Timeline for Lowerbounding expectation value of binomial tail
Current License: CC BY-SA 4.0
14 events
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| Jun 23, 2020 at 15:14 | comment | added | ofer zeitouni | OK, I approved the edit. Thanks. And please in the future do not use the language you used earlier in comments. Editors are volunteers. | |
| S Jun 23, 2020 at 15:14 | history | suggested | Mateus Araújo | CC BY-SA 4.0 |
Added expression for $c$ and simplified expression for $\beta$.
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| Jun 23, 2020 at 14:13 | comment | added | Mateus Araújo | There is an $\eta$ both in the numerator of the lattice formula, but it cancels out with the $\eta$ in the expression of $J_n$. It is given by $\eta=\log\sqrt{\frac{q(1-p)}{(1-q)p}}$. | |
| Jun 23, 2020 at 14:08 | comment | added | ofer zeitouni | There is an $\eta$ in the numerator, no? and the $\eta$ is a logarithm, being the tilt? | |
| Jun 23, 2020 at 13:44 | comment | added | Mateus Araújo | Yes, I used the formula for the lattice case, with $d=1$. I don't think there's a logarithm missing, $\eta$ is a logarithm, but it only appears through an exponential. | |
| Jun 23, 2020 at 13:41 | comment | added | ofer zeitouni | I just noticed a few minutes ago that you did, and was going to check it when I get to it. Did you use the formula for the lattice case? I suspect that there is a logarithm missing, isn't there? | |
| Jun 23, 2020 at 13:35 | comment | added | Mateus Araújo | I calculated the value of $c$ and added it as an edit to your answer, but some random prick deleted it saying that it does not matter. For the record, it is $c = \frac{\sqrt{pq} + \sqrt{(1-p)(1-q)}}{\sqrt{4\pi}\left(1-\sqrt{\frac{p(1-q)}{(1-p)q}}\right)\big(pq(1-p)(1-q)\big)^\frac14}.$ | |
| Jun 23, 2020 at 10:12 | review | Suggested edits | |||
| S Jun 23, 2020 at 15:14 | |||||
| Jun 22, 2020 at 20:47 | comment | added | ofer zeitouni | Now, by Chebycheff, $P_\lambda^n([0,(\epsilon+\delta)n]^c)\leq e^{-c'(\epsilon \wedge \delta)n}$ for some $c'$ that is explicit and one can compute. For $n$ such that the RHS is smaller than $1/2$, this completes the lower bound. | |
| Jun 22, 2020 at 20:45 | comment | added | ofer zeitouni | Here is a sketch of how to do it quantitative: make a tilt of the law of $W_1$ to have the mean at +ϵ. This will give you a factor$e^{-n\Lambda^*(\epsilon)}$ and then a lower bound is $e^{-n \Lambda^*(\epsilon) - n\lambda \delta} P_\lambda^n ([0,(\epsilon+\delta)n])$ which is greater or equal to $e^{-n(\Lambda^*(\epsilon)+\delta\lambda)} (1-P_\lambda^n([0, (\epsilon+\delta)n]^c)$ | |
| Jun 22, 2020 at 19:58 | comment | added | ofer zeitouni | Thanks for pointing out the $>$ vs $\geq$ - the asymptotics is actually the same. The lower bound is quantitative - that is, you can actually write $P(B\geq 0) =ce^{-n\beta}/\sqrt{n}(1+C_1/\sqrt{n})$, and with enough diligence get an upper bound on $C_1$. Note that the lower bound involves computing, under the tilted measure, an upper bound on the complementary event, | |
| Jun 22, 2020 at 19:54 | history | edited | ofer zeitouni | CC BY-SA 4.0 |
added 8 characters in body
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| Jun 22, 2020 at 15:30 | comment | added | Mateus Araújo | It's nice to know the asymptotics, but I still don't see how that helps finding a lower bound. I checked your book, and the theorem doesn't provide a bound, it only talks about the limit. Also, is there any reason you wrote $P(B>0)$ instead of $P(B\ge 0)$? Since you're taking $\Lambda^*(0)$ you must mean the latter, which is indeed what I asked about. | |
| Jun 22, 2020 at 7:07 | history | answered | ofer zeitouni | CC BY-SA 4.0 |