I have a random variable $X\sim \text{Bin}(3k, q)$ where $q < \frac{1}{2}$ and $q(1-q) \leq \frac{1}{5}$. I want to show that for $k=O(\log_2 r)$ I can bound the probability $\mathbb{P}(X\geq k)$ by $\frac{1}{2^r}$. I tried several upper bounds for the tails of binomial distributions I found online but none of them seems to do the trick.
Any help would be appreciated!
Best Answer
We have the exact formula (often used by myself) $$\mathbb{P}(X\geqslant k)=k\binom{3k}{k}\int_0^q t^{k-1}(1-t)^{2k}\,dt$$ and the asymptotics for $q<1/3$ (which holds under your conditions on $q$) $$\mathbb{P}(X\geqslant k)\underset{k\to\infty}{\large\asymp}\frac12\sqrt{\frac3{k\pi}}\frac{1-q}{1-3q}\left(\frac{27}4q(1-q)^2\right)^k$$ (obtained using Stirling's asymptotics for $\binom{3k}{k}$ and Laplace's method for the integral).
For $k=O(\log_2 r)$, this is definitely not $O(2^{-r})$, but rather $O(\color{blue}{r^{-\alpha}}\log^{-1/2}r)$ with $$\alpha=2-\log_2\big(27q(1-q)^2\big)\color{gray}{\big[{}>0\big]}.$$