Expectation of the difference of sums Let a be vector in $R^{2m}$. 
I would like to calculate $E|\sum_{k=1}^ma_{\pi(k)}-\sum_{k=m+1}^{2m}a_{\pi(k)}|^2,$
Here $\pi(\cdot)$ is a permutation on the set{1,...,2m} with uniform distribution.
Thank you for the help.
 A: The result is
$$
\frac{2mB-A^2}{2m-1},\quad \text{with}\quad A=\sum_{k=1}^{2m}a_k\quad\text{and}\quad B=\sum_{k=1}^{2m}a_k^2.
$$
To prove this, note that for every $k$ and every $\ell\ne k$,
$$
\mathbb E(a_{\pi(k)}^2)=\frac{B}{2m},\quad\mathbb E(a_{\pi(k)}a_{\pi(\ell)})=\frac{A^2-B}{2m(m-1)},
$$
and expand the square in the expectation to be computed.
A: The square contains the square of each entry once, which leads to a term $\lVert a\rVert^2$. It also contains $2m(m-1)$ positive and $2m^2$ negative terms with pairs of different entries, all of which get averaged to the same expectation value A, for a sum of $-2mA$. We can express this in terms of the variance of $a$:
$$
\begin{eqnarray}
\def\Var{\operatorname{Var}}\Var a
&=&
\langle a_i^2\rangle-\langle a_i\rangle^2
\\
&=&
\frac1{2m}\sum_ia_i^2-\frac1{(2m)^2}\left(\sum_ia_i\right)^2
\\
&=&
\frac{2m-1}{(2m)^2}\sum_ia_i^2-\frac1{(2m)^2}\sum_{i\ne j}a_ia_j
\\
&=&
\frac{2m-1}{(2m)^2}\lVert a\rVert^2-\frac1{(2m)^2}2m(2m-1)A
\\
&=&
\frac{2m-1}{(2m)^2}\left(\lVert a\rVert^2-2mA\right)
\end{eqnarray}$$
Thus your expectation value is $(2m)^2/(2m-1)$ times the variance of $a$.
