Estimates of the sum of the biggest terms in a binomial expansion.

Question

Perhaps this is an easy and standard result in the binomial distribution. Consider the expansion $(a+b)^n$ where $a=(\frac{1}{2} +\delta)$ and $b=(\frac{1}{2}-\delta)$ for some arbitrarily small $\delta$. Given a fixed small $0<c<1$, one would expect for all $n$ large enough, the existence of $t=t(\delta,n,c)< \frac{n}{2}$ so that with the initial $t$ terms of the binomial expansion we have $\sum\limits_{j=0}^{t} {n\choose j} a^j b^{n-j}>c$.

Answer 1

Let $S_n$ be a random variable with binomial distribution with parameters $n$ and $b$. Then your sum $$\sum_{j=0}^t {n \choose j} a^j b^{n-j} = \mathbb P(S_n \le t)$$ By the Law of Large Numbers, $ \mathbb P(S_n \le r n) \to 1$ as $n \to \infty$ for any $r > b$. In particular, taking $b < r < 1/2$ and $t = r n < n/2$ we have $\mathbb P(S_n \le t) > c$ for sufficiently large $n$.

Your expectation that $t$ is a decreasing function of $n$, let alone rapidly decreasing, is not at all true. The same Law of Large Numbers says $\mathbb P(S_n \le r n) \to 0$ as $n \to \infty$ if $0 < r < b$. And for $t = b n$, the Central Limit Theorem says $\mathbb P(S_n \le bn) \to 1/2$ as $n \to \infty$.