using independence definition prove $x+y$ and $x-y$ independent Given $x$ and $y$ independent and $N(0,1)$, show that $x+y$ and $x-y$ are independent.
I started like this $x+y=u$ and $x-y=v$ :
I want to prove this :
$$P[u \leq t,v \leq s]=P[u \leq t][v \leq s]$$
However, I have some trouble starting this calulation since i don't know what is $f_{u,v}$ the density of $u$ and $v$. Anyone can help? thank you
 A: Hint: 
If $u,v$ have a joint normal distribution then they are independent if their covariance equals $0$.
Apply this on $u=x+y$ and $v=x-y$.

edit (here I use capitals for random variables)
In general if $\vec{Z}\sim N\left(\vec{\mu},\Sigma\right)$ then $$A\vec{Z}\sim N\left(A\vec{\mu},A\Sigma A^{T}\right)$$
for matrix $A$.
Note that $$\left(\begin{array}{c}
X\\
Y\end{array}\right)\sim N\left(\left(\begin{array}{c}
0\\
0\end{array}\right),\left(\begin{array}{cc}
1 & 0\\
0 & 1\end{array}\right)\right)$$ and $$\left(\begin{array}{c}
U\\
V\end{array}\right)=\left(\begin{array}{cc}
1 & 1\\
1 & -1\end{array}\right)\left(\begin{array}{c}
X\\
Y\end{array}\right)$$ leading to $$\left(\begin{array}{c}
U\\
V\end{array}\right)\sim N\left(\left(\begin{array}{c}
0\\
0\end{array}\right),\left(\begin{array}{cc}
2 & 0\\
0 & 2\end{array}\right)\right)$$
Actually this shows that $\left(\begin{array}{c}
U\\
V\end{array}\right)$ has the same distribution as $\sqrt{2}\left(\begin{array}{c}
X\\
Y\end{array}\right)$ so an immediate conclusion about independence is at hand.
If you insist on proving the line mentioned in your question then based on this you can also find $f_{U,V}$.
