Which of this two estimators of $\mu$ is better (Exponential distribution)? The problem goes like this:
"Suppose we have some electronic device which duration follows an Exponential distribution of an unknown mean $\mu$. We want to estimate $\mu$ and two teams will take care of it. Each team tests $n$ different devices:


*

*First team measures the time $T$ in which the first device fails, and provides the estimation $T_1=nT$.

*Second team measures each device's duration and estimate $T_2$ as the mean of their durations.
Which team uses a better estimator of $\mu$?"
I know that the second team is obviously using a better estimator of $\mu$, but i'm not sure how to prove it. I've tried using the definition of allowable estimator, that says that every allowable estimator is asymptotically centered ($(\hat{\theta}_n)$ is asymptotically centered as a $\theta$ estimator if $\lim\limits_{n \rightarrow \infty}\tau_n(\theta)=\theta$, being $\tau_n(\theta)=\text{E}[\hat{\theta}_n(X_i)]$).
Calculating that limit for the team two's estimator gives me $\mu$, because the sample mean goes to $\mu$ when the number of devices sampled go to infinity.
My real problem is with $T_1$. I don't get why it multiplies the time $T$ of the first devide by $n$, and evaluating that limit gives $\infty$ if i'm not wrong:
$$\lim\limits_{n \rightarrow \infty}\tau_n(n\cdot T)=\text{E}[\infty \cdot T]=\infty.$$
Is my reasoning correct? Or what am i doing wrong?
 A: You asked a similar question more recently, but I am going to provide a separate answer to this question since the comments attached to this question are mistaken.
What the other commenters in this question have misunderstood is that the $n$ devices are being tested simultaneously and not sequentially; therefore, the time to failure of the first device to fail is exponential with mean $\mu/n$, which is what you calculated in the more recent linked question.  Multiplying this by $n$ yields an estimator for the mean time to failure; however, as I have pointed out there, the asymptotic variance of this estimator is not a decreasing function of the sample size.  This makes it an inferior estimator compared to the one used by the second team.
It is worth noting that $$\lim_{n \to \infty} \operatorname{E}[nT_{(1)}] \ne \operatorname{E}[\infty \cdot T_{(1)}].$$  This is of course an invalid interchange of limits and expectations as well as not recognizing that $T_{(1)}$ has a limiting expectation of $0$.  Instead, $$\operatorname{E}[n T_{(1)}] = n \operatorname{E}[T_{(1)}] = n (\mu/n) = \mu.$$
