Consider a random variable $X \sim B(n,p)$. Take the difference in conditional expectations: $DE := E[X|X \geq n/2]-E[X|X < n/2]$
Q1. Does DE has simple form in terms of $n$?
Q2. Does $(n+1)/DE$ increases or decrease as $n$ increases?
I'm looking for properties for $n < 30$. So, applying approximations for large $n$ won't help.
Thanks
Recall that for an event $E$ and a random variable $X$, the conditional expectation $\mathbb E[X\mid E]$ is equal to $\frac{\mathbb E[X\mathsf 1_E]}{\mathbb P(E)}$. So we have \begin{align} \mathbb E\left[X\mid X\geqslant\frac n 2\right] &= \frac{\mathbb E[X\mathsf 1_{\{X\geqslant n/2\}}]}{\mathbb P(X\geqslant n/2)}\\ &= \frac{\sum_{k=\lceil n/2\rceil}^n k\binom nkp^k(1-p)^{n-k}}{\sum_{k=\lceil n/2\rceil}^n \binom nkp^k(1-p)^{n-k}}\\ &= \frac{(1-p)^{-\left\lceil \frac{n}{2}\right\rceil +n-1} p^{\left\lceil \frac{n}{2}\right\rceil } \left(p \binom{n}{\left\lceil \frac{n}{2}\right\rceil +1} \, _2F_1\left(2,-n+\left\lceil \frac{n}{2}\right\rceil +1;\left\lceil \frac{n}{2}\right\rceil +2;\frac{p}{p-1}\right)-(p-1) \left\lceil \frac{n}{2}\right\rceil \binom{n}{\left\lceil \frac{n}{2}\right\rceil } \, _2F_1\left(1,\left\lceil \frac{n}{2}\right\rceil -n;\left\lceil \frac{n}{2}\right\rceil +1;\frac{p}{p-1}\right)\right)}{\binom{n}{\left\lceil \frac{n}{2}\right\rceil } (1-p)^{n-\left\lceil \frac{n}{2}\right\rceil } p^{\left\lceil \frac{n}{2}\right\rceil } \, _2F_1\left(1,\left\lceil \frac{n}{2}\right\rceil -n;\left\lceil \frac{n}{2}\right\rceil +1;\frac{p}{p-1}\right)}, \end{align} where ${}_2F_1$ denotes the hypergeometric function: $${}_2F_1(a,b;c;z) = \sum_{n=0}^\infty \frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}$$ and $(q)_n$ denotes the rising Pochammer symbol, defined by $$ (q)_n = \begin{cases} 1,& n=0\\ \prod_{i=0}^{n-1} (q+i),& n>0. \end{cases} $$ Similarly, \begin{align} \mathbb E\left[X\mid X< n/2\right] &= \frac{\mathbb E[X\mathsf 1_{\{X< n/2\}}]}{\mathbb P(X< n/2)}\\ &= \frac{\sum_{k=0}^{\lfloor n/2\rfloor} k\binom nkp^k(1-p)^{n-k}}{\sum_{k=0}^{\lfloor n/2\rfloor} \binom nkp^k(1-p)^{n-k}}\\ &= \frac{p (1-p)^n \left(n \left(\frac{1}{1-p}\right)^n-\frac{\Gamma (n+1) (1-p)^{-\left\lfloor \frac{n}{2}\right\rfloor -1} p^{\left\lfloor \frac{n}{2}\right\rfloor } \left(n p \Gamma \left(\left\lfloor \frac{n}{2}\right\rfloor +1\right) \, _2\tilde{F}_1\left(1,-n+\left\lfloor \frac{n}{2}\right\rfloor +1;\left\lfloor \frac{n}{2}\right\rfloor +2;\frac{p}{p-1}\right)-p+1\right)}{\Gamma \left(n-\left\lfloor \frac{n}{2}\right\rfloor \right) \Gamma \left(\left\lfloor \frac{n}{2}\right\rfloor +1\right)}\right)}{(1-p)^n \left(\left(\frac{1}{1-p}\right)^n-\binom{n}{\left\lfloor \frac{n}{2}\right\rfloor +1} (1-p)^{-\left\lfloor \frac{n}{2}\right\rfloor -1} p^{\left\lfloor \frac{n}{2}\right\rfloor +1} \, _2F_1\left(1,-n+\left\lfloor \frac{n}{2}\right\rfloor +1;\left\lfloor \frac{n}{2}\right\rfloor +2;\frac{p}{p-1}\right)\right)}, \end{align} where ${}_2\tilde F_1$ denotes the regularized hypergeometric function: $$ {}_2\tilde F_1(a,b;c;z) = \frac1{\int_0^\infty t^{c-1}e^{-t}\ \mathsf dt}\sum_{n=0}^\infty \frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}. $$ Then we have \begin{align} DE &= \mathbb E\left[X\mid X\geqslant \frac n 2\right] - \mathbb E\left[X\mid X<\frac n 2\right]\\ &= p (1-p)^n \left(n \left(\frac{1}{1-p}\right)^n-\frac{\Gamma (n+1) (1-p)^{-\left\lfloor \frac{n}{2}\right\rfloor -1} p^{\left\lfloor \frac{n}{2}\right\rfloor } \left(n p \Gamma \left(\left\lfloor \frac{n}{2}\right\rfloor +1\right) \, _2\tilde{F}_1\left(1,-n+\left\lfloor \frac{n}{2}\right\rfloor +1;\left\lfloor \frac{n}{2}\right\rfloor +2;\frac{p}{p-1}\right)-p+1\right)}{\Gamma \left(n-\left\lfloor \frac{n}{2}\right\rfloor \right) \Gamma \left(\left\lfloor \frac{n}{2}\right\rfloor +1\right)}\right) - \frac{p \left(n \left(\frac{1}{1-p}\right)^n-\frac{\Gamma (n+1) (1-p)^{-\left\lfloor \frac{n}{2}\right\rfloor -1} p^{\left\lfloor \frac{n}{2}\right\rfloor } \left(n p \Gamma \left(\left\lfloor \frac{n}{2}\right\rfloor +1\right) \, _2\tilde{F}_1\left(1,-n+\left\lfloor \frac{n}{2}\right\rfloor +1;\left\lfloor \frac{n}{2}\right\rfloor +2;\frac{p}{p-1}\right)-p+1\right)}{\Gamma \left(n-\left\lfloor \frac{n}{2}\right\rfloor \right) \Gamma \left(\left\lfloor \frac{n}{2}\right\rfloor +1\right)}\right)}{\left(\frac{1}{1-p}\right)^n-\binom{n}{\left\lfloor \frac{n}{2}\right\rfloor +1} (1-p)^{-\left\lfloor \frac{n}{2}\right\rfloor -1} p^{\left\lfloor \frac{n}{2}\right\rfloor +1} \, _2F_1\left(1,-n+\left\lfloor \frac{n}{2}\right\rfloor +1;\left\lfloor \frac{n}{2}\right\rfloor +2;\frac{p}{p-1}\right)}\\ &= p \left((1-p)^n-\frac{1}{\left(\frac{1}{1-p}\right)^n-\binom{n}{\left\lfloor \frac{n}{2}\right\rfloor +1} (1-p)^{-\left\lfloor \frac{n}{2}\right\rfloor -1} p^{\left\lfloor \frac{n}{2}\right\rfloor +1} \, _2F_1\left(1,-n+\left\lfloor \frac{n}{2}\right\rfloor +1;\left\lfloor \frac{n}{2}\right\rfloor +2;\frac{p}{p-1}\right)}\right) \left(n \left(\frac{1}{1-p}\right)^n-\frac{\Gamma (n+1) (1-p)^{-\left\lfloor \frac{n}{2}\right\rfloor -1} p^{\left\lfloor \frac{n}{2}\right\rfloor } \left(n p \Gamma \left(\left\lfloor \frac{n}{2}\right\rfloor +1\right) \, _2\tilde{F}_1\left(1,-n+\left\lfloor \frac{n}{2}\right\rfloor +1;\left\lfloor \frac{n}{2}\right\rfloor +2;\frac{p}{p-1}\right)-p+1\right)}{\Gamma \left(n-\left\lfloor \frac{n}{2}\right\rfloor \right) \Gamma \left(\left\lfloor \frac{n}{2}\right\rfloor +1\right)}\right). \end{align} So no, $DE$ does not have a simple form in terms of $n$.