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Consider for given mean $m\in \mathbb R$ the Gaussian product model $(\mathbb R^n, \mathcal{B}(\mathbb R^n),\mathcal{N}_{m,\theta}:\vartheta >0)$ where $\vartheta$ denotes the variance, and a statistic $T=\sqrt\frac{\pi}{2} \frac 1n \sum_{i=1}^n \mid X_i-m\mid$. Show that the variance of $T$ does not reach the Cramer-Rao-bound $G'(\vartheta)^2/I(\mu_\vartheta)$, where $G(\vartheta)=\sqrt\vartheta$ and $\mu_\vartheta=\cal N^{\otimes n}_{m,\vartheta}$

I have shown that $T$ is unbiased for $G(\vartheta)$. I am struggling to calculate the Fisher information, i.e.

$$I(\mu_\vartheta)=\operatorname{Var}\bigr(\frac{\partial}{\partial\vartheta}\log(\frac{1}{\sqrt {2\pi\theta}}\exp(\frac{(-x-m)^2)}{2\theta}) \bigl)\\=\operatorname{Var}\bigr( -\frac{\sqrt{2\pi}}{2\sqrt\theta}\frac{1}{\sqrt{2\pi\theta}}+\frac{(x-m)^2}{2\theta^2}\bigl)\\=\operatorname{Var}\bigr(\frac{-1}{2\theta}+\frac{(x-m)^2}{2\theta^2}\bigl)\qquad\qquad$$ I think $(x-m)^2$ is $\chi^2$ distributied, so my final result would be simply $$\operatorname{Var}\bigr(\frac{-1}{2\theta}+\frac{(x-m)^2}{2\theta^2}\bigl)=\operatorname{Var}\bigr(\frac{(x-m)^2}{2\theta^2} \bigl)=\frac{1}{2\theta^4}$$ is this reasoning correct?

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1 Answer 1

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Indeed, when $X\sim{\cal N}(m,\theta)$, $$Z=\dfrac{(X-m)^2}{\theta}\sim\chi^ 2_1$$which implies$$\mathbb{E}[Z^{1/2}]=\sqrt{2/\pi}\quad\mathbb{E}[Z]=1\quad\text{var}(Z)=2$$and hence $$\text{var}(Z/2\theta^2)=1/2\theta^2$$

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    $\begingroup$ yes I see.. perfect, I forgot to rescale the distribution, that's why I have had $1/2\theta^4$ instead of $1/ 2\theta^2$ $\endgroup$
    – user209752
    Commented May 28, 2018 at 10:03

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