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

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!

• Here's the paper where the question popped out: arxiv.org/abs/2105.11302 see Lemma 6.2 and Theorem 6.3 May 25, 2021 at 8:16

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.

• Nice. Lemma 2 follows just from an identity $|C+a|+|-C+a|=2\max(|C|, |a|)$. May 23, 2021 at 19:52
• Beautiful proof, thanks a lot, that's exactly what I was looking for. Thanks Fedor for simplifying Lemma 2 May 24, 2021 at 8:11
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.

1. 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.
• I would have loved to split the "accepted answer" token between yours and Steve's, but unfortunately MO enforces a choice :) May 24, 2021 at 8:13
• I would also choose Steve's: ) May 24, 2021 at 8:44