Q:

The mean and variance of the sample median

The population and the median of a sample sized $2k+1$ should have the same mean and variance. Why is that? Will the result still be so tidy for a sample sized $2k$?

A:

The population and the median of a sample sized $2k+1$ should have the same mean and variance.

Should they, really? To check this, let us compute the mean of the median $M$ of a sample of size $3$ from the distribution with density $f:x\mapsto ax^{a-1}\mathbf 1_{0\lt x\lt1}$, for some positive $a$, and with CDF $F$. 
The distribution of $M$ has density $6fF(1-F)$ and $F(x)=x^a$ for every $x$ in $(0,1)$ hence
$$
E[M]=\int 6xf(x)F(x)(1-F(x))\,\mathrm dx=\int_0^1 6a(x^{2a}-x^{3a})\,\mathrm dx=\frac{6a^2}{(2a+1)(3a+1)},
$$
while the mean of the population is
$$
E[X]=\int xf(x)\,\mathrm dx=\int_0^1ax^{a}\,\mathrm dx=\frac{a}{a+1},
$$
hence $E[M]\ne E[X]$ unless $a=1$.

