Given four independent, identically distributed Gaussian random variables with zero mean and unit variance $x\_1$, $x\_2$, $y\_1$, $y\_2$, consider 


\begin{equation}
u \equiv \max(x\_1+C\, y\_1, x\_2+C \, y\_2) - \max(x\_1-C \, y\_1, x\_2-C \, y\_2),
\end{equation}


where $C$ is a real number.


Do you know how to compute the PDF of $u$, or a least its variance?



Difference between maxima of random variables
Using $\max(a,b) = \dfrac{a+b}{2} + \left| \dfrac{a-b}{2}\right|$, write $u = w\_1 + |w\_2| - |w\_3|$ where 
$$ \eqalign{ w\_1 &= C (y\_1 + y\_2) \cr
 w\_2 &= \dfrac{1}{2} (x\_2 - x\_1 + C (y\_2 - y\_1))\cr
 w\_3 &= \dfrac{1}{2} (x\_2 - x\_1 - C (y\_2 - y\_1))\cr}$$
are jointly normal with mean $0$ and covariance matrix
$$ V = \pmatrix{2 C^2 & 0 & 0 \cr 0 & (1+C^2)/2 & (1-C^2)/2\cr
 0 & (1-C^2)/2 & (1+C^2)/2\cr}$$
In particular, $w\_1$ is independent of $w\_2$ and $w\_3$.
Thus $\text{Cov}(w\_1, |w\_2|) = 0$ and $\text{Cov}(w\_1, |w\_3|) = 0$.
The only nontrivial computation is $ \text{Cov}(|w\_2|, |w\_3|)$. If $|C| < 1$ get
$$ \eqalign{ {\mathbb E} [|w\_2| |w\_3|] &= \dfrac{1-C^2}{\pi} \arctan\left(\frac{2|C|}{C^2-1}\right) + \dfrac{1-C^2}{2} + 2 \frac{|C|}{\pi}\cr
{\mathbb E}|w\_2| &= {\mathbb E} |w\_3| = \sqrt{\dfrac{C^2+1}{\pi}}\cr
\text{Cov}(|w\_2|,|w\_3|) &= \dfrac{1-C^2}{\pi} \arctan\left(\frac{2|C|}{C^2-1}\right) + \dfrac{1-C^2}{2} - \dfrac{(1-|C|)^2}{\pi}
} $$
and then 
 $$ \text{Var}(w\_1 + |w\_2| - |w\_3|) = 2\dfrac{C^2-1}{\pi} \arctan\left(\frac{2|C|}{C^2-1}\right) + 4 C^2 - \frac{4|C|}{\pi} $$


EDIT: That formula was only for $|C|<1$ because of branch problems with the arctan. Here's one that should work for all $C \ne 0$:


$$ \text{Var}(w\_1 + |w\_2| - |w\_3|) = 2 \dfrac{C^2-1}{\pi} \arctan \left(\frac{1-C^2}{2|C|}\right) + 3 C^2 - \frac{4|C|}{\pi} + 1 $$


