How to see a function is Gaussian

Question

I have this function $$f(x, y) = \frac {1}{2\pi}\exp(−0.5(x^2-2xy+9y^2))$$

I proceed like this:

First I compute $\Sigma^{-1}$ which is \begin{bmatrix} 1 & -1 \\ -1& 9\end{bmatrix} Then I see the determinant is positive (which is good) but I don't know how to proceed.

My principal question is this: The value of the det $\Sigma^{-1}$ has to be between precise values or it just has to be $>0$ in order to be a Gaussian density??

Answer 1

My principal question is this: The value of the $\det \Sigma^{-1}$ has to be between precise values or it just has to be >0 in order to be a Gaussian density??

A non-zero determinant is a necessary but not sufficient condition. The main thing you need to check for is that $\Sigma^{-1}$ is positive definite so that it is invertible. Also to be complete, we need a scaling factor of $\frac{1}{\sqrt{(2 \pi)^k \det \Sigma}}$, multiplying the exponential to have a Gaussian PDF.