Bivariate normality and statistical work when the joint density is not defined Let $X$ follow a Normal distribution, and let $Y = a+bX$ (not both constants zero). The normal distribution is closed under shifting and scaling, so $Y$ also follows a normal distribution.  
If I recall correclty, there exists "Cramer's condition" for bivariate normality, where we get it if all linear combinations of two random variables have a univariate normal distribution. In our case it is easy to see that any $Z = \delta_1X + \delta_2 Y$ will follow a Normal distribution too (not both constants zero). So it appears that $(X,Y)$ are jointly normal, with correlation coefficient equal to unity, in absolute value.
The question is the following: the joint support of $(X,Y)$ is a line in $\mathbb R^2$, and as a $(n-1)-$dimensional object in a $n-$dimensional space, it has Lebesgue measure zero. This I think, implies that the joint probability density function does not exist.  
Q1: Is the above correct? 
Q2: What are the main practical consequences of not being allowed to define the joint density (like, indicatively, what kind of statistical work we cannot do, and what kind we still can do, what statistical methods are no longer applicable, and which are still applicable)?
I guess the situation generalizes to any distribution that is closed under affine transformations, and when we want to assume in addition that the bivariate vector follows the 2-D analogue of the marginal distributions.
 A: Let $P_{XY}(A) := P\left((X , Y) \in A\right)$. There exists a set $A \in \mathfrak B^2$ (the Borel $\sigma$-algebra over $\mathbb R^2$) such that $\lambda^2(A) = 0$ yet $P_{XY}(A) > 0$, so $P_{XY}$ is not dominated by $\lambda^2$ and therefore it has no Lebesgue pdf (with respect to $\lambda^2$, the Lebesgue measure on $\mathbb R^2$), as you said. That's not up for debate. But, as @Dilip Sarwate says in the comments, we can consider $(X , Y)$ to have a degenerate bivariate Gaussian distribution. I think the way to reconcile this is that having a pdf proportional to $\exp\left(-\frac 12 (z - \mu)^T \Sigma^{-1} (z - \mu)\right)$ with $z \in \mathbb R^n$ is sufficient, but not necessary, for being a Gaussian over $\mathbb R^n$. That's one key advantage to the definition of a multivariate Gaussian being $Z \sim \mathcal N_n(\mu, \Sigma)$ iff $a^T Z \sim \mathcal N_1(a^T \mu, a^T \Sigma a)$ for all $a \in \mathbb R^n$ rather than specifying a pdf for $Z$. 
Now let $h(x) = a + bx$ so $h$ is measurable. We have
$$
P(X \in A_1, Y \in A_2) = P(X \in A_1, X \in h^{-1}(A_2)) = P(X \in A_1 \cap h^{-1}(A_2)) = P_X(A_1 \cap h^{-1}(A_2)).
$$
We do have $P_X \ll \lambda$ (assuming $X$ is not also degenerate) so the 1-dimensional Lebesgue pdf $\frac{\text dP_X}{\text d\lambda}$ is defined. Thus we have
$$
P_{XY}\left(A_1 \times A_2\right) = \int_{A_1 \cap h^{-1}(A_2)} \frac{\text dP_X}{\text d\lambda} \,\text d\lambda
$$
so even though the 1-dimensional Gaussian pdf $\frac{\text dP_X}{\text d\lambda}$ isn't the pdf of $P_{XY}$, it is still all we need to perfectly describe $P_{XY}$.
To conclude, if $(X, Y)$ is degenerate it won't have a pdf with respect to $\lambda^2$ but it is still considered a bivariate Gaussian, and we can compute $P((X, Y) \in A_1 \times A_2)$ by integrating the pdf of $X$ with respect to to Lebesgue measure on $\mathbb R$. This agrees with @whuber's comment because there's no mystery about the behavior of $(X, Y)$, it just doesn't have a 2-D pdf.
