# If you have a really big N, is using a regular vs. an m-n bootstrap ever a good idea?

I know that the regular bootstrap (assuming that the necessary assumptions are valid) is more efficient than the m-n bootstrap.

However, supposing that you have access to really large data (such as a 1% sample of Google's daily searches), does it ever make sense to use the regular bootstrap whose assumptions may be violated?

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@user68 and dchandler - please do join in the discussion over on meta: meta.economics.stackexchange.com/questions/40/… –  EnergyNumbers Oct 16 '11 at 8:28
Judging from this article this questions is more suitable for mathoverflow.net. As far as I understand the answer is yes. Figure out whether the conditions are violated and if you need efficiency go with normal bootstrap. If you only care about consistency use m out of n. –  mpiktas Oct 16 '11 at 13:21
@dchandler, if the issue is computational vs. statistical (i.e., being unable to efficiently sample $10^9$ observations with replacement), you might want to state so. Michael Chernick gave an exhaustive answer to the statistical question, but it may not answer the computational question. –  StasK May 3 '12 at 19:23

## migrated from economics.stackexchange.comMay 3 '12 at 18:21

This question came from our site for economists and graduate-level economics students.

There are very few assumptions to violate with the regular bootstrap. The issue with the bootstrap is whether or not it provides consistent estimates for the particular problem. When the ordinary (i.e., the usual $n$-out-of-$n$) bootstrap is consistent, $m$-out-of-$n$ is less efficient and there is no reason to use it over the ordinary bootstrap.
The $m$-out-of-$n$ bootstrap comes in handy for the few exceptional cases where the ordinary bootstrap fails. In cases such as estimating extremes the ordinary bootstrap fails, but $m$-out-of-$n$ has been shown to be consistent as long as m goes to infinity at a slower rate than $n$. Then $m$-out-of-$n$ should be used.
A practical question might be what to take for $m$. I would suggest something simple like square root of $n$.