12
$\begingroup$

In case of robust estimators, What does Gaussian efficiency means? For example $Q_{_n}$ has 82% Gaussian efficiency and 50% breakdown point.

The reference is: Rousseeuw P.J., and Croux, C. (1993). “Alternatives to median absolute deviation.” J. American Statistical Assoc., 88, 1273-1283

$\endgroup$
6
  • $\begingroup$ please add more context. Reference where you found this would be very helpful. $\endgroup$
    – mpiktas
    Commented May 6, 2011 at 9:51
  • 8
    $\begingroup$ My guess: If the sample follows a Gaussian distribution, then the asymptotic relative efficiency of the robust estimator in 95%. $\endgroup$
    – cardinal
    Commented May 6, 2011 at 12:08
  • $\begingroup$ the reference is: Rousseeuw P.J., and Croux, C. (1993). “Alternatives to median absolute deviation.” J. American Statistical Assoc., 88, 1273-1283. $\endgroup$
    – K-1
    Commented May 7, 2011 at 4:51
  • 3
    $\begingroup$ @cardinal Your interpretation is almost always what is intended, especially in discussions of robust estimators. I would elevate your comment from "guess" to "near certainty." $\endgroup$
    – whuber
    Commented May 17, 2011 at 2:48
  • 2
    $\begingroup$ @cardinal: your comment is the right answer. Please post it as so (i just saw this question). $\endgroup$
    – user603
    Commented Jan 30, 2012 at 7:02

2 Answers 2

1
$\begingroup$

I agree with @cardinal and @whuber that this is about asymptotic relative efficiency and here I write an answer if other are interested in the concept.

In statistics, it is very common to measure how efficient an estimator is using its asymptotic variance. Then, in this context, the asymptotic relative efficiency of $d_n$ relative to $s_n$ is defined by $$ARE=\lim_{n \to \infty}\frac{var(s_n)/E(S_n)^2}{var(d_n)/E(d_n)^2}.$$ In robust statistics, the law with respect to which we take the expectation in ARE is often a corrupted distribution.

For instance, I present an example from the book Robust Statistics by Huber and Ronchetti (see page 3). Suppose $X_1,\dots, X_n$ are from a corrupted gaussian distribution $$F(x)=(1-\varepsilon)\Phi(x)+\varepsilon \Phi(\frac{x}{3}) $$ so that with high proba ($1-\varepsilon$) the data are standard normal and with small proba ($\varepsilon$) the data are normal with a higher variance. Then, the ARE between $$d_n=\frac{1}{n}\sum_{i=1}^n |X_i-\overline{X}|\text{ and }s_n=\left(\frac{1}{n}\sum_{i=1}^n (X_i-\overline{X})^2\right)^{1/2}$$ is $\simeq 0.876$ when $\varepsilon = 0$ but as soon as $\varepsilon>0.005$, we have $ARE>1$ and for instance for $\varepsilon=0.01$, we have $ARE \simeq 1.44$. We conclude that the mean absolute deviation is more efficient than the std when the data are corrupted.

$\endgroup$
-3
$\begingroup$

I guess Gaussian efficiency is something related to computation cost.

The efficiency of Gaussian adaptation relies on the theory of information due to Claude E. Shannon. When an event occurs with probability P, then the information −log(P) may be achieved. For instance, if the mean fitness is P, the information gained for each individual selected for survival will be −log(P) – on the average - and the work/time needed to get the information is proportional to 1/P. Thus, if efficiency, E, is defined as information divided by the work/time needed to get it we have: E = −P log(P). This function attains its maximum when P = 1/e = 0.37. The same result has been obtained by Gaines with a different method.

I may simply conclude that the higher the Gaussian Efficiency is, less resources (RAM) is needed for computing something like a robust scale estimator of a large sample. Since CPUs are much faster than the rest of computer we prefer to run a trial/error algorithm for times rather doing it at once with saying 128GB of RAM. when the Gaussian Efficiency is high the job will be done in a shorter time.

$\endgroup$
2
  • 2
    $\begingroup$ This interpretation is sort of on the right track, at least at the beginning. I'm not sure who Gaines is or how it relates to this problem. But, see my hint, which provides you the answer. If needed, I can expand on it a bit. I would definitely not equate asymptotic relative efficiency to resources used, as you have tried to do in your last paragraph. $\endgroup$
    – cardinal
    Commented May 17, 2011 at 2:17
  • $\begingroup$ @ Cardinal: Could you please explain more about Gaussian efficiency? For example what's the difference between Qn which benifits from 82% Gaussian efficiency and MAD with 37%? Actually my background is Coastal engineering far from statistics! $\endgroup$
    – K-1
    Commented May 19, 2011 at 7:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.