Find a confidence interval using as pivotal quantity a function of the MLE 
Let $X_1,\ldots,X_n$ be a random sample from $f(x\mid\theta)=\theta x^{\theta -1}$ for $0<x<1$. Find a confidence interval for $\theta$ using as pivotal quantity a function of the maximum likelihood estimator for $\theta$. 

Well, using the logaritmic version of the likelihood function, I got that the MLE of $\theta$ is $T_1=\frac{-n}{\sum\limits_{i=1}^n \log x_i}$. Is this correct? And how could I use a function of it as a pivotal quantity?
I know that a pivotal quantity is a function of the sample and the parameter, whose distribution doesn't depend on $\theta$.
 A: Assuming you want a confidence interval for $\theta$, you may use the CLT or a known chi distribution for variances and some integrals to get an interval for the parameter. First, we have that
$$
\hat{\theta}=-\frac{n}{\sum ln\ x_i}.
$$
For the expected value we have the integral
$$
E[X] = \int_0^1 \theta x^{(\theta-1)}x\ \text{d}\theta = \frac{\theta}{1+\theta}
$$
And for the second moment
$$
\mu_2 = \int_0^1 \theta x^{\theta-1}x^2\ \text{d}\theta = \frac{\theta}{2+\theta}
$$
So the variance is
$$
Var(X) = \frac{\theta}{2+\theta} - \left( \frac{\theta}{1+\theta} \right)^2 = \\
= \frac{\theta}{(\theta + 1)^2(\theta+2)}.
$$
Now then, we know that the following quotient has a chi-squared distribution, so in this case
$$
C = \frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1}
\text{ and } C = \frac{(n-1)S^2}{\frac{\theta}{(\theta + 1)^2(\theta+2)}}
$$
where $S^2 = \frac{1}{n-1}\sum(X_i-\overline{X})^2$. Finally we can work with the random variable $C$ distributed as a chi squared. We want a confidence interval of $(1-\alpha)$. Then
$$
P \left( \chi^2_{n-1}(\alpha/2) < C < \chi^2_{n-1}(1-\alpha/2)\right) = 1-\alpha
$$
denoting that $\chi^2_{n-1}(\beta)$ is the $\beta$ percentile of the $\chi^2$ which may be found in reference tables.
From here we work with the interval only substituting the percentiles with $\frac{1}{D_l}$ and $\frac{1}{D_s}$, lower and superior.
$$
\frac{1}{D_l} < C < \frac{1}{D_s} \implies
\frac{1}{D_l} < \frac{(n-1)S^2}{\frac{\theta}{(\theta + 1)^2(\theta+2)}} < \frac{1}{D_s} \\\ \\\ \\\
D_l > \frac{\frac{\theta}{(\theta + 1)^2(\theta+2)}}{(n-1)S^2} > D_s \\\ \\\ \\\
D_l((n-1)S^2) > \frac{\theta}{(\theta + 1)^2(\theta+2)} > D_s((n-1)S^2) \\\ \\\ \\\
(D_l((n-1)S^2)) > \frac{\theta}{(\theta + 1)^2(\theta+2)} > (D_s((n-1)S^2)) \\\ \\\ \\\
(\hat{\theta} + 1)^2(\hat{\theta}+2)\left( D_l((n-1)S^2) \right)
> \theta >
(\hat{\theta}+ 1)^2(\hat{\theta}+2)\left( D_s((n-1)S^2) \right) \\\ \\\ \\\
$$
where, taking the before definition and derivation, $\hat{\theta} = -\frac{n}{\sum ln\ x_i}$ and $S^2 = \frac{1}{n-1}\sum(X_i-\overline{X})^2$. When we sent the $\theta$ to the other side, we simply used its estimate to work around the problem. (However a mathematician should confirm this is valid!) Either way, if you find a better known distribution that may include $\theta$, then try it and tell us if it is simpler!
Hope it helps!
Note: made some edits due to an error :/
