Let $X$ be a random variable following a $\mathrm{Binomial}(n,p)$ distribution, and let $$Y=\min\{X,n-X\}.$$ Ispired by the problem posed by C. Clement on https://math.stackexchange.com/questions/1696256/expectation-and-concentration-for-minx-n-x-when-x-is-a-binomial, I want to ask whether there exists some constant $c>0$ such that $\mathbb{E}(Y)\geq c\cdot\min\{p,1-p\}\cdot n$ for all $0<p<1$. If this is not true, can we find some $p_0$ with $\frac{1+\sqrt{5}}{4}\leq p_0<1$ such that if $X\sim \mathrm{Binomial}(n,p_0)$ then $\mathbb{E}(Y)\geq (1-p)[(1+4p)n-8(1+p)]$?
Probability – Lower Bound for Expectation of $\min\{X,n-X\}$ in Binomial Distribution
expectationpr.probabilityprobability distributions
Best Answer
Since $\min(a,b)=(a+b-|a-b|)/2$, your question is really about upper bounding $E|2X-n|$, or, equivalently, $E|X-n/2|$:
$$E\min(X,n-X)=n/2-2E|X-n/2|.$$
You can upper bound $E|X-n/2|$ using Jensen's inequality: $E|X-n/2|\le\sqrt{E(X-n/2)^2}$.
The latter, if I'm not mistaken, evaluates to $$ n\sqrt{ p(1-p)/n+p^2-p+1/4 } =:nF(p).$$
For $n$ sufficiently large and $p$ sufficiently small (certainly, $p\le 0.65$; the exact value can be easily computed), we have $2F(p)\le c(1-p)$ for some universal $c>0$. That means that $n/2-2F(p)n\ge cnp$, so your conjecture holds for this range of $p$. [I'm confident that with a bit more care you can extend the result to all $p$, perhaps with a worse constant. Update: this "confidence" has proven misplaced, see Dmitry Krachun's answer below!]