Let $X_1,...,X_n$ be a sample with distribution

$p_{\theta}(x)=\theta x^{-2}$, for $x \geq \theta$

and $0$ for $x < \theta$ with $\theta > 0$ unknown.

Determine the maximum likelihood-estimator of $\theta$.


We can do this with the log-likelihoodfunction which we take the derivative of and set equal to $0$ in order to find a maximum. By doing this we get

$\frac{n}{\theta} = 0$

I have no idea where to go from here since this doesn't have a solution for $\theta$.

  • 1
    $\begingroup$ Some funny stuff happens because your parameter defines the range. en.wikipedia.org/wiki/… $\endgroup$ – T.J. Gaffney Nov 20 '16 at 23:46
  • $\begingroup$ This means it has no maximum? The answers say that $X_1$ is what we are looking for. I have no clue why though. $\endgroup$ – Vydai Nov 20 '16 at 23:49
  • $\begingroup$ I think you should look for the minimum of $X_i$, rather than $X_1$ $\endgroup$ – Med Nov 20 '16 at 23:55
  • $\begingroup$ The solution set probably mentions $X_{(1)}$, not $X_1$. $\endgroup$ – Did Nov 21 '16 at 0:01
  • $\begingroup$ @Did You are right. What does $X_{(1)}$ mean in this context? $\endgroup$ – Vydai Nov 21 '16 at 0:08

We want to maximise this probability.


A formula is used to get


If the experiments are independent, then the last result can be simplified to


Taking the logarithm, we get


Now, there is one thing to pay attention to. If $\theta$ is chosen to be greater than even one of $X_i$, then we have a term $logP(X_i|\theta)=log0$ in the sum, according to the probability model that you defined. Therefore, we would like to have $\theta$ to be less than all $X_i$. In this case the summation can be modified as

$\sum_{i}logP(X_i|\theta)=\sum_{i}log (\theta X_i^{-2})=nlog\theta+\sum_{i}logX_i^{-2}$

If there was no limit on $\theta$, it could have gone to infinity and therefore, it had no maximum. But it was assumed that $\theta $ is lees than the minimum of $X_i$, for all $i={1,2,...,n}$. So, $\theta=min{X_i}$


Your likelihood function is $$ L(\theta) = \theta^n \left[ \prod_{i=1}^n x_i^{-2} \cdot \mathbb{1}_{[x_i \ge \theta]} \right] =\theta^n \cdot \mathbb{1}_{[x_{(1)} \ge \theta]} \left[ \prod_{i=1}^n x_i^{-2} \right] $$ Now, $L(\theta)$ is increasing in $\theta$ as long as $\theta \le x_{(1)}$, and so the MLE is $\hat \theta_{MLE} = X_{(1)}$.


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