Timeline for Tail bound for maximum of independent (but not identical) binomial random variables
Current License: CC BY-SA 3.0
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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 |