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Let

$g(\boldsymbol{\theta},\boldsymbol{\theta_0}) = trace [ \boldsymbol{\Omega{(\boldsymbol{\theta})}}^{-1} \boldsymbol{\Omega{(\boldsymbol{\theta_0})}}]-ln[det(\boldsymbol{\Omega{(\boldsymbol{\theta})}}^{-1} \boldsymbol{\Omega{(\boldsymbol{\theta_0}))}}]-N $

where $\boldsymbol{\theta} \in \boldsymbol{\Theta}$ with $\boldsymbol{\Theta}$ a compact subset of $R^{n}$, $n$ and $N$ are fixed numbers, and $\boldsymbol{\theta_0}$ belongs to the interior of $\boldsymbol{\Theta}$.

Denote the eigenvalues of the symmetric matrix $\boldsymbol{\Omega{(\boldsymbol{\theta})}}^{-1} \boldsymbol{\Omega{(\boldsymbol{\theta_0})}}$ by $\lambda_s$ $(s=1,2,...,N)$ where $\lambda_s>0$ for all $s$.

We then have

$g(\boldsymbol{\theta},\boldsymbol{\theta_0}) = \sum_{s=1}^N [\lambda_{s}-ln(\lambda_{s})-1] \geq 0,$ with $g(\boldsymbol{\theta},\boldsymbol{\theta_0}) =0$ if and only if $\lambda_{s}=1$ for every $s$.

Am I correct in thinking that $g(\boldsymbol{\theta},\boldsymbol{\theta_0}) =0$ if and only if $\lambda_{s}=1$ for every $s$ does not necessarily imply that this equality is true only for $\boldsymbol{\theta}=\boldsymbol{\theta_0}$ and that it can happen also for $\boldsymbol{\theta}\ne\boldsymbol{\theta_0}$?

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  • $\begingroup$ indeed: $g(\theta,\theta_0)=0$ implies $\Omega(\theta)=\Omega(\theta_0)$; there could be $\theta\neq\theta_0$ where this holds. $\endgroup$ Commented Jun 11, 2020 at 17:40
  • $\begingroup$ Thank you Carlo for the confirmation and for putting it this way. $\endgroup$
    – user0735
    Commented Jun 11, 2020 at 19:57

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