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Given $c>0$. Let $\gamma_c:{\cal M}_{k \times k}^+\mapsto {\cal M}_{k \times k}^+$ is a function defined by \begin{equation} \gamma_c(\Omega)=\frac1{\sqrt{(2\pi)^{k}|\Omega|}}\int_{\mathbb{R}^k}\{(-c)\vee x\wedge c\}\{(-c)\vee x\wedge c\}^Te^{-\frac12x^T\Omega^{-1}x}dx_1\cdots dx_k \end{equation} with $x=(x_1,\dots,x_k)^T$ and ${\cal M}_{k \times k}^+$ denotes the set of all symmetric positive definite $k\times k$ matrices and $(-c)\vee x\wedge c$ represent the vector in which its $j-$th element is $\max\{-c,\min\{x_j,c\}\}$ for every vector $j=1,\dots,k$. We can see $\gamma_c$ as the second moment of $(-c)\vee X\wedge c$ where $X\sim N_k(0,\Omega)$. Now, suppose $n-$vectors $v_1,\dots,v_n\in\mathbb{R}^k$ are given such that $V=\frac1n\sum_{i=1}^n{\{(-c)\vee v_i\wedge c\}\{(-c)\vee v_i\wedge c\}^T}$ is non singular.

I'd like to have that there exists $\Omega\in {\cal M}_{k \times k}^+$ such that $\gamma_c(\Omega)=V$.

I have investigated for $k=1$. In this case we know $\gamma_c$ is continuous strictly increasing with range $(0,c^2)$. Therefore, $V\in (0,c^2)$ and so we can choose $\omega\in\mathbb{R}^+$ such that $\gamma_c(\omega)=V$.

However, if $k>1$, it is really difficult to find the range of $\gamma_c$. Can anyone help me? Thank you in advance.

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    $\begingroup$ Is there not a sign problem in the integral? In dimension $1$, $$ \gamma_c(a) \geq \frac{1}{\sqrt{2\pi a}}\int_{c}^\infty c^2 exp(x^2/a) dx=\infty.$$ $\endgroup$
    – username
    Feb 23, 2014 at 18:42
  • $\begingroup$ Oh, I did a typo. Thanks @AthanagorWurlitzer for reminding me. I've edited my formula $\endgroup$
    – Jlamprong
    Mar 1, 2014 at 19:11

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