Any idea for the following question?

  1. 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?

  • $\begingroup$ tell us more about what you tried? $\endgroup$ Commented Oct 31, 2021 at 5:46
  • $\begingroup$ thanks for your suggestion. I modified the question a little bit $\endgroup$
    – Alex
    Commented Oct 31, 2021 at 5:52

1 Answer 1


There are no issues in this exercise. The absolute value, related to the data, does not affect $\theta$ estimation

Your likelihood is

$$L(\theta)\propto \theta^n\prod_i|x_i|^\theta$$

$$l(\theta)=n\log \theta+\theta\sum_i\log|x_i|$$


That is


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...

  • $\begingroup$ thank you for your reply @tommik. However, the value of information matrix does depend of the $\theta$, am I right? $\theta$ could be an even or an odd number? $\endgroup$
    – Alex
    Commented Oct 31, 2021 at 6:45
  • $\begingroup$ @MathildaFang : $\theta>0$ is a real parameter: $\theta \in \mathbb{R}^+$ $\endgroup$
    – tommik
    Commented Oct 31, 2021 at 7:00
  • $\begingroup$ I still have some problems of solve questions and I edited the question. feel free to check and many thanks in advance $\endgroup$
    – Alex
    Commented Oct 31, 2021 at 7:05
  • $\begingroup$ @MathildaFang : Yes, I can surely solve your problem but I think that is better for you to do more efforts by yourself. Hints: the variance of $X_1$ is simply the variance of your rv X...as per asymptotic CI is concerned, you can observe that MLE are asymptotically gaussian... $\endgroup$
    – tommik
    Commented Oct 31, 2021 at 7:09
  • $\begingroup$ thanks @tommik. I solve the MLE of $\theta$ is 0.9488 and therefore $var(\theta)$ doesn't exist. Do I miss something here? $\endgroup$
    – Alex
    Commented Oct 31, 2021 at 7:12

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