Prove that distribution of sample median (for even sample) is symmetric Let $\mathbf{X}=(X_1,X_2,\ldots,X_{2k})$ be a sample from normal distribution $\mathcal{N}(0,1)$.
Prove that distribution of sample median (for even sample) is symmetric around 0.
$me(\mathbf{X}) = \frac{X_{k:2k}+X_{(k+1):2k}}{2}$, where $X_{i:2k}$ is $i$-th order statistic.
I was trying to use formula $f_{X_{k:n}}(x) =\frac{n!}{(k-1)!(n-k)!}[F_X(x)]^{k-1}[1-F_X(x)]^{n-k} f_X(x)$, but without any success (maybe there is a way to do it differently than finding $me(\mathbf{X})$ distribution?
 A: To those comfortable with the mathematics of probability and random variables the following argument is a one-liner, because you will take for granted most of the manipulation (and will find it obvious).  To cover all bases, though, I provide the details.
The setting--and a generalization
Let $X = (X_1, X_2, \ldots, X_n)$ be a vector-valued random variable with a distribution $F$ that is "symmetric" about the origin in the sense that for all events $E\subset\mathbb{R}^n,$
$${\Pr}_F(E) = {\Pr}_F(-E).$$
The notation "$-E$" refers to $\{-x\ |\ x \in E\}$.  No assumptions are made about the parity of $n$ or about independence of the components $X_i$.  Note that the conditions of the problem--viz, $n$ even and $X_i$ iid Normal$(0,1)$--are a special case.
Observe--because this is the crux of the matter--that median$(-X)$ = $-$median$(X)$ no matter what $X$ may be.  (For this to be true, it is essential that we define the median of an even number of elements to be the average of their two middle values.)
The one-line proof
The distribution of the median is symmetric because the distribution of $X$ is symmetric and, as we just observed, the median commutes with the symmetry operation $X \to -X,$ QED.
The details
We are asked to show that $f(X)$ = median$(X)$ has a symmetric distribution.  To this end, let $D\subset\mathbb{R}$ be measurable.  The chance that $f(X)$ lies in $D$ is--by definition--given by
$${\Pr}_F(f(X)\in D) = {\Pr}_F(X\in f^{-1}(D)) = {\Pr}_F(f^{-1}(D))$$
The first equality is valid because $f$ is a measurable function.  To prove symmetry, we need to deduce that ${\Pr}_F(f^{-1}(D)) = {\Pr}_F(f^{-1}(-D)).$
Using the definitions and the key observation (which justifies the third equality below), notice that
$$\eqalign{
f^{-1}(D) &= \{X\in\mathbb{R}^n\ |\ f(X)\in D\}  \\
          &= \{X\in\mathbb{R}^n\ |\ -f(X)\in -D\} \\
          &= \{X\in\mathbb{R}^n\ |\ f(-X)\in -D\} \\
          &= \{-X\in\mathbb{R}^n\ |\ f(X)\in -D\} \\
          &=-f^{-1}(-D).
}$$
The symmetry of $F$, applied to the event $E = f^{-1}(D),$ implies the second equality below:
$${\Pr}_F(f^{-1}(D)) = {\Pr}_F(-f^{-1}(-D)) = {\Pr}_F(f^{-1}(-D)).$$
But the latter is exactly the chance that the median lies in $-D$, proving the median has a symmetric distribution.
