Does $(X,X)'$ follow a bivariate normal distribution? I'm fairly new to multivariate distributions. I'm trying to figure out if $(X,X)^{'}$ follow a bivariate normal distribution (the prime = transposed). If $X\sim N(\mu, \sigma^{2})$ where $\mu \in \mathbb{R}$ and $\sigma^{2}\geq 0$. What I know is that if $\sigma^{2}=0$ then $X\sim N(\mu,0)$ and then we have a distribution that has no density, hence degenerate normal distribution. But how do I conclude if it has a bivariate normal distribution in this case? And assuming that $\sigma^{2}>0$, then we have a normal distribution. I think my main question is how do I show that $(X,X)^{'}$ either has/do not has a bivariate normal distribution. Can somebody help or point me in the right direction of what I have to do?
 A: Let $A = {1\choose 1}$ and note that $A$ is a linear transformation from $\mathbb R$ to $\mathbb R^2$. Then if $X\sim\mathcal N(\mu,\sigma^2)$ we'll have
$$
{X\choose X} = AX \sim \mathcal N(A\mu, \sigma^2 AA^T)
$$
since linear transformations of Gaussians are also Gaussian. The covariance matrix is rank $1$ when $\sigma>0$ and rank $0$ when $\sigma = 0$ so either way this distribution will not have a Lebesgue pdf. It'll be entirely supported on $\{x \in \mathbb R^2 : x_1 = x_2\}$ when $\sigma>0$ and if $\sigma = 0$ we'll have $AX$ as a point mass at $A\mu = {\mu\choose \mu}$. But it is a valid bivariate Gaussian.
If you want to be more sure, we can use  characteristic functions (CFs). Let $\varphi_Y(t)$ be the CF of a random variable $Y$ evaluated at $t$. Then
$$
\varphi_{AX}(t) = \text E(\exp(it^TAX)) = \varphi_{X}(t^TA) \\
= \exp\left(i\mu t^TA - \frac 12\sigma^2 (t^TA)^2 \right) \\
= \exp\left(it^T(A\mu) - \frac 12\sigma^2 t^TAA^Tt \right)
$$
which is the CF of a $\mathcal N(A\mu, \sigma^2 AA^T)$ random variable.
This is also a reason for defining a multivariate Gaussian as a random variable $X\in\mathbb R^n$ where $a^TX\sim\mathcal N(a^T\mu,a^T\Sigma a)$ for all $a\in\mathbb R^n$ rather than by specifying the pdf, since this case includes Gaussians with singular covariance matrices for which there is no Lebesgue pdf.
