Let $X$ a random variable with density $$f(x)=\lambda\theta^\lambda x^{-(\lambda+1)},$$ with $x\geq \theta$, $\lambda>0$ and $\theta>0$.

Let $S=\min\{X_1,\cdots, X_n\}$.

I have the following two questions:

  1. Is $S$ a sufficient and minimal estimator of the parameter $\theta$?
  2. Is $S$ a unbiased estimator?

I can prove that $S$ is sufficient and that $S$ is the estimator that comes from the maximum likelihood method. I don't know how to prove if $S$ is sufficient or correct.

I also found the distribution of $S$ that, if I am correct, is $P(S\leq x)=\Big(1-\frac{\theta^\lambda}{x^\lambda}\Big)^n$.

  • 1
    $\begingroup$ What is a 'correct' estimator? $\endgroup$ Sep 2, 2020 at 21:21
  • $\begingroup$ An unbiased estimator. $\endgroup$
    – user268193
    Sep 3, 2020 at 7:37

1 Answer 1


If $X \ge \theta$, then $\Pr[X < \theta] = 0$, and similarly, $S = \min\{X_1, \ldots, X_n\}$ implies $S \ge \theta$ and $\Pr[S < \theta] = 0$. Moreover, $\Pr[S > \theta] > 0$: why? What does this tell you about $\operatorname{E}[S]$?

As for minimal sufficiency, use the property that $S$ is minimal sufficient if $$\frac{f_\theta(\boldsymbol x)}{f_\theta(\boldsymbol y)}$$ is independent of $\theta$ if and only if $S(\boldsymbol x) = S(\boldsymbol y)$ for samples $\boldsymbol x$ and $\boldsymbol y$.

  • $\begingroup$ Thank you for your answer. I agree that $P(S<\theta)=0$ and I thing that $P(S>\theta)=(P(X>\theta))^n=1$. Therefore $E(S)>\theta$ and consequently the estimator $S$ is not unbiased. Is this correct? $\endgroup$
    – user268193
    Sep 3, 2020 at 9:44
  • $\begingroup$ If this is correct I have a question. $S$ is the estimator that I find also using the maximum likelihood method. Is not true that the maximum likelihood estimators are unbiased? $\endgroup$
    – user268193
    Sep 3, 2020 at 9:47
  • $\begingroup$ About the minimality I can conclude that $S$ is sufficient but not minimal because that ratio in my case is always independent if $\theta$. Is this correct? Can I ask the name of this criteria for the minimality? $\endgroup$
    – user268193
    Sep 3, 2020 at 9:57
  • $\begingroup$ @user268193 $S$ is biased as you correctly deduced in your first comment. This also illustrates that maximum likelihood estimators are not always unbiased. As another example, the MLE for the variance of a normal distribution with known mean $\mu$ is a biased estimator of $\sigma^2$, hence the existence of Bessel's correction. $\endgroup$
    – heropup
    Sep 3, 2020 at 10:08
  • $\begingroup$ @user268193 As for minimality, you need to remember that the density is actually $$f_\theta(x \mid \lambda) = \lambda \theta^\lambda x^{-(\lambda+1)} \color{red}{\mathbb 1 (x \ge \theta)}.$$ Thus the joint density for a sample has a dependence of $\theta$ on the sample $\boldsymbol x$ and you cannot ignore it when you compute the quotient. But because this dependence is reducible to an equivalent statement about the sufficient statistic $S$, it becomes clear that minimality is satisfied. $\endgroup$
    – heropup
    Sep 3, 2020 at 10:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .