Stanislav Minsker addresses the following question:

Problem: Given X_{1}, …, X_{k} that are i.i.d. random variables, how can we estimate the mean of the distribution so that the confidence interval still has a coverage probability ≥ 1 – O(e^{-t})? Note: the smaller the confidence interval, the better the estimator.

1. Assuming the variables from normal distribution N(μ, σ^{2}). If the estimator is constructed as

For non-normal distribution, the confidence interval is much different:

Since the length of this interval depends on e^{t}, can we do better?

2. If we split the sample into k = floor(t) + 1 groups G_{1}, …, G_{k} each of size approximate n /t, then median-of-means estimator has the following confidence interval (result due to Nemirovskii and Yudin, 1983):

Where C is a constant. The length of confidence interval for median-of-means is much smaller than for the naive mean estimator.

3. Minsker’s work extends this to multi-dimensional distributions in Banach space using *geometric median *to obtain confidence intervals of sub-exponential length with high coverage.

More in the paper here.

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