Consider $n$ independent draws from a symmetric, continuous and non-degenerate distribution F with expectation $\text{E}[X]$. Let us order these draws so that $X_1 \leq X_2 \leq ... \leq X_n$, so $X_k$ is the kth order statistic of this random variable (e.g. $X_n$ is the maximum).
Intuitively, I would expect that any order statistic that is higher than the median would have an expected value that is strictly greater than $\text{E}[X]$. In other words, $\text{E}[X_k] > \text{E}[X]$ if $k > n/2$. Similarly, I would expect that any order statistic that is lower than the median would have an expected value that is strictly smaller than $\text{E}[X]$. That is, $\text{E}[X_k] < \text{E}[X]$ if $k < n/2$. However, I cannot prove this - does anyone have any ideas or pointers?
My efforts so far: The CDF of the maximum (nth order statistic) is $F()^n$. By Fubini's theorem, we can write its expected value as
$E[X_n] = \int_{0}^{\infty} 1 - F(x)^n dx - \int_{-\infty}^{0}F(x)^n dx > \int_{0}^{\infty} 1 - F(x) dx - \int_{-\infty}^{0}F(x) dx = \text{E}[X]$
assuming that $n > 1$ and $F(x)<1$ for some $x$.
The argument also establishes that the expected value of the minimum is less than $\text{E}[X]$. However, I am struggling to make it work for other order statistics (e.g. second highest).
Note: if the general case is difficult, it would still be interesting to see if this could be shown for a specific distribution, e.g. the normal distribution.
Your intuition is correct. Assume $\Bbb E[|X|]<\infty$. Observe that if $Y_{(1)}\le Y_{(2)}\le \cdots \le Y_{(n)}$ is the order statistics of $(X_i)_{i\le n}$, then $$ 2\Bbb E[X] -Y_{(n)}\le2\Bbb E[X] -Y_{(n-1)}\le\cdots\le2\Bbb E[X] -Y_{(1)} $$ is the order statistics of $(2\Bbb E[X]-X_i)_{i\le n}$. Since we are assuming symmetry of the law of $X$, we find that $$ (X_i)_{i=1}^n \stackrel{d}= \left(2\Bbb E[X] -X_i\right)_{i=1}^n. $$ This implies $$ Y_{(r)} \stackrel{d}= 2\Bbb E[X]-Y_{(n+1-r)} $$ for all $r=1,2,\ldots,n$. Thus $$ \Bbb E[Y_{(r)}] = 2\Bbb E[X]-\Bbb E[Y_{(n+1-r)}],\quad \forall 1\le r\le n. $$ We have $Y_{(r)}\ge Y_{(n+1-r)}$ for $r\ge\frac{n+1}{2}$, hence $$ \Bbb E[Y_{(r)}] \ge \frac{1}{2}\left(\Bbb E[Y_{(r)}]+\Bbb E[Y_{(n+1-r)}]\right)\ge \Bbb E[X],\quad \forall r\ge \frac{n+1}{2}. $$ We also obtain $$ \Bbb E[Y_{(n+1-r)}]=2\Bbb E[X]-\Bbb E[Y_{(r)}] \le \Bbb E[X]. $$ If we assume non-degenerate distribution of $X$ and $r>\frac{n+1}{2}$, then since $Y_{(r)}>Y_{(n+1-r)}$ a.s., we have strict inequality $$ \Bbb E[Y_{(r)}] >\Bbb E[X]>\Bbb E[Y_{(n+1-r)}]. $$