Let's suppose we have some electronic device which duration follows an Exponential distribution of unknown mean $\mu$. Some research team wants to estimate $\mu$ and uses a sample of $n$ devices to do it. They use the estimator $T=n\cdot \min\{T_1,...,T_n\}$, to be said, $n$ times the time in what the first device fails when they put to work all at the same time.

I need to prove that this estimator $T$ is not allowable. That would mean that $T$ isn't asymptotically centered or that $T$ isn't consistent.

I've found the distribution function of $\min\{T_1,...,T_n\}$ (I'll call it $F$): $$F(t)=P(\min\{T_1,...,T_n\}\leq t)=1-P(\min\{T_1,...,T_n\}> t) =$$ $$= 1-P(T_1>t,T_2>t,...,T_n>t)=1-(P(T_1>t))^n=1-e^{-nt/\mu}.$$ I'm stucked here. I need to prove it's NOT allowable.


You are on the right track but you need to keep going a bit further.

The distribution of the first order statistic $$T_{(1)} = \min (T_1, T_2, \ldots, T_n)$$ is indeed exponential with CDF $$\Pr[T_{(1)} \le t] = 1 - e^{-nt/\mu}.$$ The next step is to compute the distribution of the scale transformed first order statistic: this is your actual $T$: $$T = n T_{(1)} = n \min (T_1, T_2, \ldots, T_n).$$ This is not hard to do; I leave it as an exercise to show $$T \sim \operatorname{Exponential}(\mu)$$ with CDF $$\Pr[T \le t] = 1 - e^{-t/\mu}.$$ What this tells us is that $n$ times the minimum observation is exponentially distributed with the same mean parameter, and our intuition should lead us to observe that if this is the case, this statistic is no better than simply observing one observation: in fact, because the CDF of $T$ is independent of $n$ entirely, this means its characteristics as an estimator of $\mu$ is also independent of the sample size! So for instance, the asymptotic variance does not tend to $0$; we explicitly have $$\operatorname{Var}[T] = \mu^2$$ hence $$\lim_{n \to \infty} \operatorname{Var}[T] = \mu^2 > 0.$$ This is an undesirable characteristic of an estimator for the parameter because it says that the precision of the estimate does not decrease with increasing sample size, so there is no information to be gained about the parameter by collecting more data if you use this estimator.

Stepping back, ask yourself why this happens. While it is true that the first order statistic $T_{(1)}$ has a decreasing variance with increasing sample size, the problem is that the rate of this decrease is not stronger than the increase in variance that occurs when we multiply $T_{(1)}$ by $n$. As you know, for a scalar constant $c$ and a random variable $X$ with finite variance, we have $$\operatorname{Var}[cX] = c^2 \operatorname{Var}[X].$$ This means that scale transformations of a random variable have a squaring effect on its variance. So if we must scale the first order statistic by $n$ in order to get an estimate, then the variance of this statistic must decrease faster than $n^2$ in order to compensate for the increase due to scaling it. And this does not occur; in fact, the two effects balance each other out exactly in this case.

I should point out here that "not allowable" is a bit strong. I would prefer to characterize the estimator $T$ to be "poor" or "undesirable." After all, it is an estimator of $\mu$--just not a good one.

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  • $\begingroup$ Thank you for your answer! I solved it by myself a few hours ago and my solution is the same. Your explanation is very clear, now I understand it a lot better, thank you so much. $\endgroup$ – Alejandro Bergasa Alonso May 22 at 18:57

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