Prove that the two polynomials intersect each other only at a single point Here are the polynomials:
$$D^K_1(\theta)=\sum_{i=\lceil{K/2}\rceil}^K \binom{K}{i}\theta^i(1-\theta)^{K-i}$$
and
$$D^K_2(\theta)=\frac{1}{2}\sum_{i=\lceil{K/2}\rceil}^K \binom{K}{i}\left(\frac{\theta}{1-\theta}\right)^i\left(1-\frac{\theta}{1-\theta}\right)^{K-i}$$

Question: For every $K$ (odd) is it true that $D^K_1(\theta)$ intersects $D^K_2(\theta)$ only at a single point on $\theta\in(0,\,0.5)$?

Note: Second polynomial is a polynomial in $\theta/(1-\theta)$ thanks to GerryMyerson.

My own work: I worked on the problem and obtained the following:

$D^K_1(0)=D^K_2(0)=0$ and $D^K_1(1/2)=D^K_2(1/2)=1/2$
Both $D^K_1(\theta)$ and $D^K_2(\theta)$ are monotone increasing functions of $\theta$ on $0<\theta<1/2$ for all $K$.
What remains to show is that $D^K_1(\theta)-D^K_2(\theta)$ is neither positive nor negative for all $0<\theta<1/2$ given a $K$

Namely, if two polynomials are increasing, starting and ending at the same points, either one is less or greater than the other at all points or it must intersect the other polynomial at a single point. Below, I give some figures for $K=3$, $K=5$ and $K=7$
Added (29.09): If I multiply both $D^K_1(\theta)$ and $D^K_2(\theta)$ by $2$, then they will be cumulative distribution functions of two different random variables $X$ and $Y$ on $[0,0.5]$. Then it will suffice to show that $X$ doesnt stochastically dominate $Y$ or vice versa.
Added (29.09)_ver2: I think there is a much easier way for the proof. Only two things are required.
$1$-$)$ For $\theta=0.45$ or some other number very close to $0.5$ but still less than that, $D^K_2(\theta)>D^K_1(\theta)\forall K$, it is obvious from the figures that red line is above the blue for all $\theta$ large enough in $(0,0.5)$
$2$-$)$ For $\theta=\epsilon$, namely keeping it as small as possible, in other words whenever $\epsilon\rightarrow 0$ then $D^K_1(\theta)>D^K_2(\theta)\forall K$.
$1$-$)$ seems to be easy I guess but for $2$-$)$ I have currently no idea.




With regards to your question #2 about showing that $D_{2}^{K}(\theta) < D_{1}^{K}(\theta)$ for all $K$ when $|\theta| < \epsilon$ is sufficiently small, I observe the following:
$$\lim_{|\theta| = \epsilon \to 0} \frac{D_{1}^{K}(\epsilon) - D_{2}^{K}(\epsilon)}{ \epsilon^{\lceil K/2 \rceil}} = \lim_{\epsilon \to 0} \sum_{i=\lceil K/2 \rceil}^{K}\binom{K}{i}\epsilon^{i-\lceil K/2 \rceil} \left[ (1-\epsilon)^{K-i} - \frac{1}{2}\frac{1}{(1-\epsilon)^{i}}\left(1 - \frac{\epsilon}{1-\epsilon}\right)^{K-i}\right]\\
= \binom{K}{\lceil K/2 \rceil}\frac{1}{2} > 0$$
Thus, by continuity we have that for any $K$ there exists an $\epsilon_{K} > 0$ small enough such that $D_{1}^{K}(\theta) > D_{2}^{K}(\theta)$ for all $\theta \in (0, \epsilon_{K})$.
