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Mar 16, 2015 at 20:40 history edited Kevin P. Costello CC BY-SA 3.0
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Mar 16, 2015 at 20:34 comment added TMM Maybe I should have added this, but $I_n(p)$ scales as $1/n$ on all of $[0,1]$. Using binary logarithms, for $p \approx 1.34/n$ we have $I_n(p) \approx 0.84/n$ and for other values of $p$ (say $p = 1/2$) we have $I_n(p) \approx 0.72/n$. So finding an upper bound with terms that only weakly converge to $0$ is not enough. (And since the values of $I_n(p)$ for $0 \ll p \ll 1$ are only $16\%$ smaller than the value at the maximum, you'll need to use very tight bounds to prove the result.)
Mar 16, 2015 at 19:58 history answered Kevin P. Costello CC BY-SA 3.0