Check whether $f(x,y)$ is the density of a gaussian vector I would like to know whether the following function:
$f(x, y) = \frac {1}{2\pi}\exp(−0.5(x^2-2xy+9y^2))$
can be the density of a Gaussian vector.
My attempt is that because of the term $\frac {1}{2\pi}$ and because the vector of means is $\mu=0$ (since there are no terms of degree zero), then the correlation coefficient $p$ must be $0$. But this would imply that the term $xy$ does not appear in the equation above, so that $f(x,y)$ can not be the distribution of a gaussian vector..
 A: If it is really a Gaussian distribution, there should exist a positive definite matrix ${\bf A} \in \mathbb{R}^{2 \times 2}$ and a vector ${\bf m} \in \mathbb{R}^2$ such that:
$$f({\bf u}) = \frac{1}{2\pi \sqrt{|\det{{\bf A}}|^{-1}}}e^{-\frac{1}{2}{(\bf u - \bf m)}^{\top} {\bf A} {(\bf u - \bf m)}},$$
where ${\bf u} = \begin{bmatrix}x\\y\end{bmatrix}.$
Let $${\bf A} = \begin{bmatrix}a & b\\b & d\end{bmatrix} ~\text{and}~ {\bf m} = \begin{bmatrix}m_x \\ m_y\end{bmatrix}.$$
Then:
$$({\bf u} - {\bf m})^{\top} {\bf A} ({\bf u} - {\bf m}) = ax^2 + dy^2 -2 x(a m_x+bm_y) - 2y(bm_x +d m_y) + 2bxy + (am_x^2 + 2bm_x m_y + dm_y^2) = x^2 - 2xy + 9y^2.$$
As a straight consequence, $a = 1$ ($ax^2 = x^2$), $b = -1$ ($2bxy = -2xy$) and $d = 9$ ($dy^2 = 9y^2$).
Notice that the eigenvalues of 
$${\bf A} = \begin{bmatrix}a & b\\b & d\end{bmatrix}$$
are both positive since both $\text{tr}({\bf A}) = a + d = 10$ and $\det{\bf A} = ad-b^2 = 8$ are positive. Hence, ${\bf A}$ is positive definite.
Now, we need to check that $|\det{{\bf A}}|^{-1} = 1.$
Recall that $\det({\bf A}) = ad - b^2$, and hence:
$$|\det{{\bf A}}|^{-1}  = \frac{1}{|ad - b^2|} = \frac{1}{|9 - 1|}  = \frac{1}{8} \neq 1.$$
In conclusion, the function you wrote is not a Gaussian distribution.
