Q:

Why is it easier (in a statistical sense) to estimate standard deviation than the mean?

Way back when I was still a student, I was listening to one of my stats lecturers talk about robust statistics. He showed (on the board) a series of transformations/derivations which concludes that it is easier to measure standard deviation, than it is to measure the mean. I can't remember the details, but the conclusion has stuck with me. It might have had something to do with influence functions or something similar.
Anyone have any idea about what his reasoning could have been?

A:

I found some lecture notes from the course I asked about. I will relate the core idea here, as best I can.
First, I am going to assume any reader is familiar with the standard error of the sample mean: $$se(\mu) =\frac{\sigma}{\sqrt n}$$
Second, from here we get that the standard error of the sample variance for any distribution is: $$se(\sigma^2)=\sqrt{\frac 1 n (\mu_4-\frac{n-3}{n-1}\sigma^4)}$$
where $\mu_4$ is the fourth central moment: $E(X-\mu)^4$.
For a normal distribution $\mu_4 = 3\sigma^4$ so this simplifies to
 $$se(\sigma^2)=\frac{\sqrt 2 \sigma^2}{\sqrt{n-1}}$$
Lastly, use the delta method to derive an approximation of the standard error of a transformed parameter with a known standard error. Here the parameter with the known standard error is variance, and the transformation is $g(\theta)=\sqrt \theta $. The link in the second point uses the delta method for just this purpose, and arrive at a result of 
$$se(\sigma)\approx\frac 1 {2\sigma}se(\sigma^2)$$
Substitute in the value of $se(\sigma^2)$ and we get
$$se(\sigma)\approx\frac \sigma {\sqrt {2(n-1)}}$$
So the conclusion is that the standard error of the sample standard deviation is smaller than the standard error of the sample mean by a factor of $\frac 1 {\sqrt 2}$, for large n, for a normal distribution, and so is in this (very limited) sense, easier to estimate.
So the result isn't as general as what my memory suggested, but interesting nonetheless.
PS: the result holds for any distribution and sample size where kurtosis is small enough. Specifically when
$$\frac{\mu_4}{\sigma^4} < 4 + \frac{n-3}{n-1}$$

