What is the effect of taking the mean of a moments-based estimate? Suppose I estimate the second and fourth moments of a signal as
$M_2 \approx \frac{1}{N} \sum_{n=0}^{N-1} | y_n |^2$
and
$M_4 \approx \frac{1}{N} \sum_{n=0}^{N-1} | y_n |^4$
and then I use these to estimate the SNR of a signal. For example, Pauluzzi and Beaulieu 1 describe an estimator for an M-ary PSK signal with complex noise:
$\rho = \frac{\sqrt{2 M_2^2 - M_4}}{M_2 - \sqrt{2 M_2^2 - M_4}}$
What would be the effect of taking, say, $N'$ estimates of $\rho$ and computing the mean of these estimates, versus just using $N\times N'$ samples to estimate $M_2$ and $M_4$ in the first place?
My intuition says that the latter is more meaningful, but I can't figure out how to explain why.
1 Pauluzzi, D.R.; Beaulieu, N.C., "A comparison of SNR estimation techniques for the AWGN channel," in Communications, IEEE Transactions on , vol.48, no.10, pp.1681-1691, Oct 2000
doi: 10.1109/26.871393 (Link)
 A: As a general rule of thumb, it's better to use a single estimator that incorporates all the rather than the average of several estimators that uses subsets of the data (side note: bagging could be thought of as an alternative, but that's really a special case). 
Here's a common example of why you would want pool all your data rather than just using the average of stratified estimators, which I believe applies to your case (although not the only reason). Many estimators are biased but consistent; I'm fairly certain your case is an example of that. Suppose we can then write the following about the expected value:
$\mathbb{E}[\hat \theta] = \theta g(n) $, 
with  $|g(n) - 1| > |g(n+1)-1|$ 
and $\lim_{n \rightarrow \infty}g(n) = 1$
(Note: there's cases of consistent estimators that don't meet this criteria, but there are plenty more that do. And I do believe that your estimator fits this criteria). 
Suppose then we have $\hat \theta_1$ and $\hat \theta_2$, where these are estimators used from cutting our data in half and separately estimating $\theta$ from each data set. Meanwhile, $\hat \theta_.$ is the estimator created from using all the data. 
Then we get that 
$\mathbb{E}[ (\hat \theta_1 + \hat \theta_2)/2] = (\theta g(n/2) + \theta  g(n/2) )/2 =\theta g(n/2)$
While on the other hand
$\mathbb{E}[ (\hat \theta_.) ] = \theta g(n)$, 
Since $|1 - g(n)|$ < $|1 - g(n/2)$|, 
this implies that $\hat \theta_.$ is less biased than $(\hat \theta_1 + \hat \theta_2)/2$
More generally, many estimators have nice asymptotic properties. By cutting up your full sample into smaller samples and averaging across, you may be reducing the asymptotic effect.
