*4.2. MSE and Parameter-Effects Curvature*

Let *LK*(*x*) denote the log of the expected value of *K*(*x*ˆ) in (57). Then

$$L\_K(\mathfrak{x}) := \log\_{10} \left\{ E[K(\mathfrak{X})] \right\}. \tag{82}$$

In order to compute *LK*(*x*), we first approximate the expectation in (82) by assuming *σx*<sup>ˆ</sup> *x*, which holds for the case investigated in our paper,

$$E\{K(\hat{\mathfrak{x}})\} = \frac{(n-1)}{na}E\left(\frac{1}{\hat{\mathfrak{x}}^n}\right) \approx \frac{(n-1)}{na}\frac{1}{[E(\hat{\mathfrak{x}})]^n} \approx \frac{(n-1)}{na}\frac{1}{\mathfrak{x}^n}.\tag{83}$$

The last step of the above equation follows from an assumption that the estimator is nearly unbiased. Now, taking the logarithm, we have

$$L\_K(\mathbf{x}) = \log\_{10}\left(\frac{n-1}{na}\right) - n\log\_{10}\mathbf{x}.\tag{84}$$

Now, from Equations (84) and (81), we can see that there is an affine mapping between *L*CRLB(*x*) and *LK*(*x*). That is,

$$L\_{\rm CRLB}(\mathbf{x}) = \boldsymbol{\alpha}\_1^K L\_K(\mathbf{x}) + \boldsymbol{\alpha}\_0^K \tag{85}$$

where

$$\begin{aligned} a\_1^K &= \frac{2(n-1)}{n}, \\ a\_0^K &= \log\_{10}\left(\frac{\sigma^2}{Nn^2a^2}\right) - \frac{2(n-1)}{n}\log\_{10}\left(\frac{n-1}{na}\right). \end{aligned} \tag{86}$$

We observe that *α<sup>K</sup>* <sup>1</sup> is positive and, hence, *LK*(*x*) and *L*CRLB(*x*) have the same sign of the non-zero slopes. As a result, *K*(*x*ˆ) and CRLB have the same sign of the non-zero slopes.
