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Mar 31, 2014 at 2:26 history edited Douglas Zare CC BY-SA 3.0
Added Poisson limit.
Mar 30, 2014 at 1:10 comment added Douglas Zare @JoelO: Yes, I'm working on an edit. Variance bounds give you better results when $b_0$ is $o(n)$. The probability that $b_t$ is large can't be too high because it is a martingale and the average stays the same.
Mar 30, 2014 at 0:41 comment added JoelO Just a quick followup question: The lower bound seems a bit insufficient for what I need. So alternatively, can we give a concentration bound on $b_i$? In particular, can we reasonably upper-bound the probability that $b_i$ is larger than some function of $n$, for any $i$? From what I understand from the Wikipedia entry, the variance of $b_i$ is given by: $Var[b_i/n] \approx \frac{b_i (n-b_i)}{n^2}(1-e^{-t/2n}) < b_i/n$, right? If that's true, we can just apply, say, Chebyshev's inequality for that purpose, right?
Mar 28, 2014 at 4:10 history answered Douglas Zare CC BY-SA 3.0