We are given $f(x|\theta) = (\theta + 1)x^{\theta} , \hspace{7mm} 0<x<1, \theta > -1.$
By considering the asymptotic distribution of $\frac{1}{\theta_{MLE}+1}$, find the asymptotic distribution of $\theta_{MLE}$.
My attempt:
I have worked out that the MLE to this question is $$\theta _{MLE} = \frac{-n}{\sum _{i=1}^{\infty}\ln(X_i)}-1$$
and I also worked out that $-\ln(X_i)$ $\sim \exp(\frac{1}{\theta + 1})$
Now, $$\frac{1}{\theta_{MLE}+1} = \frac{\sum _{i=1}^{\infty} -\ln(X_i)}{n}$$
I don't know how to proceed from here as the question asks for asymptotic distribution for $\theta_{MLE}$ yet what can I extract from $\frac{1}{\theta_{MLE}+1}$ ?
How do I compute the asymptotic distribution for a question like this?
Note that you know the exact distribution of the MLE, $\hat{\theta}$. Namely, as $-\ln X_i = Y_i$ has an exponential distribution with $\theta+1$, that is $$ f_Y(y)=(\theta+1)\exp\{ - y(\theta+1)\}, \quad y>0, $$ thus $ \sum Y_i $ distributed Gamma with $n$ and $\theta + 1$. So $(\sum Y_i)^{-1}$ has an inverse Gamma distribution with the same parameters. As such, $$ \hat{\theta}_n = n/\sum Y_i - 1 \equiv n\times \mathrm{InvGamma(n,\theta+1) -1}. $$ For the asymptotic properties define $g(x) = 1/x -1$, thus the MLE is
$$ \hat{\theta}_n = g(\bar{Y}_n) = 1/\bar{Y}_n-1, $$ by the continuous mapping theorem $$ g(\bar{Y}_n) \xrightarrow{P}g(\mathbb{E}Y)=(1/(\theta + 1))^{-1} - 1= \theta, $$ hence by the CLT $$ \sqrt{n}( g(\bar{Y}_n) - \theta)\xrightarrow{D} N(0, (g'(\theta))^2/(\theta+1)^2), $$ where $(g'(\theta))^2 = 1 / \theta ^ 2$.