What is asymptotic variance? I am struggling to understand the concept of asymptotic variance. The context is the geophysical time series processing with robust methods being employed.
Methods with a very high breakdown point usually have a smaller asymptotic relative efficiency at the Gaussian distribution than LS. This means that the higher the robustness of the estimator, the higher the asymptotic variance. In order to achieve the same parameter uncertainties by the robust procedure, more measurements are required.
Can someone explain this?
 A: A robust estimator is one that is unchanged or changes very little when new data are introduced or assumptions are violated. For example, the median is a more robust estimator than the mean because if you add a relatively large observation to your data set, your median will change very little whereas your mean will change much more.
When fitting a linear regression model, we get parameter estimates and associated standard errors of our estimates. One of the assumptions of the linear regression model is equality of variance - that is, regardless of the $x$ value, the errors will be distributed with mean $0$ and standard deviation $\sigma$. In the case where this assumption is violated, we may prefer to use robust standard errors which are generally larger standard errors that will account for any violation of our equality of variances assumption. (This violation is known as heteroscedasticity.)
When we use robust standard errors, our standard errors (and equivalently, our variances) are generally larger than they would be if we didn't use robust standard errors. Let's denote the robust standard error as $\frac{\sigma_R}{\sqrt{n}}$ and the "typical" (non-robust) standard error as $\frac{\sigma_T}{\sqrt{n}}$. It should be clear that, when the robust standard error is larger, $\frac{\sigma_R}{\sqrt{n}}>\frac{\sigma_T}{\sqrt{n}}$. It should also be clear that, asymptotically, the robust standard error will be larger than the "typical" standard error because we can cancel the $\sqrt{n}$ out on both sides.
Let's say that our "typical" standard error is $k = \frac{\sigma_T}{\sqrt{n}}$. Then $k < \frac{\sigma_R}{\sqrt{n}}$. In order for the robust standard error to equal $k$, we must make $n$ larger (a.k.a. collect more observations/sample).
Hope this makes sense!
EDIT: See the included link and the comments below for a brief discussion on when the robust standard errors will actually be larger than the "typical" (non-robust) standard errors. http://chrisauld.com/2012/10/31/the-intuition-of-robust-standard-errors/
