In Lehmann's formulation, almost any formula can be an estimator of almost any property. There is no inherent mathematical link between an estimator and an estimand. However, we can assess--in advance--the chance that an estimator will be reasonably close to the quantity it is intended to estimate.
An "estimator" or "point estimate" is a statistic (that is, a function of the data) that is used to infer the value of an unknown parameter in a statistical model.
An estimator is consistent if, as the sample size increases, the estimates (produced by the estimator) "converge" to the true value of the parameter being estimated.
What is the difference between a consistent estimator and an unbiased ...
I am a bit confused about the terminology used in the context of sampling of populations. The Horvitz-Thompson estimator, as well as the Hansen-Hurwitz estimator, for example, are examples of estim...
We don't, in general, want to get unbiased estimators. It's extremely common to use biased estimators, especially when we don't have very much data: random-effects models, Bayesian estimators, penalised regression, small-area estimation, smoothing, density estimation... However, when you have a smooth parametric model and a lot of data, we do know that the best estimator in the sense of mean ...
For example, we say that an estimator is unbiased if the expected value of the estimator is the true value of the parameter we're trying to estimate. However, if we already know the true value of the parameter, why are we trying to estimate it?
You can always compute the OLS estimator, apart from the case when you have perfect multicollinearity. In this case, you do have perfect multilinear dependence in your X matrix. Consequently, the full rank assumption is not fulfilled and you cannot compute the OLS estimator, because of invertibility issues. Technically, you do not need the other OLS assumptions to compute the OLS estimator ...