Jacobian and uniform distribution If I perform a change of variables for independent and uniform random variables $x, y, z$, which is something like the following: $$u=x-y+z/2 \\ v=z/2 - x \\ w = z/2$$ then how do I know whether $u,v,w$ are still uniformly distributed? In the particular example I gave above, $w$ is clearly still uniformly distributed, but I've heard there's a way go about this using the Jacobian. I have been unable to find information about that using google. Anyone who can help? Thank you!
 A: You need to characterise uniformly distributed. Let $m$ be the Lebesgue measure on $\mathbb{R}^n$.
The key fact here is that if $L: \mathbb{R}^n \to \mathbb{R}^n$ is a linear map, and $A$ a measurable set, then $m(LA) = |\det L| m (A)$. (In this case, the Jacobian of the transformation is $L$ itself, since it is linear.)
Suppose $p$ is a probability measure on the Borel subsets of $\mathbb{R}^n$.
Then I will call $p$ uniformly distributed on a set $\Omega$ of strictly positive measure iff $p A = {1 \over m \Omega} m(A \cap \Omega)$, where $A$ is a Borel subset of $\mathbb{R}^n$.
In the example above, $x,y,z$ are independent and uniformly distributed, so we can take $\Omega$ to have the form $\Omega = I_x \times I_y \times I_z$, where $I_x,I_y,I_z$ are non-degenerate bounded intervals. The measure on $\mathbb{R}^3$  in this case is given by  $p(A) = {1 \over m \Omega} m(A \cap \Omega)$. (We have 
$m \Omega = l(I_x)l(I_y)l(I_z)$.)
In your example above, the change of variables is represented by the linear operator $L$ whose matrix is given by
$\begin{bmatrix} 1 & -1 & {1 \over 2} \\
-1 & 0 & {1 \over 2} \\
0 & 0 & {1 \over 2}
\end{bmatrix}$. The only relevant characteristic of $L$ is that $\det L \neq 0$.
The transformation $L$ induces a probability measure on $\mathbb{R}^3$ defined by
$\mu(A) = p(L^{-1} A)$. We want to check if it is a uniform measure.
We have
\begin{eqnarray}
\mu(A) &=& p(L^{-1} A) \\
&=& {1 \over m \Omega} m((L^{-1} A) \cap \Omega) \\
&=&  {1 \over m \Omega} m((L^{-1} (A \cap (L \Omega))) \\
&=&  {1 \over m \Omega} { 1 \over |\det L| } m(A \cap (L \Omega)) \\
&=&  {1 \over m (L \Omega)} m(A \cap (L \Omega)) \\
\end{eqnarray}
Hence $\mu$ is uniformly distributed on $L \Omega$.
(Note that the variables $u,v,w$ are no longer independent.)
A: Since it's a linear transformation, the Jacobian is constant, so the joint density of $u,v,w$ will be uniform on their support, which is the image of the
support of $x,y,z$ under the transformation.  If, for example, $x,y,z$ are 
uniform on the interval $[0,1]$, then the support of $[u,v,w]$ is the polytope
with vertices $[0,0,0],  [1, -1, 0], [-1, 0, 0], [0, -1, 0], [1/2, 1/2, 1/2], [3/2, -1/2, 1/2], [-1/2, 1/2, 1/2], [1/2, -1/2, 1/2]$.
