Jensen-like inequality for random matrix: $\Bbb E[\det X^2]\ge\det\Bbb E[X^2]$

Question

Let $X\in M_n(\Bbb R)$ be a random matrix with iid elements following a continuous distribution.

What are the necessary and sufficient conditions for $$\Bbb E[\det X^2]\ge\det\Bbb E[X^2]$$ to hold? Is there still a geometric interpretation of this inequality?

Examples where the inequality is true:

• $X_{i,j}$ is exponentially distributed, as LHS/RHS equals $(n+1)!/(n^2+1)$

• $X_{i,j}$ is normally distributed with zero mean, as LHS/RHS equals $n!$

Answer 1

From the identities in @OlivierBégassat's answer, the inequality $\Bbb E\det X^2\ge\det\Bbb EX^2$ can be written as $$\small n!\sum_{f=0}^n\binom nf(-1)^{n-f-1}(n-f-1)(\Bbb V[X]+\Bbb E[X]^2)^f\Bbb E[X]^{2(n-f)}\ge\Bbb V[X]^{n-1}(\Bbb V[X]+n^2\Bbb E[X]^2).$$ As @CarloBeenakker showed, the inequality holds when $\Bbb E[X]=0$ so here we assume a non-zero mean.

Let $s=\Bbb V[X]/\Bbb E[X]^2$ be the squared coefficient of variation, so we have the equivalence $$n!\sum_{f=0}^n\binom nf(-1)^{n-f-1}(n-f-1)(1+s)^f\ge s^{n-1}(n^2+s).$$ The LHS can be expressed as \begin{align}\small-n!(1+s)^{n-2}\sum_{f=0}^n\binom nf(n-f-1)\left(-\frac1{1+s}\right)^{n-f-2}&=\small-n!(1+s)^n\frac d{ds}\sum_{f=0}^n\binom nf\left(-\frac1{1+s}\right)^{n-f-1}\\&=\small n!(1+s)^n\frac d{ds}\frac{s^n}{(1+s)^{n-1}}\\&=\small n!s^{n-1}(n+s)\end{align} so the inequality simplifies to $n!(n+s)\ge n^2+s$ which is true for all $n\ge1$ and $s>0$.

Thus $\Bbb E\det X^2\ge\det\Bbb EX^2$ holds for all square $X$ with iid elements.

Answer 2

A sufficient condition is zero mean; the variance $\sigma^2$ may be arbitrary and no normality asssumption is needed: then $\mathbb{E}[\det X^2]=n!\sigma^{2n}$ and $\det\mathbb{E}[X^2]=\sigma^{2n}$.

To see that $\mathbb{E}[\det X^2]=n!\sigma^{2n}$, for any zero-mean i.i.d. of the matrix elements, consider the $n!$ terms in $\det X$. Upon squaring and averaging the cross-terms vanish, only $n!$ terms with average $\sigma^{2n}$ remain.

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Is the zero-mean condition necessary? I checked it is not for $n\leq 4$.

Conjecture: $\Bbb E[\det X^2]\ge\det\Bbb E[X^2]$ for any $n\times n$ matrix $X$ with i.i.d. matrix elements.

Answer 3

Not an answer but too long for a comment. By multiplying out both terms one gets $$ \DeclareMathOperator{\Fix}{Fix} \newcommand{\E}{\mathbf{E}} \newcommand{\detxsq}{\det(X)^2} \left\{ \begin{array}{lcl} \E[\detxsq] & = & \displaystyle n! \cdot \sum_{f=0}^n ~ c_f \cdot m_2^f \cdot m_1^{2(n-f)} \\ \det\E[X^2] & = & \displaystyle (m_2 - m_1^2)^{n-1}\Big[m_2 + (n^2-1)\cdot m_1^2\Big] \\ \end{array} \right.$$ with $$ c_f = \sum_{\substack{\sigma \in \mathfrak{S}_n \\ \#\Fix(\sigma) = f}} (-1)^\sigma \overset{(\star)}= \begin{cases} \binom{n}{f}(-1)^{n-f-1}(n-f-1) & \text{if }f<n \\ 1 & \text{if }f=n \\ \end{cases} $$ and $m_1 = \E[x]$ and $m_2 = \E[x^2]$ with $x$ following the distribution of the entries of $X$ and $(\star)$ follows (for $f<n$) from Recounting the Odds of an Even Derangement.