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Apr 13, 2017 at 12:58 history edited CommunityBot
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Mar 9, 2017 at 4:28 history edited stats134711 CC BY-SA 3.0
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Mar 9, 2017 at 3:23 comment added usul Continuing Fedor's line of reasoning, we can use Chernoff-type bounds to obtain $P(X_i \leq c) \leq e^{-f(c,p_i,n)}$, so the probability is at least $1 - \exp\left[ - \sum_{i=1}^m f(c,p_i,n) \right]$. For example, I think Hoeffding's gives $f(c,p_i,n) = 2(c - np_i)^2/n$.
Mar 8, 2017 at 21:44 vote accept stats134711
Mar 8, 2017 at 21:21 answer added Rodrigo Ribeiro timeline score: 1
Feb 23, 2017 at 15:52 comment added stats134711 That was the first step I tried but I'm having trouble getting anywhere productive for successive calculations.
Feb 23, 2017 at 5:50 comment added Fedor Petrov Of course it depends on $p_i$, if all $p_i$ are almost 0, this probability is small. We have $P(\max_{1\leq i\leq m} X_i > c)=1-\prod_{1\leq i\leq m}P(X_i\leq c)$, this should be useful for formulating concrete results
Feb 23, 2017 at 3:44 history edited stats134711 CC BY-SA 3.0
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Feb 23, 2017 at 2:52 history edited stats134711 CC BY-SA 3.0
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Feb 23, 2017 at 2:29 history edited stats134711 CC BY-SA 3.0
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Feb 23, 2017 at 2:05 history edited stats134711 CC BY-SA 3.0
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Feb 23, 2017 at 1:52 history asked stats134711 CC BY-SA 3.0