Joint CDF of $X, Y$ dependent on $X$ Let X have a standard normal distribution and let $Y = 2X$. Determine whether (i) the cdf $F(x, y)$ of $(X,Y)$ is continuous, and (ii) the distribution of $(X, Y)$ is absolutely continuous with respect to the Lebesgue measure in the $(x, y)$ plane.
I'm not sure how to approach (i), I understand that the joint cdf of continuous random variables is 
$$
\iint f(x,y) \,dx\,dy
$$
but I'm not sure how to obtain $f(x, y)$ from the definitions of $X$ and $Y$. 
 A: When you deal with continuous r.v. it is usually much more easy and clear to start working with their cdf, and that's actually what they ask for (not the density $f_{XY}$; in fact, you can't actually assume it exists).
Remember that the CDF is the real-valued function $F_{XY}$ that takes two real variables defined as
$$F_{XY}(x,y)=P(X\le x \wedge Y\le y)$$
that in this case can be reduced to
$$F_{XY}(x,y)=P(X\le x \wedge 2X\le y).$$
Then try to write an expression that relates this function to the (one variable) CDF of a normal standard r.v. (while there is not a nice formula for it you can just call it $\Phi(t)$ and all you need to know is that its derivative is the density function of $N(0,1)$ distribution.
Once you have a formula for $F_{XY}$ (which is not straightforward), you can see if it is a continuous function (this is equivalent to say thay $(X,Y)$ is a continuous r.v.) and if there is a function $f(x,y)\ge 0$ such that
$$F_{XY}(x,y)=\iint \limits_{\{(s,t)\colon s\le x,\,t\le y\}} f(s,t)\; d\mu.$$
When this is the case, such an $f$ is called a 'probability density function' and we say that $(X,Y)$ is an absolutely continuous r.v.
Not an easy one, but is possible if you use patience and detail.
A: I just posted this answer showing the c.d.f. is continuous.
This is not absolutely continuous with respect to Lebesgue measure on the plane, because the Lebesgue measure of the line $y=2x$ is $0,$ whereas this measure assigns the number $1$ to that set.
