Talk:Fisher consistency

Examples, counterexamples, and the article is still confusing

Note: I never heard of the concept of Fisher consistency so I might say something stupid here...

Looking at this "finite population example"; what if I modify it in the following way:

Suppose our sample is obtained from a finite population Z1, ..., Zm. We can represent our sample of size n in terms of the "proportion−−" of the sample n_i / (n–1) taking on each value in the population. Writing our estimator of θ as T(n1/(n–1), ..., nm/(n–1)), the population analogue of the estimator is T(p1, ..., pm), where pi = P(X = Zi). Thus we have Fisher consistency if T(p1, ..., pm) = θ.

The numbers n_i/(n–1) still "represent" the population (as long as n>1), and still converge to p_i's; so this would imply that an estimator Sum(Xi)/(n–1) will also be Fisher consistent for μ. At the same time the referenced lecture notes claim that the estimator of sample variance SSR(Xi)/(n–1) is not Fisher-consistent, while the corresponding MLE estimate SSR(Xi)/n is.

The definition says that in order to verify this consistency property we have to find the explicit function T such that the estimator is equal to $\hat\theta = T(\hat{F}_n)$. So is it possible to give this explicit function in general? Or even in simplest cases? We had to restrict ourselves to finite population to make the sample average example work, but what do we do in continuous case? It seems to me that this definition makes sense only for iid data, and only for estimators which are symmetric wrt permutations of the sample. But then how is estimator $\hat\theta=X_1$ Fisher consistent (the example given later in the article), what is the corresponding functional T?

Thoroughly confused, ... stpasha » talk » 02:34, 17 July 2009 (UTC)

It is not necessary restrict the sample mean to a finite population, but I don't know why the example is only for this case.

For the mean the required functional T(F) is

$T(F)=\int xdF(x)$.

You may be starting from the wrong end of things. If you make the distinction between a "population parameter" which doesn't depend on any given family of distributions (such as the mean) and a parameter of a family of distributions, the "Fisher consistency" says that you take the definition of the population parameter in terms of any general distribution function (ie the definition which you already know, which is equivalent to the functional), and apply it to the sample distribution function to create the estimate of the population parameter.

Melcombe (talk) 14:08, 17 July 2009 (UTC)

I think the example with ${\hat {\theta }}=X_{1}$ is instructive, but it doesn't quite fall within the definition of Fisher consistency given in the article. A related example would be to draw an iid sample, X_i,  i = 1,...n, then select one value at random, say X_j, and estimate the mean as X_j. This would fall within the definition of Fisher consistency given in the article if we allow the estimator to be a random function of the data.Skbkekas (talk) 03:39, 18 July 2009 (UTC)

This is equivalent to taking ${\hat {\theta }}=F_{n}^{-1}(U)$, where U is a uniform [0,1] auxiliary random variable. However such estimator would estimate the X itself, not its mean, so this estimator would be neither Fisher-consistent nor asymptotically consistent. ... stpasha » talk » 05:19, 18 July 2009 (UTC)