1
$\begingroup$

Data-generating process: $$ y = f(X) + \epsilon $$ The goal is to estimate $E[y|X]$. One estimator is simply $\bar{y}$. Many other estimators are better, inasmuch as the variance of $\hat{\epsilon}$ will be less than the variance of $y$.

Is there a term for such estimators? Basically all estimators that are better than the sample mean and aren't necessarily optimal in any sense.

Note: I'm not even thinking about in-sample/out-of-sample here. Just estimators that yield a residual with a lower variance than $y$.

Improve this question
$\endgroup$
5
  • $\begingroup$ Like, something snazzier than "not-ridiculous"? $\endgroup$
    – generic_user
    Jul 6, 2017 at 21:28
  • $\begingroup$ Would it not be better to just refer to the relative efficiency of your estimator and the sample mean? Can't think of any term that refers specifically to the sample mean myself. $\endgroup$
    – Patty
    Jul 7, 2017 at 3:08
  • 1
    $\begingroup$ Given that in the context of linear models (and indeed, some non-linear ones) the model for the mean that doesn't vary with changing $x$'s: $E(y|X)=\mu$ would be called a null-model, some obvious names would be "non-null estimators" or perhaps "estimators which are non-trivially related to X", but it's not clear whether this is the kind of thing you seek. $\endgroup$
    – Glen_b
    Jul 7, 2017 at 3:11
  • $\begingroup$ @Glen_b I'm looking for a simple term that I can use without defining, so that I don't need to throw some writing off-focus in order to define a term. "non-null estimators" certainly makes sense, but I'd probably have to define what that means. $\endgroup$
    – generic_user
    Jul 7, 2017 at 15:01
  • $\begingroup$ Conditional mean? $\endgroup$
    – Dave
    Jan 22, 2020 at 15:50

1 Answer 1

Reset to default
3
$\begingroup$

$\bar y$ is not a bad estimator at all, compared to something like the estimator: 0. However, I think that either "nontrivial" or "non-naive" would be good terms to describe the category of estimators better than $\bar y$. There's no official term (or at least not that I know of) for an estimator better than $\bar y$, so if you're writing a paper on it, you should probably define your own.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Thanks! "Non-trivial" certainly fits. But I think I'm forced to simply say something long-winded, like "any estimator that is better than $\bar{y}$ in the sense that $var(\epsilon)<var(y)$. So no sense defining a term unless I want to use that term over and over. $\endgroup$
    – generic_user
    Jul 7, 2017 at 14:59

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.