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$**X**=\left(X_1,...X_n\right) $is a simple random sample from an exponential distribution $p\left(x;\theta\right)$, where $\theta$ is the parameter. $g_n\left(**X**\right)$ is an estimator of $\theta$. And $$\sqrt{n}\left(g_n\left(**X**\right)-\theta\right)\to N\left(0,V\left(\theta\right)\right).$$

So what is the relation between $V\left(\theta\right)$ and the Fisher information $I\left(\theta\right)$?

Caran-d'Ache
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user44618
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2 Answers2

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According to the Cramer-Rao lower bound,

$$ V(g(\mathbf{X}))\geq \frac{1}{\mathcal{I}[g(\mathbf{X})]} $$

If the estimator is asymptotically efficient it means that it achieves the bound, that is

$$ V(g(\mathbf{X})) \to \frac{1}{\mathcal{I}[g(\mathbf{X})]} $$

This means that in the case of asymptotic efficiency, the variance is equal to the inverse of the Fisher information.

hejseb
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  • Thank you for your response. But the problem is that whether the estimator is asymptotically efficient is unknown. – user44618 Jul 08 '13 at 12:35
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Well. By saying that $\sqrt{n}\left(g_n\left(X\right)-\theta\right)\xrightarrow d N\left(0,V\left(\theta\right)\right)$ you assume asymptotic normality of the estimator $g_n(\theta)$. If (in addition to normality) you say that $g_n(\theta)$ is a MVUE (or maybe just BLUE) (that means it is the best in some class of estimators), then hejseb is correct and you estimator will be asymptotically efficient and $V(g(\mathbf{X})) \to \frac{1}{\mathcal{I}[g(\mathbf{X})]}$. Otherwise there will be an estimator with the variance smaller than yours. But since your estimator is (at least) asymptotically unbiased then it will not be smaller than the one, given by the Cramer-Rao bound: $V(g(\mathbf{X}))\geq \frac{1}{\mathcal{I}[g(\mathbf{X})]}$.

Caran-d'Ache
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