# 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 $K\to\infty$,

\begin{equation} \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 \end{equation}

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). – Seyhmus Güngören Mar 9 '16 at 21:24