# estimators with singular covariance matrix

Suppose I have 2 vectors of random variables $\boldsymbol\theta_1 \in \mathbb{R^n}$ and $\boldsymbol\theta_2 \in \mathbb{R^m}$ with asymptotic covariance $\Sigma_1$ and $\Sigma_2$ respectively. I want to compare two quadratic forms $x^{\top}\Sigma_1x$ and $y^{\top}\Sigma_2y$ such that I know $\left(y^{\top}\Sigma_2y-x^{\top}\Sigma_1x\right) > 0$ must hold true. $\Sigma_1 \succ 0$ but $\Sigma_2$ is singular. It is clear that an estimator using $\boldsymbol \theta_2$ will not have a valid distribution and $\Sigma_2^{-1}$ does not exist. What other deductions can I make?

What consequences does the singularity of $\Sigma_2$ have on its quadratic form? Any other pointers would be great

• Could you please edit once more: Are $\theta_1$ and $\theta_2$ parameters and $x$ and $y$ the random varibales or vice versa? Also, do you just want to know if the difference of the quadratic forms in the parameters is positive (making your question effictively a testing problem) or are you really interested in the values of the quadratic forms themselves? – Horst Grünbusch Jun 12 '15 at 12:57
• $\boldsymbol\theta_1 \in \mathbb{R^n}$ and $\boldsymbol\theta_2 \in \mathbb{R^m}$ are random vectors. x and y are real vectors. I am interested in properties of $\Sigma_2$ that can be exploited to make any meaningful conclusions. – Rohit Arora Jun 14 '15 at 13:57
• Are $x$ and $y$ some parameters of the distribution of $\theta_1$ and $\theta_2$? Or are they known? – Horst Grünbusch Jun 19 '15 at 16:14

Since $Σ^{−1}_2$ does not exist, meaning that the lines of $Σ$ are redundant, the global minimum for the quadratic form does not exits either, instead there should be a "line" which is, depending of the $Σ$ values, horizontal to plane $X_1$ , $X_2$ or continuous falling down to one of the corners of that plane.
$Σ=\matrix{ 6 , -3\\ -6 , 3 }$ 