2
$\begingroup$

Many times I've read that the following expresion $(x_1+x_2+...+x_n)/n$ is an estimator of the population mean. I also read, for example here: that the previous expresion defines a sequence of estimators, indexed by the sample size n.

Is the difference a matter of definitions or due to language abuse? What is the proper way to refere to it? According to Wikipedia an estimator "is a rule for calculating an estimate of a given quantity based on observed data" and "is a function that maps the sample space to a set of sample estimates". It seems to me that according to the second definition it is due to language abuse (because we will have functions of different number of variables), but I am not sure.

How does this generalize to the case when another statistic is applied to it? Take, for example, the variance of the mean. For instance, if $\hat T_n=(x_1+x_2+...+x_n)/n$ we would have an $\hat S^2_n(\hat T_n)$. Are each $\hat S^2_n$ different estimators, and $S^2=(S^2_1,S^2_2,...,S^2_n)$ a sequence of estimators? Or can we just call $S^2$ an estimator.

$\endgroup$

1 Answer 1

1
$\begingroup$

If you fix $n$, there is an estimator. If you consider all the indices in the index set $I=\{1, \ldots, n\}$, you have a sequence of estimators. When we say that expression defines a sequence of estimators, we are then talking about the sequence $$\left( \frac{x_1 + \ldots + x_n}{n} \right)_{i = 1}^n.$$

If you take $n$ to be some random value like $5$, you can maybe see more concretely that $\hat{T}_5$ is an estimator and $(\hat{T}_i)_{i=1}^5 = (\hat{T}_1, \hat{T}_2, \ldots, \hat{T}_5)$ is a sequence of estimators. And each element of the sequence is an individual estimator. Similarly, for your second question, you have an estimator $\hat{S}^2_n$ and a sequence of estimators over the index set.

And if we really wanted to be pedantic, we could say $$(\hat{T}_i)_{i \in I}, \quad I = \{n\}$$ or something like that is a sequence of estimators of length one, but that is not very useful.

Usually, it is useful to consider the sequence of estimators we can define with an expression based on $n$ when we want to think about the asymptotic properties of the estimator. But for a typical real-life sample, we have a constant $n$ so we are only talking about one estimator. So it's typically clear from context whether we are talking about a single estimator or a sequence of estimators.

$\endgroup$
3
  • $\begingroup$ +1 Thank you. Then, when it is clear from the context, is it allowable to refer generically to an "estimator"? My intention is to present a sequence of estimators in a non-mathematical field article. I do not want to be pedantic, but neither too imprecise. $\endgroup$ Commented Sep 7, 2023 at 14:43
  • 1
    $\begingroup$ estimator usually ( I don't know the percentage of time ) refers to the "kind of" estimate formula being used in the calc of the estimate. So, IMHO, I think sequence of estimates is better than sequence of estimators. For example, the sample variance, $\sum_{i=1}^{n} \frac{(X_{i} - \mu)^2}{(n-1)}$, is an estimator for the population variance. When you calculate the sample variance and obtain a number, you have an estimate of the population variance. But the sample range is another estimator for the population mean. sample range and sample variance are estimators for the population variance. $\endgroup$
    – mlofton
    Commented Sep 7, 2023 at 19:53
  • 1
    $\begingroup$ @user1420303 I think if you are going to present a sequence, you should call it a sequence of estimators (or give the whole sequence a name, like you did with $S^2$ in your question and call it that). In non-math settings I think people will typically think of "the estimator" as the formula with the $n$ in it, so the last element in a sequence of this type and might get confused if you call an entire sequence "the estimator". $\endgroup$
    – wzbillings
    Commented Sep 8, 2023 at 1:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.