Asymptotic variance of an estimator of the parameter of esponential distribution. $**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 $I\left(\theta\right)$?

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. 
