In the process of computing inclusion constants for the complex matrix cube (which is a free spectrahedron), the following identity was proven: for all $n \geq 1$,
$$\mathbb E \Big| \sum\_{i=1}^{2n} x\_i^2 - \sum\_{j=1}^{2n} y\_j^2 \Big| = \mathbb E \Big| \sum\_{i=1}^{2n} x\_i^2 - \sum\_{j=1}^{2n-2} y\_j^2 \Big| = 4^{1-n}n \binom{2n}{n},$$
where $x\_i$,$y\_j$ are i.i.d. standard real Gaussian random variables. The proof we have at this moment is by using the explicit form of the density of the difference of two $\chi^2$ random variables, see [here](https://math.stackexchange.com/questions/85249/distribution-of-difference-of-chi-squared-variables) and also *Klar, Bernhard*, [**A note on gamma difference distributions**](http://dx.doi.org/10.1080/00949655.2014.996566), [ZBL07183251](https://zbmath.org/?q=an:07183251).


Since the result is so simple, there should be a more direct and more insightful proof of it.


**Question 1:** give an easy, conceptual proof of the identity above.


Consider now the function
$$k \mapsto \mathbb E \Big| \sum\_{i=1}^{2k} x\_i^2 - \sum\_{j=1}^{2(n-k)} y\_j^2 \Big|$$
from $\{0,1,\ldots, n\} \to \mathbb R\_+$; as above, the $x\_i$ and $y\_j$ are standard i.i.d. Gaussians.


**Question 2:** Show that the function above is unimodular, and that its minimum is attained at $k = \lfloor n/2 \rfloor$.


There is a pretty involved proof of the second fact above in *Helton, J. William; Klep, Igor; McCullough, Scott; Schweighofer, Markus*, [**Dilations, linear matrix inequalities, the matrix cube problem and beta distributions**](http://dx.doi.org/10.1090/memo/1232), Memoirs of the American Mathematical Society 1232. Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-3455-7/pbk; 978-1-4704-4947-6/ebook). vi, 106 p. (2019). [ZBL1447.47009](https://zbmath.org/?q=an:1447.47009).


**Question 3:** Does the claim in ``Question 2'' hold for arbitrary probability distributions?


Any help or insight about these questions would be appreciated!



A remarkable identity involving $\chi^2$ random variables
I think I found an elementary proof of Question 2/3 for arbitrary probability distributions. In fact, it is not required that the components in the sums are squares, but general i.i.d. non-negative random variables work. Further, the requirement that both sums have an even number of terms ($2k$ and $2(n-k)$ in the question) is not required.


Let $X\_1, ..., X\_N$ be i.i.d. non-negative, integrable random variables.
Let $T\_k := \mathbb{E}[|\sum\_{i=1}^k X\_i - \sum\_{i=k+1}^N X\_i|]$.


**Lemma 1:** For $k \geq \lfloor N/2\rfloor$ it holds $T\_{k+1} \geq T\_k$.


By symmetry (for $k \leq \lfloor N/2\rfloor$) this Lemma yields the question.
To prove Lemma 1, we require the two following supplementary statements


**Lemma 2:** Let $C$ be a symmetric, integrable random variable, and $a, b\in\mathbb{R}$, $|b| \geq |a|$. Then $\mathbb{E}[|C+b|] \geq \mathbb{E}[|C+a|]$.


Hereby, $C$ being symmetric means that $C$ and $-C$ have the same distribution.


**Lemma 3:** Let $C$ be a symmetric, integrable random variable and $A$, $B$ non-negative and integrable random variables and assume that $A, B, C$ are independent. Then $\mathbb{E}[|C+A-B|] \leq \mathbb{E}[|C+A+B|]$.


***Proof of Lemma 2:*** Without loss of generality, by symmetry of $C$, let $0\leq a\leq b$. Then one calculates
\begin{align\*}
\mathbb{E}[|C+a|] &= \mathbb{E}[\mathbb{1}\_{a \geq |C|} (C+a) + \mathbb{1}\_{a < |C|} (\mathbb{1}\_{C > 0} + \mathbb{1}\_{C < 0}) |C+a|]\\
&= \mathbb{P}(|C| \leq a) a + \mathbb{E}[\mathbb{1}\_{a < |C|} (\mathbb{1}\_{C > 0} (C+a) + \mathbb{1}\_{C < 0} (-C-a)]\\
&=\mathbb{P}(|C| \leq a) a + \mathbb{E}[\mathbb{1}\_{|C| > a} |C|] \\
&=\mathbb{E}[\max\{a, |C|\}]
\end{align\*}
and this term is obviously increasing in $a$. (in general, it simply holds $\mathbb{E}[|C+a|] = \mathbb{E}[\max\{|C|, |a|\}]$, which can also be proved via the identity $|C+a| + |-C+a| = 2\max\{|C|, |a|\}$ as pointed out by Fedor.)


***Proof of Lemma 3:***
Let $C \sim \mu, A \sim \nu, B \sim \theta$. Then
$$
\mathbb{E}[|C+A-B| - |C+A+B|] = \int \int \Big(\int |c+a-b| - |c+a+b| \mu(dc)\Big)\nu(da)\theta(db) \leq 0
$$
since the term inbetween the large brackets is non-positive since $|a+b| \geq |a-b|$ (since $a, b \geq 0$) and by Lemma 2.


***Proof of Lemma 1:***
Let $C := \sum\_{i=1}^{N-k-1} X\_i - \sum\_{i=k+1}^N X\_i$ and note that $C$ is symmetric, $A := \sum\_{i=N-k}^k X\_i$ (where the sum is understood to be 0 if $k < N-k$), and $B := X\_{k+1}$. Note that, for $k \geq \lfloor \frac{N}{2}\rfloor$, it holds $T\_k = \mathbb{E}[|C+A-B|]$ and $T\_{k+1} = \mathbb{E}[|C+A+B|]$. The claim now follows from Lemma 3.


