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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?

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    $\begingroup$ You probably meant the log-likelihood $\endgroup$
    – Firebug
    Dec 1, 2023 at 12:54
  • $\begingroup$ Please provide the expression you derived for the (log-)likelihood $\endgroup$
    – Firebug
    Dec 1, 2023 at 13:12
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    $\begingroup$ of course! The log-likelihood that i've found is: $$l (\theta \mid X) = n \log{3} + 3n \log{\theta} - 4\log {\sum_{i = 1}^{n} X_i}$$ But i fell like im missing something, in this approach i completely ignored $\mathbf{1}_{[\theta,\infty)}(x)$ that is just a long way to say that the density is defined at the points where $x \ge \theta$ $\endgroup$
    – Occhima
    Dec 1, 2023 at 13:18
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    $\begingroup$ You shouldn't ignore that term $\endgroup$
    – Firebug
    Dec 1, 2023 at 13:43
  • $\begingroup$ Best to edit the new information into the question as comments are not permanent. $\endgroup$
    – mdewey
    Dec 1, 2023 at 15:53

1 Answer 1

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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?

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