My question is, can the following function:
$f(x, y) = \frac {1}{2π}exp(−0.5(x^2+xy))$
be the density of a Gaussian vector?
I do not think so, since $y^2$ does not even appear, and it looks like there is no correlation $\rho$ compatible with this distribution. Am I wrong or not?
Edit, here is my attempt: If f(x,y) was the density of a gaussian vector, then: $\iint_{\mathbb{R}^2} f(x,y) dxdy = 1$ (But would this be sufficient? I do not think so.), then
$$\iint_{\mathbb{R}^2} f(x,y) dxdy = \int_{-\infty}^\infty\int_{-\infty}^\infty \frac {1}{2π}exp(−0.5(x^2+xy))\ dxdy = \frac{1}{2π} \int_{-\infty}^\infty e^{-0.5x^2}dx\int_{-\infty}^\infty e^{-0.5xy}\ dy $$
We can check that $$\int_{-\infty}^\infty e^{-0.5xy}\ dy$$ is $+\infty$ if $x<0$, while $-\infty$ if $x>0$. So that the $\iint_{\mathbb{R}^2} f(x,y) dxdy $ diverges. Am I right??
Well forget the scaling factor (the thing multiplying your exponential) for a moment. A multivariate Gaussian distribution constitutes of the term $$\exp \Big( -\frac{1}{2} (z-\mu)^T \Sigma^{-1} (z-\mu) \Big)$$ where $\mu = 0$ (zero-mean) in your case and $z =[x,y]^T$. By inspection, we can note that $$\Sigma^{-1} = \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 0\end{bmatrix} $$ Note that $\Sigma$ is the covariance matrix. We can see that $\Sigma^{-1}$ has a negative determinant, hence not a PSD (positive semi-definite). So, by contradiction, this is not a Gaussian distribution since a covariance matrix has to be PSD.