# Sufficient, minimal and correct estimator

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 $$P(S\leq x)=\Big(1-\frac{\theta^\lambda}{x^\lambda}\Big)^n$$.

• What is a 'correct' estimator? Sep 2, 2020 at 21:21
• An unbiased estimator. Sep 3, 2020 at 7:37

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 $$\boldsymbol y$$.
• 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? Sep 3, 2020 at 9:44
• 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? Sep 3, 2020 at 9:47
• 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? Sep 3, 2020 at 9:57
• @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. Sep 3, 2020 at 10:08
• @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. Sep 3, 2020 at 10:12