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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 $n$ is even, so that we can have an integer as threshold when $p=(1-\delta)/2$.

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  • $\begingroup$ We get the same value for $p'=1-\delta-p$ instead of $p$, i.e. the graph is symmetric with respect to $p=\frac{1-\delta}{2}$. What would be sufficient to prove is that there are no more local minimizers. I guess we have convexity. $\endgroup$
    – Markus Sprecher
    Commented Jun 5, 2017 at 20:22
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    $\begingroup$ $p=(1-\delta)/2$ cannot possibly be right for odd $n$ because then $(p+\delta/2)n$ is not an integer, and modifying $p$ so that you hit one of the adjacent integers is an obvious improvement. $\endgroup$
    – Christian Remling
    Commented Jun 5, 2017 at 22:30
  • $\begingroup$ @MarkusSprecher Thanks for the comment. (1) That 's right, the quantity is symmetric w.r.t $p=\frac{1-\delta}2$. (2) Actually we don't have convexity in this case, an obvious reason would be that when $(p+\delta/2\cdot n)$ is an integer, the quantity would be discontinuous $\endgroup$
    – Oliver
    Commented Jun 5, 2017 at 22:32
  • $\begingroup$ @ChristianRemling, thanks for the comment! You're right, what if $n$ is assumed to be even? Do you have any ideas for proving this? $\endgroup$
    – Oliver
    Commented Jun 5, 2017 at 22:34
  • $\begingroup$ I think "tail probabilities" is the technical term for what you're looking at here. See for example this paper by our own @BrendanMcKay, who may be able to say something useful about this: users.cecs.anu.edu.au/~bdm/papers/littlewood2.pdf $\endgroup$
    – Christian Remling
    Commented Jun 5, 2017 at 23:05

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