allocation may be undesirable in real applications due to a non-negligible complexity during the channel training phase. Considering this problem, we propose a suboptimal power allocation in a closed-form. Since the power allocation problem (18) is convex-concave, we can interchange the minimum and maximum operators of the problem (18) according to [31] (Lemma 36.2)  
as follows:  
\[\max_{\mathbf{d}_{T},\mathbf{d}_{Q}} \min_{\mathbf{d}_{P}} \zeta(\mathbf{d}_{P},\mathbf{d}_{T},\mathbf{d}_{Q})\] (19a) s.t. \[\mathbf{d}_{P}\in\mathcal{D}_{P},\ \ \mathbf{d}_{T}\in\mathcal{D}_{ \mathcal{T}},\ \ \mathbf{d}_{Q}\in\mathcal{D}_{Q}. \tag{19b}\]  
Defining  
\[\mathbf{x}_{T}=[\lambda_{T,1}+d_{T,1},\cdots,\lambda_{T,M_{t}}+d_{T,M_{t}}]^{T}\]  
and  
\[\mathbf{x}_{Q}=[\lambda_{Q,1}+d_{Q,1},\cdots,\lambda_{Q,L}+d_{Q,L}]^{T},\]  
problem (19) can be equivalently reformulated as  
\[\max_{\mathbf{x}_{T},\mathbf{x}_{Q}} \min_{\mathbf{d}_{P}} g(\mathbf{d}_{P},\mathbf{x}_{T},\mathbf{x}_{Q})\] (20a) s.t. \[\mathbf{d}_{P}\in\mathcal{D}_{P},\ \ \mathbf{x}_{T}\in\mathcal{X}_{ \mathcal{T}},\ \ \mathbf{x}_{Q}\in\mathcal{X}_{Q}, \tag{20b}\]  
where the cost function \(g\) in (23a) is  
\[g(\mathbf{d}_{P},\mathbf{x}_{T},\mathbf{x}_{Q})=\beta\sum_{i=1}^{v}\biggl{(} \frac{1}{x_{T,i}}+\frac{d_{P,i}}{x_{Q,i}}\biggr{)}^{-1}+\beta\sum_{i=v+1}^{M_ {t}}x_{T,i}, \tag{21}\]  
and the constraint sets \(\mathcal{X}_{\mathcal{T}}\) and \(\mathcal{X}_{\mathcal{Q}}\) are defined as  
\[\mathcal{X}_{\mathcal{T}}=\bigl{\{}\mathbf{x}_{T}:\|\mathbf{x}_{T}-\boldsymbol {\lambda}_{T}\|^{2}\leq\epsilon_{T},\ \ x_{T,i}\geq 0,\ \ i=1,\cdots,M_{t}\bigr{\}}\]  
and  
\[\mathcal{X}_{\mathcal{Q}}=\bigl{\{}\mathbf{x}_{Q}:\|\mathbf{x}_{Q}-\boldsymbol {\lambda}_{Q}\|^{2}\leq\epsilon_{Q},\ \ x_{Q,i}\geq 0,\ \ i=1,\cdots,L\bigr{\}},\]  
respectively. It is still difficult to obtain the optimal solution to problem (20) in a closed-form. To overcome this difficulty, in the following, we instead find a suboptimal solution. Definition 2: [32] For any \(\mathbf{a}\in\mathbb{R}^{n}\), let \(a_{[1]}\geq\cdots\geq a_{[n]}\) denote the components of \(\mathbf{a}\) in decreasing order. Definition 3: [32] (Ch.1.A.1) The vector \(\mathbf{a}\in\mathbb{R}^{n}\) is majorized by \(\mathbf{b}\in\mathbb{R}^{n}\), denoted by \(\mathbf{a}\prec\mathbf{b}\), if \(\sum_{m=1}^{l}a_{[m]}\leq\sum_{m=1}^{l}b_{[m]}\), \(1\leq l\leq n-1\), and \(\sum_{m=1}^{n}a_{[m]}=\sum_{m=1}^{n}b_{[m]}\). Definition 4 (Schur-concave function): [32] (Ch.3.A.1) A real-valued function \(\varphi:\mathcal{A}\subset\mathbb{R}^{n}\to\mathbb{R}\) is said to be Schur-concave on \(\mathcal{A}\) if \(\mathbf{a}\prec\mathbf{b}\Rightarrow\varphi(\mathbf{a})\succeq\varphi( \mathbf{b})\). **Lemma 2**. : _Let \(g^{\star}(\mathbf{x}_{T},\mathbf{x}_{Q})=\underset{\mathbf{d}_{p}\in\mathcal{D}_{p} }{min}\;\;g(\mathbf{d}_{p},\mathbf{x}_{T},\mathbf{x}_{Q})\) be the minimum MSE. Then, \(g^{\star}(\mathbf{x}_{T},\mathbf{x}_{Q})\) is Schur-concave in \(\mathbf{x}_{T}\) and \(\mathbf{x}_{Q}\). In other words, if there exists the vectors \(\tilde{\mathbf{x}}_{T}\) and \(\tilde{\mathbf{x}}_{Q}\) such that \(\tilde{\mathbf{x}}_{T}\prec\mathbf{x}_{T}\) and \(\tilde{\mathbf{x}}_{Q}\prec\mathbf{x}_{Q}\), then \(g^{\star}(\tilde{\mathbf{x}}_{T},\tilde{\mathbf{x}}_{Q})\geq g^{\star}( \mathbf{x}_{T},\mathbf{x}_{Q})\)._  
Proof.