Skip to main content
7 events
when toggle format what by license comment
May 10 at 16:21 vote accept Betty
May 10 at 11:44 answer added Iosif Pinelis timeline score: 1
May 9 at 14:32 history edited Betty CC BY-SA 4.0
added 21 characters in body
May 9 at 14:32 comment added Betty Thanks for the comments, I have updated the question. For $c=1+1/m$ with $m\geq n$, how should I derive the concentration bound?
May 9 at 12:40 comment added Iosif Pinelis If $c\ge2$ (say) or, more generally, if $c\ge1+t_n$ with $nt_n\to\infty$, then the weighted sum of the $n$ random variables $c^{i-1}x_i$ will be close to the sum of a few $c^{i-1}x_i$'s with the largest values of $i$, and then there will no concentration. That is, you can get concentration only if $c=1+O(1/n)$.
S May 9 at 2:23 review First questions
May 9 at 5:47
S May 9 at 2:23 history asked Betty CC BY-SA 4.0