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Let $a_1,a_2,\dots,a_n$ be independent identically distributed random vectors in $\mathbb R^n$. I need a bound for $E[|\det A|^{-1}]$, where $A$ is the matrix composed out of these vectors.

More specifically, these vectors take their values on a curve.

And more generally, I will be happy even if there is an estimate for a "non-square" determinant, precisely, for $E[G^{-1/2}]$, where $G$ is the Grammian determinant of $n$ iid vector in $\mathbb R^m$ and $m>n$.

Update: yes, I need an upper bound.

Precisely, I have something like $a_i = (f_1(\xi_i),f_2(\xi_i),\dots,f_n(\xi_i))$, where $f_j(x) = |x|^{-\alpha_i}\sin(\beta_i x)$ with $\alpha_i\in(1,2)$ and look for an estimate in terms of $|\alpha_i-\alpha_j|$ and $|\beta_i-\beta_j|$.

I'm flexible with the choice of distribution for $\xi$, but it should be the same for all sets of $\alpha$'s and $\beta$'s.

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  • $\begingroup$ Do you want an upper bound, a lower bound, or both? $\endgroup$
    – Mark Meckes
    Nov 4, 2010 at 17:56
  • $\begingroup$ Can you write more precisely how do you choose the probability measure? Is your curve compact, and then you restrict the standard Lebesgue measure from $R^n$ to this curve? If this is the case then the answer certainly depends on the curve - how big a proportion of it lies near the point $(0,\ldots, 0)$. $\endgroup$
    – Łukasz Grabowski
    Nov 4, 2010 at 18:01

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Perhaps my paper "How good is Hadamard's inequality for determinants", Canad. Math. Bull. 27 (1984) 260-264 may help.

A rough statement of the main theorem is the following. Assume that we have a probability distribution on the set of all real $n \times n$ matrices $A$ such that: (1) the density at $A$ only depends on the lengths of the columns of $A$; and (2) the probability that $\det A$ is nonzero is 1. Define the Hadamard ratio $h(A)$ to be $|det A|$ divided by the product of the lengths of the columns.

Then for nearly all $A$ the value of $h(A)^2$ is close to $n!/n^n$. (The theorem gives more precise results.)

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    $\begingroup$ Note quite what I needed. Still I voting it up so you will be able to add a comment rather than posting an answer next time. $\endgroup$
    – zhoraster
    Jul 6, 2012 at 21:56

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