A remarkable identity involving $\chi^2$ random variables

Question

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 and also Klar, Bernhard, A note on gamma difference distributions, ZBL07183251.

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, 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.

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

Any help or insight about these questions would be appreciated!

Answer 1

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.

Answer 2

1. The proof of $$ \mathbb E \Big| \sum_{i=1}^{2n} x_i^2 - \sum_{j=1}^{2n} y_j^2 \Big| = 4^{1-n}n \binom{2n}{n}. $$

Denote $z_i=(x_i-y_i)/\sqrt{2}$, $w_i=(x_i+y_i)/\sqrt{2}$. Then the vectors $z$ and $w$ are i.i.d. standard Gaussian, and $$\left|\sum_{i=1}^{2n} (x_i^2-y_i^2)\right|=2\left|\langle z,w\rangle\right|.$$ For finding the expectation of $|\langle z,w\rangle|$ we may fix $z$, if $|z|=a$, then $|\langle z,w\rangle|$ is distributed as $a|X|$ for standard Gaussian $X$, the expected value over $w$ is $a\sqrt{2/\pi}$ (I use that mean absolute value of $X$ equals $\sqrt{2/\pi}$). So, it remains to compute $\mathbb{E} |z|$ which is also well known.

2. The proof of $$\mathbb E \Big| \sum_{k=1}^{2n} x_k^2 - \sum_{j=1}^{2n} y_j^2 \Big|=\mathbb E \Big| \sum_{k=1}^{2n} x_k^2 - \sum_{j=1}^{2n-2} y_j^2 \Big|.$$ We apply the integral representation $$ \frac{\pi}2\left(|b|-|a|\right)=\int_0^\infty \frac{\cos at-\cos bt}{t^2}dt. $$ Thus by Fubini theorem it suffices to prove that for each specific $t$ we have $$\mathbb E \cos t\Big( \sum_{k=1}^{2n} x_k^2 - \sum_{j=1}^{2n} y_j^2 \Big)=\mathbb E \cos t\Big( \sum_{k=1}^{2n} x_k^2 - \sum_{j=1}^{2n-2} y_j^2 \Big). \quad\quad\quad (\heartsuit) $$ For this we write $2\cos x=e^{ix}+e^{-ix}$ and apply the independence of $x_k$'s and $y_j$'s. We have for real $t$ and standard gaussian $X$ $$ \mathbb E e^{itX^2}=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{it X^2-X^2/2}dt=\frac1{\sqrt{1-2it}} $$ (the square root branch in the right half-plane is natural). Therefore $(\heartsuit)$ reads as $$ \frac2{(1-2it)^n(1+2it)^n}=\frac1{(1-2it)^n(1+2it)^{n-1}}+ \frac1{(1-2it)^{n-1}(1+2it)^n} $$ that is true.