Determinant of the covariance matrix in a normal distribution Suppose a $p \times 1$ vector $x \sim N_p(\boldsymbol 0, \boldsymbol \Sigma_1)$. Now, There is another covariance matrix $\boldsymbol \Sigma_2$. We know that $|\boldsymbol \Sigma_2| < |\boldsymbol \Sigma_1|$, where $|\cdot|$ is the determinant. Is there any relationship between  $\boldsymbol x^\prime \boldsymbol \Sigma_2 \boldsymbol x$ and $\boldsymbol x^\prime \boldsymbol \Sigma_1 \boldsymbol x$? 
Or $\boldsymbol x^\prime \boldsymbol \Sigma_2^{-1} \boldsymbol x$ and $\boldsymbol x^\prime \boldsymbol \Sigma_1^{-1} \boldsymbol x$? Or at least, $\text{E} (\boldsymbol x^\prime \boldsymbol \Sigma_2^{-1} \boldsymbol x)$ and $\text{E}(\boldsymbol x^\prime \boldsymbol \Sigma_1^{-1} \boldsymbol x)$? 
Thank you.
 A: $|\boldsymbol \Sigma_2| < |\boldsymbol \Sigma_1|$, does not imply anything about the ordering of the quadratic forms  $\boldsymbol x^\prime \boldsymbol \Sigma_2 \boldsymbol x$ and $\boldsymbol x^\prime \boldsymbol \Sigma_1 \boldsymbol x$, in general.  (This ordering is sometimes called the Loewner ordering, written as $\boldsymbol \Sigma_2 \prec \boldsymbol \Sigma_1$ and it is implied if and only if $\boldsymbol \Sigma_2 - \boldsymbol \Sigma_1$ is positive-definite).
To see this, use the fact the determinant is the product of the eigenvalues, and consider diagonal matrices 
$$\Sigma_1 = \begin{bmatrix} 11 & 0 \\ 0 &  1 \end{bmatrix}$$ and
$$\Sigma_2 = \begin{bmatrix} 5 & 0 \\ 0 &  2 \end{bmatrix}$$
then for $x=\left[0, 1 \right]$ provides you with your counter-example. This example also demonstrates that $\boldsymbol \Sigma_1^{-1}$ and $\boldsymbol \Sigma_2^{-1}$ are not ordered by taking $x=\left[1, 0\right]$.
As far as expectations of the quadratic forms, you can use the higher-dimensional analog to the law of total variance to get that.  From wiki,
$$E(x^T \Lambda x) = \mbox{tr} (\Lambda \Sigma_1) + \mu^T \Lambda \mu.$$
So you would have that 
$E(\boldsymbol x^\prime \boldsymbol \Sigma_1^{-1} \boldsymbol x) = p$, and $E(\boldsymbol x^\prime \boldsymbol \Sigma_2^{-1} \boldsymbol x) = \mbox{tr}(\Sigma_1^{-1}\Sigma_2)$.  Turning again to our counter example, this would be equal to $2 \frac{5}{11}>2$, so we see again that the answer is no.
