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!


2 Answers 2


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.

  • 2
    $\begingroup$ Nice. Lemma 2 follows just from an identity $|C+a|+|-C+a|=2\max(|C|, |a|)$. $\endgroup$ May 23, 2021 at 19:52
  • $\begingroup$ Beautiful proof, thanks a lot, that's exactly what I was looking for. Thanks Fedor for simplifying Lemma 2 $\endgroup$ 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.
  • $\begingroup$ I would have loved to split the "accepted answer" token between yours and Steve's, but unfortunately MO enforces a choice :) $\endgroup$ May 24, 2021 at 8:13
  • 3
    $\begingroup$ I would also choose Steve's: ) $\endgroup$ May 24, 2021 at 8:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.