Independence of variables Q:Two variables X and Y have same mean and variance.If U=X+Y and V=X-Y then, are U and V independent and correlated?
I found that U and V are uncorrelated. But don't know how to check for independence. 
Kindly help. 
 A: Assuming that $X$ and $Y$ are jointly continuous random variables, the
 joint density of $(U,V) = (X+Y,X-Y)$ is related to the joint density of $X$ and $Y$ as
$$f_{U,V}(\alpha,\beta) 
= \frac 12 \cdot f_{X,Y}\left(\frac{\alpha+\beta}{2},\frac{\alpha-\beta}{2}\right)$$
which corresponds to a rotation by $\pi/4$ and a dilation of axes.
I always
like to think of this geometrically and visually. The joint density
is a solid (of volume $1$) sitting on the plane, and what happens is that the solid gets rotated by $\pi/4$. It also gets flattened down 
by a factor of $\frac 12$ and spread out by a 
factor of $\sqrt{2}$ along both axes (so as to keep the volume equal to $1$). 
Example: If $(X,Y)$ is uniformly distributed on the interior
of the square with vertices $(0,1)$, $(1,0)$, $(0,-1)$, and $(-1,0)$, 
then its
joint density is a rectangular parallelopiped with that square base
(of area $2$) and height $\frac 12$.  The joint density of $(U,V)$
can be found by rotating this parallelopiped by $\pi/4$ (which makes
the sides of the square base parallel to the axes) and then squashing it
down by a factor of $2$ to height $\frac 14$ and spreading out 
the base by a factor of $\sqrt{2}$ in each direction which converts
it into the square with vertices $(\pm 1, \pm 1)$, which is
of area $4$ and so the volume stays $1$ as it should be.
Now, to check
whether $U$ and $V$ are independent, you need to check whether
$$f_{U,V}(\alpha,\beta) = f_U(\alpha)f_V(\beta)$$ for all
real numbers $\alpha$ and $\beta$. As you have found,
equal variances suffice to prove that $U$ and $V$ are uncorrelated,
but for independence, you need to know the joint density.
A canonical example of independent $U$ and $V$ is when
$X$ and $Y$ are independent $N(0,\sigma^2)$ random variables,
and it is readily shown that $U$ and $V$ are
independent $N(0.2\sigma^2)$ random variables. In this case,
joint density solid has circular symmetry which means that
rotation by any angle, not just by $\pi/4$, leaves the solid
unchanged. And, of course, the squashing and scaling does not
affect normality, only the variance.
However, it is possible that $U=X+Y$ and $V=X-Y$ are independent 
even when $X$ and $Y$ are not.
Example (continued): It is easily shown
that $X$ and $Y$ are dependent random variables, e.g. if we
know that $X=\frac 12$, then $Y$ is necessarily restricted
to $(-\frac 12, \frac 12)$ whereas independence would require
that $Y$ continue to enjoy the range $(-1,1)$ regardless of the
value of $X$. But, $U$ and $V$ are independent $U(-1,1)$ random
variables.
Conversely, it is possible that $X$ and $Y$ are independent
but $(X+Y,X-Y)$ are not. Just reverse the idea of the example,
starting out with independent $(X,Y)$ uniformly distributed 
on interior of the square with
vertices $(\pm 1, \pm 1)$ to arrive at dependent $(X+Y,X-Y)$
uniformly distributed on the interior of the square with vertices
$(0,2)$, $(2,0)$, $(0,-2)$, and $(-2,0)$.
