Deriving the distribution function of $P(c(X+Y) \mid X>0,Y>0)$ $X$ and $Y$ are independent standard normal random variables.
Let $Z=X+Y$. Find $P(Z\leqslant z \mid X>0,Y>0)$.
My professor gave us some hints as follows:


*

*Let $U=\frac{X+Y}{\sqrt{2}}, V=\frac{X-Y}{\sqrt{2}}$

*$\displaystyle P(Z\leqslant z \mid X>0,Y>0)=P(|U|\leqslant \frac{z}{\sqrt{2}},|V|\leqslant \frac{z}{\sqrt{2}})$


I got a question about the equation above. The area surrounded by $\left(\displaystyle |U|\leqslant \frac{z}{\sqrt{2}},|V|\leqslant \frac{z}{\sqrt{2}}\right)$ is a rectangle with vertices $\displaystyle \left(\frac{z}{\sqrt{2}},\frac{z}{\sqrt{2}}\right), \left(-\frac{z}{\sqrt{2}},\frac{z}{\sqrt{2}}\right), \left(-\frac{z}{\sqrt{2}},-\frac{z}{\sqrt{2}}\right), \left(\frac{z}{\sqrt{2}},-\frac{z}{\sqrt{2}}\right)$
However, I think $P(Z\leqslant z\mid X>0,Y>0)$ is corresponding to the $1/4$ of the diamond (because conditioning on $X>0$, $Y>0$). Why is the equation of the hint valid? 
Thanks a lot!
 A: This question should use the circular symmetry of the joint density; joint
normality is not really needed.
For $z \geq 0$, 
$$P\{X+Y \leq z\mid X > 0, Y > 0\} 
= \frac{P\{X+Y \leq z, X > 0, Y > 0\}}{P\{X > 0, Y > 0\}}
= 4 P\{X+Y \leq z, X > 0, Y > 0\}$$
since $P\{X > 0, Y > 0\} = \frac 14$.  We concentrate on calculating
$4 P\{X+Y \leq z, X > 0, Y > 0\}$. Note that
$$P\{X+Y \leq z, X > 0, Y > 0\} = P\{(X,Y) \in \text{triangle with vertices
at}~(0,0), (z,0), (0,z)\}.$$
Now, the axes divide the square with vertices at $(z,0), (0,z), (-z,0), (0,-z)$
into $4$ triangles (one in each quadrant), and because of the circular
symmetry, the probability that $(X,Y)$ lies in a specific one of these 
$4$ triangles is exactly the same as the probability that $(X,Y)$ lies 
in another specific triangle. More succinctly, our desired probability 
$4 P\{X+Y \leq z, X > 0, Y > 0\}$ is just 
$$P\{(X,Y) \in \text{square with vertices} ~(z,0), (0,z), (-z,0), (0,-z)\}.$$
Use circular symmetry again to rotate the square by $\pi/4$
and argue that our desired probability $4 P\{X+Y \leq z, X > 0, Y > 0\}$
is the same as 
$P\left\{|X| \leq \frac{z}{\sqrt{2}}, |Y| \leq \frac{z}{\sqrt{2}}\right\}$.
Your professor's hint rotates the axes by $\pi/4$ so as to transform $(X,Y)$
into $(U,V)$ and the sides of the square are now parallel to the (rotated)
axes. As you correctly computed in your comment on Batman's answer,
$U$ and $V$ are also independent standard normal random variables
(just as $X$ and $Y$ are), and
thus his hint works out to be 
$$P\{X+Y \leq z\mid X > 0, Y > 0\} = P\left\{|U| \leq \frac{z}{\sqrt{2}}, |V| \leq \frac{z}{\sqrt{2}}\right\}\tag{1}$$
whereas the above approach gets to 
$$P\{X+Y \leq z\mid X > 0, Y > 0\} = P\left\{|X| \leq \frac{z}{\sqrt{2}}, |Y| \leq \frac{z}{\sqrt{2}}\right\}\tag{2}$$
without needing to define $U$ and $V$ at all. In closing, I point out
that joint normality is not needed at all in the above calculations: if the joint density of $(X,Y)$ has circular symmetry,  then $f_{X,Y}(\cdot, \cdot)$
is the same function as $f_{U,V}(\cdot,\cdot)$ and so $(1)$ and $(2)$ 
are the same result. But circularly symmetric $X$ and $Y$ are independent 
only if they are jointly normal with identical variance, and in this
case, so are $U$ and $V$ independent normal random variables with the same
variance.
He says rotato, I say ro-tato, let's call the whole thing off!
A: Hint: What is the joint distribution of U, V?
