If $X \sim N(0,1)$, does $(X, X)^\prime$ have a bivariate normal distribution? If $X \sim N(0,1)$, does $Y=(X,X)^\prime$ have a bivariate normal distribution? If so (actually also if not), what is the joint density?
 A: Two copies of the same normal variable stacked up in a vector yield a degenerate bivariate normal. 
See wikipedia on the degenerate case of the multivariate normal
While it is a special case of the multivariate normal it doesn't have a bivariate density.
The variance-covariance matrix of $Y=(X,X)^\prime$ ($\text{Var}(Y)=\Sigma$) for a standard normal $X$ will be all ones.
As such you can't invert $\Sigma$ and must instead use a generalized-inverse in the exponent; you'll also need to redefine the determinant -- the ordinary determinant will be 0 -- to a pseudo-determinant, which in this case will give $1$.
The density will be zero everywhere but the line $y_1=y_2$ (but if you condition on being on that line it looks like a standard normal density)
If you consider the bivariate distribution in the orthogonal direction to that line you're looking at the density of the variable $Y_1-Y_2=X-X=0$  -- you have a degenerate "normal" with mean 0 and variance 0; it's in this direction we see that the density disappears (it's just a spike at 0).
