*2.4. Cramér-Rao Lower Bound*

The CRLB [2,41] for the MSE in the current problem is given by

$$\text{CRLB}\_x = \left[\frac{d\mathbf{h}'(x)}{dx}\mathbf{R}^{-1}\frac{d\mathbf{h}(x)}{dx}\right]^{-1}.\tag{31}$$

**Remark 5.** *Calculation of the variance σ*<sup>2</sup> *<sup>x</sup> and* CRLB*<sup>x</sup> are similar. For σ*<sup>2</sup> *<sup>x</sup> , we use the estimate x*ˆ *while calculating the Jacobian of the measurement function, whereas, for* CRLB*x, we use the true x while calculating the Jacobian of the measurement function.*

Using similar procedure, we obtain

$$\text{CRLB}\_x = \frac{\sigma^2}{N(anx^{n-1})^2},\tag{32}$$

$$\sqrt{\text{CRLB}\_{\text{X}}} = \frac{\sigma}{\sqrt{N} a n x^{n-1}}.\tag{33}$$

From (30) and (33), we find that, for a given *x*, the standard deviation (SD) and square root of CRLB are inversely proportional to the power *n*. Secondly (33) shows that, for a given power, the square root of CRLB decreases as *x* increases.
