Relative efficiency: mean deviation vs standard deviation I have difficulties following a seemingly elementary claim from Tukey (1960):

It is well known that, in large samples, the relative efficiency as a measure of scale of the mean deviation compared with the standard deviation is 88% when the underlying population is normal.

I'm trying to simulate that by generating samples from the normal distribution and arrive at value of $\approx 0.798$ for $RE = \frac{1/n \sum \lvert Y- \overline{Y} \rvert}{\sqrt{1/(n-1) ( Y - \overline{Y} )^2}}$, i.e. not surprisingly, I see $\sqrt{\frac{2}{\pi}}$.
The aforementioned 88% is then picked up in the Ripley lecture notes (2004). It says $ARE(\tilde{\sigma}^2, s^2) = 0.876$, where $\tilde{\sigma}^2 = d^2 \pi / 2$ ($d$ is the absolute mean deviation) and $s$ is deviation of the optimal estimator. With these details, it seems to me that $ARE$ should converge to $1$ since $d \rightarrow \sqrt{2/\pi} \sigma$ and $s \rightarrow \sigma$. I think still that both are unlikely to be typos and I simply misunderstand what they are trying to say. 
Could someone, please, point me where the "well known" 88% comes from?
References:


*

*Tukey, John W. "A survey of sampling from contaminated
distributions." Contributions to probability and statistics 2 (1960):
448-485. 

*Ripley, B.D. "Robust Statistics", lecture notes, (2004) 
(https://www.stats.ox.ac.uk/pub/StatMeth/Robust.pdf)

 A: You don't compute relative efficiency by taking the ratio of the estimators.
For unbiased estimators, relative efficiency is the ratio of their precision (inverse of variances). Equivalently, the efficiency of the first to the second would be the variance of the second estimator divided by the variance of the first (so estimators with larger variance are less efficient).
More generally, relative efficiency either looks at the ratio of their mean square errors 
$$e(T_1,T_2)=\frac {\mathrm{E} \left[ (T_2-\theta)^2 \right]} {\mathrm{E} \left[ (T_1-\theta)^2 \right]}$$
or sometimes they divide by $\theta^2$ instead -- taking ratios of coefficients of variation squared. 
Since you're computing asymptotic relative efficiency for consistent estimators, you could take the ratio of the variances.
In the case of the scaled mean deviation vs the standard deviation as an estimate of $\sigma$ in the normal, Fisher derived the ARE to be $\frac{1/2}{(\pi/2)-1} = \frac{1}{\pi-2} \approx 0.87597$. This will be the result Tukey refers to. 
Some details of the derivation (and other references) are given in Pham-Gia and Hung (2001)[1].
Note, however, that Ripley is talking not about comparing a multiple of $d$ with $s$ as estimates of $\sigma$ but comparing a multiple of $d^2$ with $s^2$ to estimate $\sigma^2$; that's a different calculation, and it surprises me that the asymptotic relative efficiency should come out the same. (There may be some obvious reason why that should work out that way that I have missed.)
[1] Pham-Gia, T. and T.L. Hung (2001),
"The mean and median absolute deviations"
Mathematical and Computer Modelling, Vol 34, No 7–8 (Oct), pp 921-936
(a pdf link to the paper is available on the paper's web-page here)
