Sum of Binomial random variable CDF Suppose there are two independent Binomial random variables
$$
X\sim Binomial(n,p)\\
Y\sim Binomial(n,p+\delta)
$$
where $\delta$ is considered to be fixed, and $p$ can vary in $(0,1-\delta)$.
Now consider the following value (sum of two probabilities):
$$
\mathbb{P} \left( X\geq(p+\frac{\delta}2)\cdot n \right)
+\mathbb{P} \left( Y\leq(p+\frac{\delta}2)\cdot n \right)
$$
My question is: Which value of $p$ maximizes the above quantity?
My intuition and simulation both says that when $p+\frac{\delta}2=1/2$, the value takes its maximum. It seems to be a very easy problem, but I could not prove it. Any thoughts or ideas would be greatly appreciated!
Edit: Let's suppose that nn is even, so that we can have an integer as threshold when $p=(1−\delta)/2p=(1-\delta)/2$.
 A: Your conjecture is close, but not quite correct.  In the case where $n$ is even, yes, it is true.  But in the case where $n$ is odd, you can achieve a slightly probability.  Let $$h(n,p,\delta) = \Pr[X \ge (p+\delta/2)n] + \Pr[Y \le (p+\delta/2)n].$$  So for example, let $n = 3$, $\delta = 1/2$, then $$\begin{align*} h(3,p,1/2) &= \Pr[X \ge 3p + 3/4] + \Pr[Y \le 3p + 3/4]  \\ &= \sum_{k=\lceil 3p + 3/4\rceil}^3 \binom{3}{k} p^k (1-p)^{3-k} + \sum_{k=0}^{\lfloor 3p + 3/4\rfloor} \binom{3}{k} (p+1/2)^k (1/2-p)^{3-k} \\
&= \begin{cases} 
\frac{1}{8} \left(-12 p^2+18 p+1\right), & 0 < p < 1/12 \\
\frac{1}{2} \left(6 p^3-6 p^2+3 p+1\right), & p = 1/12 \\
\frac{1}{2} \left(6 p^2-3 p+1\right), & 1/12 < p < 5/12 \\
\frac{1}{8} \left(-24 p^3+12 p^2-6 p+7\right), & p = 5/12 \\
\frac{1}{8} \left(-12 p^2-6 p+7\right), & 5/12 < p < 1/2.
\end{cases} \end{align*}$$  It is clear that the maximum is attained when $p \in \{1/12, 5/12\}$, whereas $p = (1-\delta)/2 = 1/4$ does not yield a maximum.  The key in the odd case is to choose a value close to $(1-\delta)/2$, but where $$\lceil n(p+\delta/2) \rceil = \lfloor n(p+\delta)/2 \rfloor$$  I leave it to you to find out what these $p$ are.
