For instance, we would could have as a distributions $f_X(x) = \theta e^{\theta x}$. In this case it is a exponential distribution, and would our parameter be the usual $\beta = \frac{1}{\theta}$, $\operatorname{Var}(\widehat{\beta_\text{MLE}}) = \frac{\sigma^2}{n}$ and it is easy to calculate, but because here $\hat{\theta} = \frac{1}{\bar{X}}$, the inversion of $X$ is what makes it tricky... (I have no idea how to calculate $\operatorname{Var}(\frac{1}{\bar{X}})$)
Thank you very much !
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Assuming that $\bar{X}$ is the mean of your data, with CLT assumptions we have that $\bar{X}$ is asymptotically normal with mean $\mu$ and variance $\frac{\sigma^2}{n}$. You can then approximate:
$$\frac{1}{\bar{X}}\approx \frac{1}{\mu}-\frac{1}{\mu}(\bar{X}-\mu)+\frac{1}{\mu^2}(\bar{X}-\mu)^2+...$$
Now proceed to calculate the variance by squaring the above and keeping terms up to order $X^2$.
$$E[\frac{1}{\bar{X}^2}]\approx \frac{1}{\mu^2}+\frac{1}{\mu^2}\frac{\sigma^2}{n}+\frac{1}{\mu^3}\sigma^2/n+O(1/n^2),$$
$$E[1/\bar{X}]\approx 1/\mu+\frac{\sigma^2/n}{\mu^2},$$
which will give you an approximation to $\mbox{Var}(\theta)$. Note that if $\bar{X}$ is exactly normal, then the expectation and variance of $1/\bar{X}$ will actually blow up, so you need some strong assumptions on either $\bar{X}$ being bounded away from zero, or decaying sufficiently fast near 0.
If $(Y_k)$ is i.i.d. standard exponential, that is, with PDF $$f(y)=e^{-y}\mathbf 1_{y>0}$$ then, for every $n\geqslant1$, $T_n=Y_1+\cdots+Y_n$ has PDF $$f_n(t)=\frac{t^{n-1}}{(n-1)!}e^{-t}\mathbf 1_{t>0}$$ hence, for every $n\geqslant2$, $$E\left((T_n)^{-1}\right)=\frac1{n-1}$$ and, for every $n\geqslant3$, $$E\left((T_n)^{-2}\right)=\frac1{(n-1)(n-2)}$$ hence $$\mathrm{Var}\left((T_n)^{-1}\right)=\frac1{(n-1)^2(n-2)}$$ Now, each of your random variables $X_k$ is distributed like $Y_k/\theta$ hence your $\bar X_n$ is distributed like $$\frac1{n\theta}T_n$$ hence, for every $n\geqslant3$, $$\mathrm{Var}\left((\bar X_n)^{-1}\right)=\frac{n^2\theta^2}{(n-1)^2(n-2)}$$