Convergence in distribution to the standard normal using Cramer-Rao 
Given is a a random sample $X_1,\ldots,X_n$ from a distribution with a pdf $f(x;\theta) = 2\theta^{-1} x^{(2/\theta)-1}$ for $0<x<1$, zero otherwise. We know that maximum likelihood estimator is $\hat{\theta} = -\dfrac 2 n \sum_i \log X_i$. Determine $c$ such that $c\left(\dfrac{n}{-2\sum_i \log X_i} - \dfrac{1}{\theta}\right) \xrightarrow{\text{d}} Z  \sim N(0,1)$, i.e. converges in distribution.

So in the correction sheet it says the answer is $\dfrac{1}{\sqrt{CRLB}}$ where CRLB stands for the Cramer Rao Lower Bound. I don't really understand this. I have found in my book that for large $n$ $\hat{\theta}_n \sim N(\theta, CRLB)$, but how can we normalise this result to a $N(0,1)$?
 A: $X_1, X_2, \ldots, X_n$ is a sample from distribution with pdf $f(x; \theta) = 2\theta^{-1}x^{(2/\theta)-1}$ for $x \in (0,1)$ and zero otherwise.
Now, the distribution of $Y=-2\log X$ is derived as follows.
$$X=e^{-\frac{Y}{2}}$$
$$\frac{dX}{dY} = -\frac{1}{2}e^{-\frac{Y}{2}}$$
The pdf of $Y$ is given by,
$$g(y;\theta) = f(e^{-\frac{Y}{2}};\theta)\left | \frac{dX}{dY} \right | = \frac{1}{\theta}e^{-\frac{y}{\theta}}$$
Therefore, $Y_1, Y_2, \ldots, Y_n$ is transformed sample with $Y_i = -2\log X_i$ with $EY_i = \theta$ and $VarY_i = \theta^2$.
Now, one can use central limits theorem to get,
$$\sqrt{n}\frac{(\overline{Y_n} - \theta)}{\theta} \rightarrow \mathcal{N}(0,1)$$
Let $h(z) = \frac{1}{z}$. Using the first order taylor approximation to estimate the limiting distribution of $h(\overline{Y_n})$ (which is $\frac{1}{\hat{\theta}}$ as per your notation), we get,
$$h(\overline{Y_n}) \approx h(\theta) + h^{'}(\theta)(\overline{Y_n}-\theta)$$
$$\sqrt{n}\frac{(h(\overline{Y_n}) - h(\theta))}{|h^{'}(\theta)|\theta} \approx \sqrt{n}\frac{(\overline{Y_n} - \theta)}{\theta} \rightarrow \mathcal{N}(0,1)$$
$$\sqrt{n}\theta(h(\overline{Y_n}) - h(\theta)) = \sqrt{n}\theta(\frac{1}{\hat{\theta}} - \frac{1}{\theta})\rightarrow \mathcal{N}(0,1)$$
Note that I used $\frac{|h^{'}(\theta)|\theta}{\sqrt{n}}$ in the denominator because it is the standard deviation of $h(\overline{Y_n})$ which is the square root of the variance $\frac{h^{'}(\theta)^2\theta^2}{n}$. That's why || came up.
So the $c$ you are looking for is $\sqrt{n}\theta.$
Edit 1. If $X \sim \mathcal{N}(\mu,\sigma^2)$ then $\frac{X-\mu}{\sigma} \sim \mathcal{N}(0,1)$
