Skip to main content
7 events
when toggle format what by license comment
Apr 20, 2017 at 18:52 comment added TOM I indeed meant that the Euclidean distance.
Nov 14, 2014 at 3:23 comment added Mark Meckes Ah, or by $\|X_i\|_2 \le 1$ did you mean that the Euclidean norm of $X_i$ is almost surely at most 1? I thought you were referring to the $L^2$ norm of a random variable. In that case, you can e.g. use a vector-valued version of Talagrand's inequality to get the same type of result as the one dimensional case.
Nov 14, 2014 at 1:25 comment added Mark Meckes Yes, in 1 dimension this is Hoeffding's inequality if you have boundedness, but that was never mentioned.
Nov 13, 2014 at 20:18 comment added TOM In 1 dimension this is exaclty hoeffding's inequality, unless I missed something obvious (we have boundedness, zero mean and independence).
Nov 12, 2014 at 3:42 comment added Mark Meckes You need additional assumptions, even in the one-dimensional case, for Hoeffding's inequality to hold.
Nov 10, 2014 at 22:33 history edited TOM CC BY-SA 3.0
added 252 characters in body
Nov 10, 2014 at 17:23 history asked TOM CC BY-SA 3.0