Sample mean is always an optimal estimator of the mean? Suppose we have $T_i,i=1..n$ i.i.d. with unknown distribution and we want to estimate $E[T]$. Note that in this setting we are not estimating E[T] as a parameter of a parameter-dependent family of distributions, therefore it is difficult to attach a meaning to a likelyhood function:
$L[E[T]]=P(T_i|E[T])$
as it would be done when for example estimating $\mu$ from maximum likelyhood, knowing that the underlying distribution is Gaussian.
Again what we would probably do is computing $\overline{T}=\frac{1}{n}\sum_i T_i$. And we would be sure that the estimator would be consistent and its variance would go to zero. For example we have trivially:
$E[\overline{T}]=E[T_i]$
and:
$\sigma^2[\overline{T}]=O(1/n)$
supposing that $T_i$ has finite variance.
Here is the question: can we prove that $\overline{T}$ is optimal in some way? To me conceptes from MLE estimators or sufficient statistics are a bit difficult to apply, since $E[T]$ is not a parameter of the distribution but maybe I am missing something? Can we "derive" the sample mean estimator to be optimal according to some criterion in the general case and without making assumptions on the underlying distributions ?
 A: Though the sample mean is an unbiased estimator of the unknown population mean, it cannot be optimal in general.  Take the case of the log-normal distribution.  The maximum likelihood estimator of the mean on the original scale is a function of the sample mean and sample variance both computed on the log scale.  This prevents outliers from ruining either the mean or SD.  There is a relationship of this problem with that fact that an accurate nonparametric confidence interval for the population mean does not exist.  When one wants to have a measure that 'works' on all continuous distributions, one has to use an estimator that aligns with nonparametrics, such as the sample median.  BY doing so one pays a high efficiency price if the data are Gaussian, since the sample median in that case is an inefficient estimator of the population mean or median.
A: What you stated in your original post is that the sample mean converges to the true mean in probability (in fact, it does with probability one) if expectations and variances are finite, i.e., the Law of Large Numbers. Together with unbiasedness, I think this makes the sample mean "optimal enough". No distributional assumptions are needed except finite first and second moments. Notice that this works equally well to estimate probabilities of events, since $P(E) = E[I_E]$. Yes, if you want to claim that the sample mean is the MLE then you need distributional assumptions.
