My primary goal is to determine an upper bound of a bivariate Gaussian cdf by removing the dependency on the cross-correlation. I could do this by using a circular Gaussian with standard deviation from the eigenvalue of the covariance matrix but i'd prefer a tighter bound. Also the correlation is often small.
I tried finding a paper on this but with no luck (i don't have access to scientific journals not online)
One idea i had was to find a simple way to define a larger ellipse that is oriented along the axes that contains a rotated ellipse (i.e. find the larger contour that contains the contour of the correlated pdf)...any ideas how to do this?
Thanks
Let's say your "rotated" ellipse has equation $a x^2 + 2 b x y + c y^2 - 1 = 0$ where $b \ne 0$, and your larger ellipse is to be of the form $A x^2 + B y^2 - 1 = 0$. The resultant of those two polynomials with respect to $y$ is $$ \left( {A}^{2}{c}^{2}-2\,ABac+4\,AB{b}^{2}+{B}^{2}{a}^{2} \right) {x} ^{4}+2\, \left( ABc-A{c}^{2}-{B}^{2}a+Bac-2\,B{b}^{2} \right) {x}^{2}+ {B}^{2}-2\,Bc+{c}^{2} $$ You want this to have just two real zeros, corresponding to the two symmetrically placed intersections of the two ellipses. Thus, replacing $x^2$ by $t$, you want the discriminant of the resulting quadratic to be $0$. That discriminant turns out to be $$ -16\, \left( AB-Ac-Ba+ac-{b}^{2} \right) {B}^{2}{b}^{2} $$ so you want $$AB - A c - B a + ac - b^2 = 0$$ That is a hyperbola in $AB$ space. Which point on this curve do you want? Perhaps, for example, you want to minimize the area of the larger ellipse, which is proportional to $AB$. That can be done using a Lagrange multiplier. I get $$ \eqalign{A &= \sqrt{\frac{a}{c}} \left(\sqrt{ac} - b\right)\cr B &= \sqrt{\frac{c}{a}} \left(\sqrt{ac} - b\right)\cr}$$