## 3 Training Signal Optimization
### Suboptimal Power Allocation in Closed-Form
: See Appendix C.  
From the above lemma, we can find a suboptimal solution to problem (20), with which the value of \(g^{\star}\) increases, but is not maximized, as follows:  
\[\tilde{\mathbf{x}}_{T}=\boldsymbol{\lambda}_{T}+\delta_{T}\mathbf{1}_{M_{t}} \;\;\text{and}\;\;\tilde{\mathbf{x}}_{Q}=\boldsymbol{\lambda}_{Q}+\delta_{Q} \Big{[}\mathbf{1}_{V},\mathbf{0}_{(L-v)\times 1}\Big{]}, \tag{22}\]  
where  
\[\mathbf{d}_{P}\in\mathcal{D}_{P},\ \ \mathbf{d}_{T}\in\mathcal{D}_{ \mathcal{T}},\ \ \mathbf{d}_{Q}\in\mathcal{D}_{Q}. \tag{19b}\]0  
and  
\[\mathbf{d}_{P}\in\mathcal{D}_{P},\ \ \mathbf{d}_{T}\in\mathcal{D}_{ \mathcal{T}},\ \ \mathbf{d}_{Q}\in\mathcal{D}_{Q}. \tag{19b}\]1  
Also, \(\mathbf{1}_{n}\) denotes the \(n\times 1\) vector whose elements are all \(1\), \(\delta_{T}=\frac{1}{M_{t}}\sum_{j=1}^{M_{t}}d_{T,j}\) and \(\delta_{Q}=\frac{1}{v}\sum_{j=1}^{v}d_{Q,j}\). To satisfy the constraints \(\tilde{\mathbf{x}}_{T}\in\mathcal{X}_{T}\) and \(\tilde{\mathbf{x}}_{Q}\in\mathcal{X}_{Q}\), the values of \(\delta_{T}\) and \(\delta_{Q}\) can be chosen as \(\delta_{T}=\sqrt{\frac{c_{T}}{M_{t}}}\) and \(\delta_{Q}=\sqrt{\frac{c_{Q}}{v}}\), respectively. By substituting (22) into (20), an optimization problem for suboptimal training power allocation can be formulated as  
\[\mathbf{d}_{P}\in\mathcal{D}_{P},\ \ \mathbf{d}_{T}\in\mathcal{D}_{ \mathcal{T}},\ \ \mathbf{d}_{Q}\in\mathcal{D}_{Q}. \tag{19b}\]2 (23a) s.t. \[\mathbf{d}_{P}\in\mathcal{D}_{P},\ \ \mathbf{d}_{T}\in\mathcal{D}_{ \mathcal{T}},\ \ \mathbf{d}_{Q}\in\mathcal{D}_{Q}. \tag{19b}\]3  
It is assumed that the elements of \(\boldsymbol{\lambda}_{T}\) and \(\boldsymbol{\lambda}_{Q}\) are arranged in descending and ascending orders, respectively. Then, the solution to problem (23) can be obtained in a closed form by the Lagrange multiplier method [28], as  
\[d_{p,i}^{(\mathrm{so})}= \begin{cases}\Big{(}\mathcal{P}_{T}+\sum_{j=1}^{r} \frac{\lambda_{Qj}+\delta_{Q}}{\lambda_{Tj}+\delta_{T}}\Big{)}\frac{\sqrt{ \lambda_{Qj}+\delta_{Q}}}{\sum_{j=1}^{r}\sqrt{\lambda_{Qj}+\delta_{Q}}}-\frac{ \lambda_{Qj}+\delta_{Q}}{\lambda_{Tj}+\delta_{T}},&i=1,\cdots,r,\\ 0,&i=r+1,\cdots,v,\end{cases}  \tag{24}\]  
where \(r=\max\{i\in\{1,\cdots,v\}:d_{p,i}^{(\mathrm{so})}>0\}\) is the largest \(i\) such that \(d_{p,i}^{(\mathrm{so})}>0\). The suboptimal solution in (24) follows the conventional water-filling strategy, i.e., it assigns more training power to the larger eigenvalues of \(\hat{\mathbf{T}}\) and smaller eigenvalues of \(\hat{\mathbf{Q}}\). One difference is that there exist the equal errors \(\delta_{T}\) and \(\delta_{Q}\) in the eigenvalues of \(\hat{\mathbf{T}}\) and \(\hat{\mathbf{Q}}\), respectively. **Remark 4**. : _For the case considered in Remark 1, we obtain the suboptimal power allocation from (24) by setting \(\delta_{Q}=0\) and \(\lambda_{Q,j}=\sigma_{Q}\), \(j=1,\cdots,L\), where \(\sigma_{Q}\) denotes the interference variance._  
**Remark 5**. : _In the case considered in Remark 2, the suboptimal power allocation can be obtained from (24) by setting \(\delta_{Q}=0\) and \(\lambda_{Q,j}=\sigma_{Q,j}\), \(j=1,\cdots,L\), where \(\{\sigma_{Q,j}\}_{j=1}^{L}\) denote the eigenvalues of \(\mathbf{Q}\)._  
The suboptimal solution \(\mathbf{d}_{p}^{(\mathrm{so})}\) in (24) becomes optimum if the minimum MSE \(g^{\star}\) is an increasing function of both \(\delta_{T}=\frac{1}{M_{t}}\sum_{j=1}^{M_{t}}d_{T,j}\) and \(\delta_{Q}=\frac{1}{v}\sum_{j=1}^{v}d_{Q,j}\), which, however, does not hold as can be seen in (A16) in Appendix C. Hence, the suboptimal solution does not guarantee achieving optimal performance in general. However, \(g^{\star}\) is tightly upperbounded by an increasing function of \(\delta_{T}\) and \(\delta_{Q}\). To show this, we consider the simple case of \(M_{t}=L\).