Maximum Likelihood Estimation for a Unique Probability Density Function

Question

In the context of estimating parameters for a uniquely distributed set of independent and identically distributed random variables, I am examining the following probability density function $ f(x|\theta) = \frac{3\theta^3}{x^4} \mathbf{1}_{[\theta,\infty)}(x) $ with $ \theta > 0 $.

(a) For the estimation, I am tasked to derive the maximum likelihood estimator (MLE) for $ \theta $ based on a sample $ X_1, \ldots, X_n $. I have started by setting up the likelihood function and taking the derivative with respect to $ \theta $, but I always get that $\hat{\theta} = 0$ because:

$$\frac{\partial l(\theta \mid X)}{\partial \theta} = \frac{3n}{\theta}$$

Is this correct?

(b) After deriving the MLE, I need to determine its distribution and whether the estimator is unbiased for $ \theta $. I am aware that the unbiasedness of an estimator is linked to its expected value equalling the parameter. However, I am struggling to find the distribution of the MLE, how do I do this? Is there any theorem for the distribution of the MLE or there is a hidden "hint" in the question?

Answer 1

As said in the comments, you cannot ignore the indicator functions.

Note that, for $\theta \in \mathbb R_{> 0}, n \in \mathbb N_{\geq 1}$, we have $$ \begin{align} \prod_{i=1}^n \left[\frac{3\theta^3}{x_i^4} \mathbf 1_{\left[\theta, \infty\right)}\left(x_i\right)\right] &\overset{\theta}{\propto} \theta^{3n} \prod_{i=1}^n \mathbf 1_{\left[\theta, \infty\right)}\left(x_i\right) \\ &= \begin{cases} \theta^{3n}, &x_i\geq \theta \;\forall i \in \{1,\ldots, n\}\\ 0, &\text{otw.} \end{cases} \\ &= \begin{cases} \theta^{3n}, &\theta \leq \min\left(\left\{x_i: i \in \left\{1, \ldots, n\right\}\right\}\right)\\ 0, &\text{otw.} \end{cases} \end{align} $$ for the likelihood function of $\theta$.

Also, $\theta \mapsto \theta^{3n}$ is strictly increasing on $(0, \infty)$ for $n \in \mathbb N_{\geq 1}$.

Can you take it from here?