# Is the given binomial sum almost everywhere negative as $K\to\infty$?

The binomial sum is as follows:

$$\mathcal {L}^K(\theta)= \sum_{i=\lceil{K/2}\rceil}^K \binom{K}{i}\theta^i\left((1-\theta)^{K-i}-\frac{1}{2}(1-\theta)^{-K}(1-2\theta)^{K-i}\right)$$

which can also be written as

$$\mathcal {L}^K(\theta)= B(K/2,K,1-\theta)-\frac{1}{2}B\left(K/2,K,\frac{1-2\theta}{1-\theta}\right)$$ where $B$ is the Binomial c.d.f.

It can be found that as $K\to\infty$,

$$\quad\mathcal {L}^\infty(\theta)= \begin{cases} 0, & \mbox{if } \mbox{ \theta<\frac{1}{3}} \\ -\frac{1}{2} & \mbox{if } \mbox{ \frac{1}{3}<\theta<\frac{1}{2}} \end{cases}\nonumber$$

My question is however not about it. I want to show that $\mathcal {L}^K(\theta)$ is negative if $K$ is chosen large enough, for every $\theta\in(0,0.5)$.

This question is equivalent to finding that the zero crossing point $\theta_0\in(0,0.5)$ of $\mathcal {L}^K$ tends to $0$ as $K\to\infty$ and for every $\theta>\theta_0$, $\mathcal {L}^K$ is negative. Notice that $K$ is an odd number for all cases.

A mathematica plot confirms that it must be the case. Here is the plot:

But I dont have any idea on how to show it. Does anyone have?

Letting $p=p(\theta)=1-\theta$ and $q=q(\theta)={1-2\theta\over 1-\theta},$ you want to show that $$\sum_{i=0}^{K/2} {K\choose i} p^i (1-p)^{K-i}<{1\over 2}\sum_{i=0}^{K/2} {K\choose i} q^i (1-q)^{K-i},\tag1$$ for sufficiently large $K$.
Trivial case: For $1/3\leq \theta <1/2$, the sum on the left converges to zero and the sum on the right converges to a positive number, so the inequality $(1)$ is true for large $K$.
Remaining case: Suppose $0< \theta <1/3$. We prove the inequality of the sums working term by term. It suffices to show that $$p^i(1-p)^{K-i}<{1\over 2}q^i(1-q)^{K-i}\tag2$$ for all $0\leq i\leq K/2$, when $K$ is large enough.
Note that ${p(1-q)\over q(1-p)}={1-\theta\over 1-2\theta}> 1$ and ${p(1-p)\over q(1-q)}={(1-\theta)^3\over 1-2\theta}< 1$ (for $0<\theta<1/3$). Therefore, $$\left({1-p\over 1-q}\right)^K\left(p(1-q)\over q(1-p)\right)^i \leq \left({1-p\over 1-q}\right)^K\left(p(1-q)\over q(1-p)\right)^{K/2} = \left({p(1-p)\over q(1-q)}\right)^{K/2}.\tag3$$
The right hand side of $(3)$ can be made less than $1/2$ by taking $K$ sufficiently large, giving the inequality (2) and hence the inequality (1).
• The difference for even $K$ is that for $\theta$ very close to $1/2$ from left, one needs so huge $K$ to get sth. negative but this is not the case for odd $K$. Both odd and even $K$ converge to the same pattern eventually, and it is also normal that one doesnt see this in the proof (and actually it is also not necessary). Commented Mar 9, 2016 at 21:24