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JoelO
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Convergence of a sequence of dependent binomial trials
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?
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Convergence of a sequence of dependent binomial trials
Right. $o(1)$ was supposed to be $o_n(1)$; i.e., it approaches $1$ as $n$ grows. For starters, I want to know how fast it converges to $1-1/n$. In particular, what is the probability of stopping with $b_t=0$ for $t =O(\log n)$?
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Balls and bins -- concentration bounds pertaining to the minimal load bin
As it turns out, one could use the Poisson approximation to estimate the distribution of the minimum load. (cf. Corollary 5.11 in p. 103 in link. Asked a similar question about $n$ Poisson variables in a separate post:link
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