In my introductory statistics course, I've just started on jointly continuously distributed random variables (call them JCD's) and am trying to understand when/when not two r.v.'s that are themselves individually continuously dist. (call them ICD's) are jointly cont. dist. There's a particular example I'm trying to wrap my head around, but can't seem to.
My textbook says that all r.v.'s that together are JCD's are individually ICD's. But that it's not always true the other way around: that any two or more ICD's may not necessarily be JCD's. The example it sites is:
'Let X be a continuously distributed r.v. and let Y = 2X. Then X and Y are both ICD's, but X and Y together are not JCD's:
For A = {(x,y): 2x=y }, $\int\int_{A}f(x,y) dy$ = $\int_{-\infty}^{+\infty}(\int_{2x}^{2x} f(x,y)dy) dx = 0$
for every piecewise continuous function f on R^2. This implies that P (2X = Y) = 0 for every two JCD's X and Y. However, for our choice of X and Y, obviously P (2X = Y) = 1, thus it must be that X and Y cannot be jointly continuously distributed.'
Am I correct in understanding what the author is saying: in the X-Y plane, instead of the usual area, the two r.v.'s here trace out a 2D line of some sort (since the bounds on the inner integral are equal, rather than being some range like 0 < Y < 10X), and so doubley integrating f(x,y) over that line of course produces a result of 0, since there's 'no area' to integrate f(x,y) over. And thus since the integral over the support of any random variable(s) is supposed to equal 1, these two r.v.'s are not JCD's. Have I got that right? As an aside, are the limits to the outer integral $-\infty$ to $+\infty$ because this example uses a generic r.v. X whose particular support is undefined at the moment, and so the bounds are the whole of R to show that whatever support any actual X has is covered in the example? Or is it because this X does have a well defined support, but that it is indeed across the whole of R (if that is the case, I'm not quite sure how that is so, from the example)?
Finally, in the last part of the example it says how 'P (2X = Y) = 0' when it should = 1. But I'm confused, I thought the probability of a continuous random variable exactly equalling something is always 0, and so the probabilities of ICD's and JCD's must always be calculated over a range, e.g. P( X < 7) or P( 2X > Y)? For example, imagine an area in the U-V plane for two JCD's over which there is some PDF f(x,y); if I wanted to work out P( 2U = V), wouldn't I be integrating f(x,y) only over the straight line 2U=V that cuts through our area? Thus the answer would always be 0, whether the two variables U and V (or X and Y) are JCD or not?
Apologies for the length of this one, but I got stuck on several fronts on this example and had to break it all down. Many thanks indeed for any help you can offer, really appreciate it!
Thanks
You are overestimating the importance of the concept of the joint pdf. In the case you described there was not joint pdf. However, the joint cdf could have been calculated.
Let's the following example. Say, $X$ is uniformly distributed over $[0,1]$. And let $Y=2X$. Now, what is the joint distribution.
By definition
$$F_{X,Y}(x,y)=P\left(X<x,X<\frac y2 \right)=\begin{cases}F_X(x)&\text{ if }&x<\frac y2\\F_X\left(\frac y2\right)&\text{otherwise.}\end{cases}$$
where $F_X(x)=x$ if $0\leq x\leq 1$, $0$ otherwise. If you try to take the partial derivatives (one after the other) you will get zero. So, no pdf descibes this distribution. The joint cdf IS the desccription.