## 3 Cramer-Rao Lower Bounds  
Before deriving the performances of the proposed estimators, we derive the corresponding Cramer-Rao lower bounds (CRLB), as they represent the theoretical limit on the best achievable accuracy that can be obtained. First, we derive the CRLB for \(\hat{S}\) and \(\hat{N}\) and then the CRLB for the estimator of SNR \(\hat{p}\).  
We start from the joint probability density function  
\[f\left(y_{0},\ldots,y_{K-1}|S,N\right)=\frac{1}{\left(2\pi\sigma^{2}\right)^{ K/2}}\exp\left\{-\frac{1}{2\sigma^{2}}\sum_{k=0}^{K-1}\left(y_{k}-Ax_{k} \right)^{2}\right\}, \tag{13}\]  
where \(S=A^{2}/2\) and \(N=\sigma^{2}\). To get the bound we need to first derive the Fisher information matrix that is defined as  
\[I\left(S,N\right)\triangleq\left(\begin{array}{cc}-E\left[\frac{\partial^{2 }\ln f}{\partial S\partial^{2}}\right]&-E\left[\frac{\partial^{2}\ln f}{ \partial S\partial N}\right]\\ -E\left[\frac{\partial^{2}\ln f}{\partial S\partial N}\right]&-E\left[\frac{ \partial^{2}\ln f}{\partial N^{2}}\right]\end{array}\right). \tag{14}\]  
In Appendix B we show that in this case, the Fisher matrix is given by  
\[I\left(S,N\right)=\left(\begin{array}{cc}\frac{K}{2A^{2}\sigma^{2}}&0\\ 0&\frac{K}{\sigma^{4}}\end{array}\right), \tag{15}\]  
and the Cramer-Rao lower bounds for the estimators \(\hat{S}\) and \(\hat{N}\), obtained by inversion of the Fisher matrix, are  
\[\text{var}(\hat{S}) \geq\frac{2A^{2}\sigma^{2}}{K} \tag{16}\] \[\text{var}(\hat{N}) \geq\frac{\sigma^{4}}{K}. \tag{17}\]  
Based on the above results it is now possible to derive the CRLB for the proposed SNR estimator \(\hat{\rho}\). Since the estimator is formed by the ratio of \(\hat{S}\) and \(\hat{N}\), the bound for \(\hat{\rho}\) can be derived from the Fisher information matrix obtained previously and the derivatives of the transformation  
\[g(S,N)=\frac{S}{N}\]  
according the following relationship [7]  
\[\text{var}(\hat{\rho})\geq\left(\begin{array}{cc}\frac{\partial\xi}{ \partial S}&\frac{\partial\xi}{\partial N}\end{array}\right)I^{-1}\left(S,N \right)\left(\begin{array}{cc}\frac{\partial\xi}{\partial S}&\frac{\partial \xi}{\partial N}\end{array}\right)^{T}. \tag{18}\]  
Since  
\[\left(\begin{array}{cc}\frac{\partial\xi}{\partial S}&\frac{\partial\xi}{ \partial N}\end{array}\right)^{T}=\left(\begin{array}{c}\frac{1}{N}\\ -\frac{S}{N^{2}}\end{array}\right),\]  
we get \[\mathrm{var}(\hat{\rho})\geq\frac{A^{2}}{\sigma^{2}K}\left(2+\frac{A^{2}}{4\sigma^ {2}}\right). \tag{19}\]  
Similarly, we can derive the bound when the SNR is expressed in dB. In this case the estimator is given by the following transformation  
\[g\left(S,N\right)=10\log_{10}\left(\frac{S}{N}\right). \tag{20}\]  
The derivatives are  
\[\left(\begin{array}{cc}\frac{\partial g}{\partial S}&\frac{ \partial g}{\partial N}\\ \end{array}\right) =\frac{10}{\ln 10}\left(\begin{array}{cc}\frac{1}{S}&-\frac{1}{N} \\ \end{array}\right) \tag{21}\] \[=\frac{10}{\ln 10}\left(\begin{array}{cc}\frac{2}{A^{2}}&- \frac{1}{\sigma^{2}}\\ \end{array}\right), \tag{22}\]  
and we have that the lower bound is  
\[\mathrm{var}(\hat{\rho})\geq\left(\frac{10}{\ln 10}\right)^{2}\left(\frac{8}{K} \frac{\sigma^{2}}{A^{2}}+\frac{1}{K}\right)\,\mathrm{dB}^{2}. \tag{23}\]