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I am reading conflicting references regarding the Fisher consistency of the sample variance.

$$s^2 = \frac{\sum_i^n (x_i - \bar{x})^2}{n}$$

Could anyone explain me how to proof whether $s^2$ is Fisher consistent or not? The discussion page on Wikipedia was inconclusive as well.

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  • $\begingroup$ why the robust tag? $\endgroup$
    – user603
    Aug 20, 2013 at 15:44
  • $\begingroup$ @user603, my view is that the relationship between functionals and sample statistics can be considered part of robustness analysis, but please edit the tag if you think my reasoning is wrong. $\endgroup$
    – sets
    Aug 20, 2013 at 19:10
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    $\begingroup$ Fisher consistency is not a property of a statistic like $s^2$: it is a relationship between that statistic and an underlying family of distributions. Thus there is no proof available because the question just doesn't make sense. $\endgroup$
    – whuber
    Aug 20, 2013 at 20:07

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Drawing from C.R. Rao "Linear Statistical Inference and its Applications" (2n ed, 1973), ch. 5, I will discuss this in the basic framework, where we have an i.i.d sample of size $n$ from a population that follows a discrete distribution that takes on a finite number $m$ of values, $\{k_1,...,k_m\}$. Denote the random variables involved $X_i$. We can pool the sample according to the permissible values that the random variables can take, and calculate the relative empirical frequencies/estimated probabilities, $\hat p_j =n_j/n,\;\; j=1,...,m, \; n_j\ge 0$.

SAMPLE MEAN
The sample mean can be written

$$\bar x = \frac 1n\sum_{i=1}^nx_i = \frac 1n\sum_{j=1}^mn_jk_j = \sum_{j=1}^m\hat p_jk_j$$ In this way the sample mean is expressed as a function of the empirical distribution function $\{\hat p_1,...,\hat p_m\}$.
The sample mean is Fisher-consistent because if it so happens that $\hat p_j = p_j,\; \forall j$, then we will have $$\bar x = \sum_{j=1}^mp_jk_j = E(X_i)$$ The important thing to note here is that the criterion for Fisher-consistency is not asymptotic: it does not examine what happens to the relative empirical frequencies when "the sample size goes to infinity". I would call it "design consistency" (Rao calls it also "method consistency"): it examines whether an estimator is constructed in such a way so as not to "miss" the true value that it attempts to estimate, in case it so happens that the finite sample it uses is a perfect representative of the population, or when we "have available the whole population" (which is how Fisher originally put it back in 1922, not realizing perhaps the conceptual and methodological problems associated with the statement "the whole population is available"). But in both cases, the "redeeming power" of an "infinite" sample size does not come into the picture, since an infinite sample size cannot be made "available". But even if the "population" is treated as a construct of infinite size, still the asymptotic logic is not applicable for Fisher consistency, since the criterion is not probabilistic: a Fisher-consistent estimator equals the value it attempts to estimate when the empirical relative frequencies happen to equal the actual probabilities -it does not just converge probabilistically to this true value.

SAMPLE VARIANCE
Following the same logic as before the sample variance (without the bias correction) can be written

$$\hat \sigma^2 = \frac 1n\sum_{i=1}^n (x_i - \bar{x})^2 = \frac 1n \sum_{i=1}^n x_i^2 - \bar x^2 = \frac 1n \sum_{i=1}^m n_jk_j^2 - \left(\sum_{j=1}^m\hat p_jk_j\right)^2$$

$$\Rightarrow \hat \sigma^2 = \sum_{i=1}^m \hat p_jk_j^2 - \left(\sum_{j=1}^m\hat p_jk_j\right)^2$$

If it so happens that $\hat p_j = p_j,\; \forall j$, then we will have

$$\Rightarrow \hat \sigma^2_{\{ \hat p_j = p_j\}} = \sum_{i=1}^m p_jk_j^2 - \left(\sum_{j=1}^mp_jk_j\right)^2 = E(X_i^2) - [E(X_i)]^2 = \text {Var}(X_i)$$

and so the sample variance estimator without the bias-correction term is Fisher-consistent.

But the bias-corrected sample variance estimator is not Fisher-consistent, because

$$s^2 = \frac 1{n-1}\sum_{i=1}^n (x_i - \bar{x})^2 = \frac {n}{n-1}\cdot \hat \sigma^2 $$

If it so happens that $\hat p_j = p_j,\; \forall j$, then we will have

$$s^2_{\{ \hat p_j = p_j\}} =\frac {n}{n-1}\hat \sigma^2_{\{ \hat p_j = p_j\}}= \frac {n}{n-1}\text {Var}(X_i) \neq \text {Var}(X_i) $$

How important/desirable is Fisher-consistency? Rao writes

"It would be, indeed, anomalous if we happen to realize the true proportions but the method of estimation does not automatically lead to the true value"

While this appears very reasonable, it does not discuss how likely can be considered the case where we "happen to realize the true proportions", and so, whether "it is worth it" to sacrifice other desirable features of the estimation procedure in order to guarantee Fisher-consistency, while the former may be of value in cases much more likely to appear than the "true proportions" instance.

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  • $\begingroup$ @CagdasOzgenc It appears to vary in order to be suitable in different contexts. The authors write "In the traditional parameter estimation situation, Fisher consistency means that the estimation procedure in the population space will produce the target of the estimation.", which is what I used here. Then they offer a definition related to a "classification procedure". $\endgroup$ Feb 9, 2020 at 12:36
  • $\begingroup$ @CagdasOzgenc Hmm, no I think they just generalize the concept of Fisher consistency to be suitable for M-estimators. My post used sample means, which, in abstract, is a function of the data. They extend the concept to the minimizer of a function of the data that has the properties of a "loss function". $\endgroup$ Feb 9, 2020 at 16:24

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