I am reading some notes and having some trouble with the definition of multivariate normal distribution when the covariance matrix is not invertible. I will state my understanding below, and hopefully someone can chime in with some words of wisdom.
Suppose $\Sigma$ is an $n\times n$ matrix. When $\Sigma$ is invertible, we say that a random vector $X$ has multivariate normal distribution with mean $0$ and covariance $\Sigma$ if it has density given by
Now, when $\Sigma$ is not invertible, then clearly the above density function is not defined. The notes mention the Cramer-Wold device can be used to define $N(0,\Sigma)$ in this case, and moves on without explicitly doing so.
Could somebody please give a simplistic explanation/definition of $N(0,\Sigma)$ when $\Sigma$ is not invertible?
Remark: For the univariate normal distribution, I understand that $N(0,0)$ corresponds to the degenerate distribution $\delta_0$. By degenerate distribution, I mean it is $0$ with probability $1$. I can not see how this would work in higher dimensions though!