Take a sequence of $n$ i.i.d. random variables symmetric around zero and with zero expectation:
$$
\eta_1,\eta_2,\dots, \eta_n.
$$
Use standard order statistic notation and consider
$$
\eta_{(1)}\leq \eta_{(2)}\leq \dots \leq \eta_{(n)}.
$$
Let $L\subset \{1,\dots, n\}$ with cardinality $n_L$ and define the sets
$$
L_\eta\equiv \{i\in \{1,\dots, n\}: \eta_i
\geq \eta_{(n_L)}\} \quad U_\eta\equiv \{i\in \{1,\dots, n\}: \eta_i
\geq \eta_{(n-n_L)}\}
$$
Question: show that
$$
E\left(\sum_{i\in U_\eta} \eta_{(i)}-\sum_{i\in L_\eta}\eta_{(i)}\right)=0.
$$
The book says that symmetry is key here but I struggle to understand the result. Any help would be appreciated.
Best Answer
The following is a proof for
$$ \mathbb E\left(\sum_{i\in \color{red}{U'_\eta}} \eta_{(i)}-\sum_{i\in \color{red}{L'_\eta}}\eta_{(i)}\right)=0. $$
where $$ \color{red}{L'_\eta}\equiv \{i\in \{1,\dots, n\}: \eta_i \color{red}{>} \eta_{(n_L)}\} \quad \color{red}{U'_\eta}\equiv \{i\in \{1,\dots, n\}: \eta_i \color{red}{>} \eta_{(n-n_L)}\}. $$
Discussion:
I think there is a typo in the book or the question. Consider the following example: $n=3$, $\eta \sim U(-0.5,0.5)$. Then, the expectations of $\eta_{(1)}, \eta_{(2)}, \eta_{(3)}$ are $\frac{-1}{4}, 0, \frac{1}{4}$. Then, for the sets defined in the question and for $n_L=1$
$$ \mathbb E\left(\sum_{i\in {U_\eta}} \eta_{(i)}-\sum_{i\in {L_\eta}}\eta_{(i)}\right)=0+ \frac{1}{4} -\left (\frac{-1}{4}+ 0+ \frac{1}{4} \right) \neq 0. $$
However, for the modified sets:
$$ \mathbb E\left(\sum_{i\in {U'_\eta}} \eta_{(i)}-\sum_{i\in {L'_\eta}}\eta_{(i)}\right)=\frac{1}{4} -\left (0+ \frac{1}{4} \right) =0.$$
Proof:
As $\eta_{(1)},\eta_{(2)},\cdots, \eta_{(n)}$ are the order statistics of $\eta_i, i=1, \dots, n$, then $-\eta_{(n)}\le -\eta_{(n-1)}\le \cdots \le -\eta_{(1)}$ are the order statistics of $-\eta_i, i=1, \dots n.$ From the symmetry assumption $$-\eta_i \stackrel{d}=\eta_i, i, \dots n,$$ which implies that
$$\eta_{(i)} \stackrel{d}= -\eta_{(n+1-i)}.$$
Moreover, we have $$\sum_{i=1}^n\eta_{i}= \sum_{i=1}^{n_L}\eta_{(i)}+\sum_{i=n_L+1}^{n}\eta_{(i)},$$
which gives
$$\sum_{i\in L'_\eta}\eta_{(i)}=\sum_{i=n_L+1}^{n}\eta_{(i)}=\sum_{i=1}^n\eta_{i} -\sum_{i=1}^{n_L}\eta_{(i)}$$
Hence, using the above identity,
$$ \sum_{i\in U'_\eta} \eta_{(i)}-\sum_{i\in L'_\eta}\eta_{(i)}=\sum_{i=n-n_L+1}^{n}\eta_{(i)}-\sum_{i=1}^n\eta_{i} +\sum_{i=1}^{n_L}\eta_{(i)}=\sum_{i=1}^{n_L}\eta_{(n+1-i)} +\sum_{i=1}^{n_L}\eta_{(i)} -\sum_{i=1}^n\eta_{i}=\sum_{i=1}^{n_L} \big [\eta_{(n+i-1)}+\eta_{(i)} \big] -\sum_{i=1}^{n}\eta_{i}. $$
Now, again from the symmetry assumption and $\eta_{(i)} \stackrel{d}= -\eta_{(n+1-i)}$, we have $$ \mathbb E\left(\sum_{i=1}^{n_L} \big [\eta_{(n+i-1)}+\eta_{(i)} \big] -\sum_{i=1}^{n}\eta_{i}\right)=0. $$