I assume that the noise corrupting the sought-for signal is [Edit: Gaussian] white noise of zero mean and unknown variance. Is the optimal solution simply the mean over the realisations (the ensemble mean), or, we can have a better estimate, say, in terms of a smaller root-mean-square-error?

Edit: I found that my question have been asked already. Although the practical context is different, as in my case there is no concern that the different observations on different sensors could be correlated or that the corrupting process is slightly different. The answer to that question suggests that superposition, i.e., ensemble mean, is the best (“efficient"?) estimate. However, it occurs to me that it might not be the case, as a moving-window temporal average of the ensemble mean would yield an even smaller RMSE if the signal is sufficiently smooth. Unfortunately, there was no published reference given there. It would be great to find the answer in a textbook or paper, as it would surely clarify the conditions under which the result/theorem applies.

In any case, even if the moving average of the ensemble mean yields a better RMSE than the ensemble mean alone, I wonder if there is an even better estimate. Intuitively -- and it might be a common failure of intuition -- there might be value (information) in having the relaisations available "separately" as opposed to only their superposition, the ensemble mean. Any further help would be much appreciated.


The term you're looking for is efficient estimator. That's the least-variance (== least RMS) estimator for a given amount of observation.

Whether or not that's the average depends on your noise distribution.

For example, consider the two noises:

  1. continuous-valued uniformly distributed noise
  2. Noise that only takes values $\in \{-\sigma, 0, \sigma\}$ with equal probability.

Both can have the same variance, both are zero mean. Both can be white. But while the average sounds like a good idea for the first one, for the second one you can get a 0 RMS estimator as soon as you've observed three different values by picking the middle one.

  • $\begingroup$ Thank you, Marcus. I edited the question to clarify that the noise is normally distributed. $\endgroup$
    – Bodai
    Jan 3 at 8:33
  • $\begingroup$ Then, the first paragraph still applies: when you look for "efficient estimator gaussian noise", you should find what you're looking for. $\endgroup$ Jan 3 at 16:56
  • $\begingroup$ Hi Marcus. Is it something that i can find in a textbook? Taking a moving average in addition seems to yield a better result if the signal is sufficiently smooth. I wonder how the result/theorem is formulated precisely. $\endgroup$
    – Bodai
    Jan 4 at 5:24
  • $\begingroup$ Yes, every textbook on estimator theory. Look for efficient estimators. $\endgroup$ Jan 4 at 7:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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