I'm trying to learn about random vectors by solving some exercise. I don't know how this is done. I would appreciate it if someone could write a rigorous proof for this.
Let $X$ be a centred Gaussian vector and with $\Sigma_X$ being its symmetric $d\times d$ covariance matrix. And let $r$ be the rank of the covariance matrix, then we can write it as $\sum_{i=1}^r Y_i e_i $ where $Y_1,\ldots,Y_r$ are independent centred Gaussian random variables and $e_1,\ldots,e_d$ are the basis of $\mathbb R^d$ in which the covariance matrix is diagonal.
I want to show that the law of $X$ is absolutely continuous with respect to Lebesgue measure on $\mathbb R^n$ if and only if $r=d$ and that for the density of $X$, we have $$P_X(x) = \frac{1}{(2\pi)^{d/2} \sqrt{\det(\Sigma_X)}} \exp\left(-\frac{1}{2} \langle x, \Sigma_X^{-1} x \rangle\right)$$
For the forward direction, I know that the density is simply the Radon-Nikodym derivative. But, I don't know how to show that it's only the case when $r=d$.
For the Backward direction, I'm really not sure how to go with it.
Best Answer
Assume that $r<d$. Then, the law of $X$ is supported on the proper subspace $\textrm{span}\{e_j\}_{1\leq r\leq d}$. However, any proper subspace of $\mathbb{R}^d$ is a Lesbegue null-set. Thus, the law of $X$ cannot be absolutely continuous with respect to the Lesbegue measure.