Concavity of $f(p)=\sum_{i=0}^\frac{n-1}{2} C_{n}^{i}p^i(1-p)^{n-i}$

Question

Assuming that $f(p)=\sum_{i=0}^\frac{n-1}{2} C_{n}^{i}p^i(1-p)^{n-i}$, where $n$ is odd and $p > 0.5$. Is that true that $f$ is a convex function?

I have tried to use derivation. It is easy to show that $f$ is monotonically decreasing since $$f'(p) = \sum_{i=0}^\frac{n-1}{2} C_{n}^{i}p^{i - 1}(1-p)^{n-i-1}(i-np)$$ and $i-np < 0$.

However, concavity is not easy as $$f''(p) = \sum_{i=0}^\frac{n-1}{2} C_{n}^{i}p^{i - 1}(1-p)^{n-i-2}(i^{2}-2inp - i +2ip - p^{2}n +n^{2}p^{2}),$$ $(i^{2}-2inp - i +2ip - p^{2}n +n^{2}p^{2})$ is not always greater than $0$.

I have done some experiments and it seems that $f$ is a convex function. So the question is how to prove it.

Answer 1

Yes, that's true. Actually, with $m:=(n-1)/2$, the derivative simplifies to $$f'(p)=-\frac{n!}{m!^2}p^m(1-p)^m$$ (a proof can be adapted from here or here). Now it's easy to compute $f''(p)$, etc.