# Derivation of the formula for the asymptotic relative efficiency of two estimators with different estimands

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.

• $s_n$ 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. Sep 22, 2022 at 1:19
• @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? Sep 22, 2022 at 6:17
• 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). Sep 22, 2022 at 7:15
• @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. Sep 22, 2022 at 7:27
• 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. Sep 22, 2022 at 7:35

• The variance divided by the squared expectation is basically the squared coefficient of variation, i.e. $$\left(\frac{\sigma}{\mu}\right)^{2}$$.