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Background

In their book, Huber & Ronchetti (pp. 2-3) compare the efficiency of the mean absolute deviation $d_n$ with the standard deviation $s_n$ with the following formula: $$ \operatorname{ARE}(d_n, s_n)=\lim_{n\rightarrow\infty}\dfrac{\operatorname{var}(s_n)/(E(s_n))^2}{\operatorname{var}(d_n)/(E(d_n))^2} $$ I encountered the formula again in Gerstenberger et al. (2015): $$ \operatorname{ARE}(s_{n}^{(1)}, s_{n}^{(2)}; F) = \dfrac{\operatorname{ASV}(s_{n}^{(2)};F)}{\operatorname{ASV}(s_{n}^{(1)};F)}\left\{\dfrac{s^{(1)}(F)}{s^{(2)}(F)}\right\}^{2} $$ where $s_{n}^{(1)}$ and $s_{n}^{(2)}$ are the estimators, $s_{n}^{(1)}(F)$ and $s_{n}^{(2)}(F)$ denote their corresponding population values and $\operatorname{ASV}(s_{n}^{(i)};F)$ denotes the asymptotic variance with respect to distribution $F$.

Question

Why does dividing/standardizing the asymptotic variance of an estimator by its squared population value make the different estimators comparable? Basically: What's the justification for this formula?

References

Gerstenberger C, Vogel D (2015): On the efficiency of Gini's mean difference. Stat Methods Appl 24:569-596. (link)

Huber JP, Ronchetti EM (2009): Robust statistics. 2nd ed. Wiley & Sons.

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    $\begingroup$ $s_n$ and $d_n$ have variances that are related to their size, but because they estimate different things, they don't have the same expected size. It would not be reasonable to favour $d_n$ as a way to estimate spread simply because $E(d_n)<E(s_n)$ whenever the variance of the distribution is positive. After all, why not choose yet another estimator $c_1 d_n$ or $c_2 s_n$ where $c_i << 1$ ... the variance could then be made as small as you like, but you have gained nothing. $\endgroup$
    – Glen_b
    Sep 22, 2022 at 1:19
  • $\begingroup$ @Glen_b Thanks for your explanations. I guess my questions are then: Why is the variance of an estimator related to its size and why divide by the square specifically? $\endgroup$
    – COOLSerdash
    Sep 22, 2022 at 6:17
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    $\begingroup$ Because they're estimates of scale parameters; the bigger the scale, the noisier the data that you're using to estimate it. In typical cases, dividing by the scale removes this effect (i.e. we compare coefficients of variation or in this case, since we're interested in efficiency, their squares). $\endgroup$
    – Glen_b
    Sep 22, 2022 at 7:15
  • $\begingroup$ @Glen_b Thank you! I think it finally clicked. It seems quite obvious now but I failed to make the connection to the (squared) coefficient of variation. $\endgroup$
    – COOLSerdash
    Sep 22, 2022 at 7:27
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    $\begingroup$ Although I feel the approach in the question is reasonably intuitive, as outlined in comments, I don't feel like that what I wrote there makes an adequate answer to the posed question. If you feel there's a good answer to be made of it, please feel free to make it into one. $\endgroup$
    – Glen_b
    Sep 22, 2022 at 7:35

1 Answer 1

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I'm trying to convert @Glen_b's comments into an answer as I understood them. The two insights that I missed were:

  • Higher population values of a scale parameter mean that the asymptotic variance of the sampling distribution is also higher because the data on which they are based on are noisier.
  • The variance divided by the squared expectation is basically the squared coefficient of variation, i.e. $\left(\frac{\sigma}{\mu}\right)^{2}$.

In essence, the asymptotic relative efficieny of two estimators that have different estimands (such as the standard deviation and the mean absolute deviation) can be calculated by taking the ratio of their squared coefficient of variation.

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  • $\begingroup$ @Glen_b I have tried to convert your helpful insights into an answer as I think it's better to have an answer instead of just the comments. Please let me know if you think I misrepresented or misunderstood anything. Thanks again for your help. $\endgroup$
    – COOLSerdash
    Sep 24, 2022 at 10:43

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