Random variables independence I need to check if $Z$ and $W$ are dependent or not.
$X,Y \sim \mathrm{Exp}(2)$
Then I define: $Z=X-Y \ \text{,}\ \   W=X+Y$.
Now How I can check that $Z$ and $W$ are dependent or not ?
I know from the theory that I should show that $f(z,w) = f(z)f(w)$, but I don't see how I do it here.
 A: It is easy to determine that $W$ and $Z$ are uncorrelated since their covariance equals $\operatorname{var}(X)-\operatorname{var}(Y)$. For the question of independence, read this answer of mine on math.SE which describes an eyeball test that tells you that $X+Y$
and $X-Y$ are dependent variables in this case.  The key idea (for those not
interested in visiting another stackexchange) is that a necessary
(but not sufficient) condition
for independence is that the support of the joint distribution is a 
rectangular region
whose boundaries are parallel to the axes. Now, the support of the joint distribution of $X$
and $Y$ is the first quadrant (which satisfies the eyeball test) or a subset thereof, while the joint 
distribution of $X+Y$ and $X-Y$
is just $f_{X,Y}$ rotated by $\pi/4$ (and dilated too, but that is irrelevant here)
and so fails the eyeball test. Ergo, dependent (but uncorrelated!) 
random variables.
Note: independence of $X$ and $Y$ is not needed in reaching these conclusions.
A: As I think this is self-study, I’ll give only a few hints (this may be completed later...).
It can be useful to remark that $X$ and $Y$ only take positive values. Thus, if $W = X+Y$ is small, both $X$ and $Y$ have to small, and $Z$ has to be small as well. Finally, if you observe a very small $Z$, you can bet that $W$ is small.
I leave it to you to find a rigorous way to use this idea.

As Dilip gave a complete answer, I complete mine: as $X > 0$ and $Y > 0$ with probability $1$, we have $\mathbb P(X < c| W < c) = 1$ and $\mathbb P(Y < c| W < c) = 1$. With $0 < X < $c and $0 < Y < c$, the absolute value of $Z = X - Y$ is $< c$ as well, so 
$$\mathbb P(-c < Z < c | W < c) = 1 \ne \mathbb P(-c < Z < c),$$
which proves that $Z$ and $W$ are not independent.
This is of course exactly the same idea as in Dilip answer.
