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Let $p \in [0,1/2)$ and $\delta \in [0,1/2)$ be fixed real numbers, and let $n$ be a positive-integer valued variable.

Let $k=k(n)$ be the smallest integer for which $\delta \leq \sum_{i=0}^k {n \choose i} p^i (1-p)^{n-i}$. I would like a lower bound on $k$ in terms of $n$.

I have tried to use known tail bounds on the binomial distribution, but these appear to give quite terrible bounds (if any). Indeed, the standard tail bound is given by

$$\begin{align} \sum_{i=0}^k {n \choose i} p^i (1-p)^{n-i} &\ge {n \choose k} p^k (1-p)^{n-k}\\ &= {n \choose k} \exp\left(-n(k/n \log1/p+(1-k/n)\log1/(1-p)\right)\\ &\ge \frac{\exp(n\text{H}(k/n))}{\sqrt{8k(1-k/n)}}\, \exp(-n(\text{D}(k/n||p) + H(k/n)))\\ &= \frac{1}{\sqrt{8k(1-k/n)}}\exp(-n\text{D}(k/n||p))\\ &\ge \frac{1}{\sqrt{2n}}\exp(-n\text{D}(k/n||p)), \end{align}$$ where $\text{D}(q,r)=q \log_e\left(\frac{q}{r}\right)+(1-q)\log_e\left(\frac{1-q}{1-r}\right).$ This bound decays with $n$ no matter what we choose for $k(n)$.

Ben
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  • The CLT suggests that this bound is linear in $n$; are you basically looking for a constant of proportionality in terms of $p$ and $\delta$? – Steven Stadnicki Jun 09 '21 at 18:50

1 Answers1

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Disclaimer: a quick and dirty engineering approach.

Let $q = \min(p, 1-p)$. Then

$$\sum_{i=0}^k {n \choose i} p^i (1-p)^{n-i} \ge \sum_{i=0}^k {n \choose i} q^i q^{n-i} = q^n O(n^{k+1})$$

Let the RHS converge to a non-zero constant $C$. Taking logarithms, obtain

$$n\ln{q} + (k+1)\ln{n} \approx \ln{C}$$

which means that $k = \dfrac{n}{\ln{n}}\ln{\dfrac{1}{q}}$ is a good candidate.

user58697
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