MLE calculation with absolute value involved Any idea for the following question?


*Let $\left\{X_{1}, X_{2}, \ldots, X_{n}\right\}$ be a random sample from a distribution with the following pdf:
$$
f(x ; \theta)=\frac{\theta}{2}|x|^{\theta-1}, \quad-1 \leq x \leq 1
$$
The density is zero elsewhere, and $\theta$ is a positive parameter.
(a) Derive the maximum likelihood estimator $\hat{\theta}$ of $\theta$.

I can write the loglikelihood function but I don't how to deal with the absolute value part. Should I discuss different cases and multiply the two cases?
Also, for the following question
(b) Suppose the following data set is observed:
$\begin{array}{cccccccccc}0.21 & -0.42 & -0.15 & 0.72 & -0.64 & -0.83 & 0.55 & 0.47 & 0.06 & -0.28 \\ 0.73 & 0.64 & -0.21 & -0.47 & 0.16 & 0.05 & -0.44 & 0.35 & 0.92 & -0.61\end{array}$
Using asymptotic theories regarding maximum likelihood estimators, construct $95 \%$ confidence intervals for:
i. the parameter $\theta$;
ii. the variance of the distribution of $X_{1}$.
I calculate the information matrix $var(\theta) = \frac{2\theta} {n [\frac{1}{\theta} (-1)^{\theta} + \frac{1}{\theta}]}$
and based on the MLE, the estimate value of $\theta$ is 0.9488. which implies variance doesn't exist. Can anyone help？
 A: There are no issues in this exercise. The absolute value, related to the data, does not affect $\theta$ estimation
Your likelihood is
$$L(\theta)\propto \theta^n\prod_i|x_i|^\theta$$
$$l(\theta)=n\log \theta+\theta\sum_i\log|x_i|$$
$$l'(\theta)=\frac{n}{\theta}+\sum_i\log|x_i|$$
That is
$$\hat{\theta}_{ML}=\frac{-n}{\sum_i\log|x_i|}$$

Further qeuestions:
(i) The given data gives you the sample mean...to calculate the asymptotic interval you have to use MLE's properties; in particular, it is well known that
$$\sqrt{n\cdot I_X(\hat{\theta})}[\hat{\theta}-\theta]\xrightarrow{\mathcal{L}}N(0;1)$$
In your specific case, after some calculation about Cramér Rao's inequality you get
$$\hat{\theta}_{ML}\xrightarrow{\mathcal{L}}N\left(\theta;\frac{\theta^2}{n}  \right)$$
here are all the calculations
$$\log f(x|\theta)=\log\theta+\theta\log|x|$$
$$\frac{\partial}{\partial\theta}\log f(x|\theta)=\frac{1}{\theta}+\log|x|$$
$$\frac{\partial^2}{\partial\theta^2}\log f(x|\theta)=-\frac{1}{\theta^2}$$
$$-n\mathbb{E}\left[ \frac{\partial^2}{\partial\theta^2}\log f(x|\theta) \right]=\frac{n}{\theta^2}$$
(ii) the variance of $X_1$ is the same variance of your original rv...
