Joint density distribution and Variance I was wondering if there is a way to calculate the joint distribution of two fully correlated variables, both with known distributions, expected value and variance, without knowing the conditional distribution?
If this is not possible, is there a way of finding Var$(X,Y)$ = E$[(XY)^2]$ - E$[XY]^2$ when knowing that Cor$(X,Y) = 1$? I can't seem to find an expression for E$[(XY)^2]$...
Thanks!
 A: We have $Y=kX+l$ where $k$ and $l$ are constants, with $k$ positive.  The constants are known, since they can be found from the means and variances of $X$ and $Y$.
Any expectations, such as $E(XY)$, can then be computed using the distribution of $X$ and the fact that $Y=kx+l$.
The joint cdf can be computed using $\Pr(X\le x\cap Y\le y)=\min(\Pr(X\le x), \Pr(Y\le y))$.
Even if the densities of $X$ and $Y$ exist, the joint density does not exist in any useful sense.
A: With slightly more general details than in the
original version of André Nicolas's answer, it must be that $Y = aX+b$ where
$$a = \sqrt{\frac{\operatorname{var}(Y)}{\operatorname{var}(X)}}
\quad \text{and}\quad b = E[Y] - aE[X].\tag{1}$$
There is no joint density of $X$ and $Y$ in the sense that $X$ and $Y$ are not jointly
continuous random variables. Thus, to find $E[g(X,Y)]$, simply substitute $aX+b$ for $Y$ and find the expectation of this function of $X$ alone via the law of the unconscious statistician. In other words, $$E[g(X,Y) = E[g(X,aX+b)] = E[h(X)]$$
where $h(x) = g(x,ax+b)$ with $a$ and $b$ are as given in $(1)$.
Note: you say that you know the marginal distributions of $X$ and $Y$ (and
their means and variances). Be aware that the assumption of perfect correlation
with correlation coefficient $+1$ means that it must be that
$$F_Y(z) = F_X\left(\frac{z-b}{a}\right), \quad f_Y(z) = \frac{1}{a}f_X\left(\frac{z-b}{a}\right).\tag{2}$$
If the distributions that are known to you (or given to you) do not satisfy $(2)$, then the problem you are trying to solve has contradictory assumptions, and has no
meaningful answer.
