# Fisher consistency versus “standard” consistency

My question relates two types of consistency. In particular, how does the Fisher consistency differ from standard notions of consistency, such as convergence in probability to the generative parameter. When will these two differ?

Following https://en.wikipedia.org/wiki/Fisher_consistency Fisher consistency means that if the estimator was calculated using the complete population and not only a sample, the correct value would be obtained. Asymptotic consistency means that as the sample size goes to infinity, the estimator converges in probability to the true value. Neither concept entails the other, see above wikipedia article for examples.

I will give some other examples, not from the wikipedia article, I don't find that particularly clear!

Let $$X_1, X_2, \dots, X_n$$ be a random sample from a population described by the cumulative distribution function $$F$$. Now we will consider "functional parameters", that is, parameters which can be written as a functional of $$F$$, $$\theta = \theta(F)$$ Examples are for instance the expectation $$\mu=\mu(F)=\int_{-\infty}^\infty x\; dF(x)$$ (where this notation denotes a Riemann-Stieltjes integral, which equals $$\int_{-\infty}^\infty x f(x) \; dx$$ in the continuous case and a summation in the discrete case). Another example is the median $$m=F^{-1}(0.5)$$, the interquartile range $$\text{IQR} = F^{-1}(0.75) - F^{-1}(0.25)$$ and many other examples like the variance $$\sigma^2 = \int_{-\infty}^\infty (x-\mu)^2\; dF(x)$$.

In the following let us use the expectation $$\mu$$ as example. So we are interested in estimating from our sample $$\mu=\int_{-\infty}^\infty x\; dF(x)$$. One estimator is the arithmetic mean which can be written $$\bar{x}=\frac1{n}\sum x_i = \int_{-\infty}^\infty x\; d\hat{F}_n(x)$$ where $$\hat{F}_n$$ denotes the empirical distribution function. The value of that at the population $$F$$ is obtained by replacing $$\hat{F}_n$$ by $$F$$ and is the true expectation, showing that the arithmetic mean is Fisher consistent.

Now, maybe you are preoccupied with the "lack of robustness" of the arithmetic mean, that it is unduly influenced by outliers, for instance. So you want some more robust (or resistant) estimator. Two such are the median and winzorized means, that is arithmetic means after throwing away some of the smallest and largest observations. The empirical median, for instance, will be an unbiased estimator of the mean $$\mu$$ in symmetric families such as the normal. But, the median as an estimator of the mean, is not Fisher-consistent. The empirical median can be written $$\hat{F}_n^{-1}(0.5)$$ and evaluated at the population $$F$$ that is in general different from the mean.

The same happens with the winzorized means. Many other robust estimators as well will not be Fisher-consistent.

Fisher-consistency is about "doing the right thing at the model", while robustness is about obtaining reasonable answers also in some neighbourhood of the model, when the model is not right. Those are different goals.

The original poster says in comments: " I would expect that except pathologies, a (classically) consistent estimator is also Fisher consistent." There is no base for such expectation! A simple example is the usual unbiased estimator of variance, $$s^2 = \frac1{n-1}\sum_{i=1}^n (x_i-\bar{x})^2$$. This is unbiased and consistent in the usual sense (whatever is $$F$$, as long as it has variance!). In functional form this can be written (after some algebra ...) $$s^2=\frac{n}{n-1} \left\{ \int x^2 \; d\hat{F}_n(x) -(\int x \; d\hat{F}_n(x))^2 \right\},$$ which, evaluated at the true population $$F$$ is $$\frac{n}{n-1}\sigma^2$$ so not Fisher consistent. We see this is because the concept of Fisher consistency is not asymptotic, so the factor of $$\frac{n}{n-1}$$ do not "disappear" as it does in the case of asymptotic consistency. So a lot of usual, not at all "pathological" estimators will be asymptotically consistent, but not Fisher consistent. The opposite case is maybe more unusual? I have had problems finding a natural countrexample in that other case, Fisher consistent but not asymptotially consistent. I guess one place to search is a situation where NO asymptotically consistent estimators exist!

• Besides the wikipedia pathologies- are there any practical cases where these two differ? – JohnRos Sep 30 '15 at 11:37
• See my extended answer above! – kjetil b halvorsen Sep 30 '15 at 15:02
• The new information in very useful, but I still fail to understand: I would expect that except pathologies, a (classically) consistent estimator is also Fisher consistent. In your examples, the median is not Fisher consistent, but neither is is (classically) consistent. So this does not demonstrate when do these concepts differ. – JohnRos Sep 30 '15 at 15:13
• Assume a sample of size $n$ with an estimate of the mean = $x_{[1]}$, the first observation. Clearly this is unbiased and not consistent; it is also Fisher consistent. – jbowman Oct 4 '15 at 15:35
• Yes, but that is one of those "pathological" examples from Wikipedia the OP do not want! – kjetil b halvorsen Oct 4 '15 at 15:36