I have a question about conditional Distribution. Let $X,Y\sim N(0,1)$ and independently distributed. Then, let $Z=X+Y$. Find the p.d.f and c.d.f of $Z\mid(X>0, Y>0)$.
The hint here is to use the circular symmetry of joint density of $X$ and $Y$. No integration needed.
My attempt:
First, we know that $Z\sim N(0,2)$ then the p.d.f of $Z$ is $f_{Z}(z)=\frac{1}{2\sqrt{\pi}}e^{\frac{-z^{2}}{4}},-\infty<z<\infty.$
Next, we have that:
$$F_{Z\mid X>0,Y>0}(z\mid x>0,y>0)=P(Z\leq z\mid X>0,Y>0)=\frac{P(Z\leq z, X>0,Y>0)}{P(X>0,Y>0)}.$$
It follows that by independence:
\begin{align} & F_{Z\mid X>0,Y>0}(z\mid x>0,y>0) = \frac{P(0<Z\leq z)}{P(X>0)P(Y>0)} \\[10pt] = {} & \frac{F_Z(z)-F_Z(0)}{\frac{1}{4}} = 4F_Z(z)-\frac{4}{2}=4F_Z(z)-2. \end{align}
So, the c.d.f is: $F_{Z\mid X>0,Y>0}(z\mid x>0,y>0)=4F_Z(z)-2$.
Then, the p.d.f will be $$f_{Z\mid X>0,Y>0}(z\mid x,y)=4f_Z(z)=\frac{2}{\sqrt{\pi}} e^{\frac{-z^{2}}{4}},0<z<\infty.$$
This is what I got; however, I'm pretty sure I did something wrong since when I integrate the p.d.f over the whole region, it does not give me 1.
Any suggestion?
Here, the phrase "using the circular symmetry of the joint density of $(X,Y)$" most likely refers to the polar decomposition$$(X,Y)=^d(R\cos\Theta,R\sin\Theta)$$ where $R$ and $\Theta$ are independent, $\Theta$ is uniform on $(0,2\pi)$, and the PDF of $R$ is $re^{-r^2/2}$ on $r>0$.
Thus, one is after the conditional distribution of $Z=R(\cos\Theta+\sin\Theta)$ conditionally on $\Theta\in(0,\frac\pi2)$.
Equivalently, thanks to the identity $\cos\Theta+\sin\Theta=\sqrt2\cos(\Theta-\frac\pi4)$, this is also the unconditional distribution of $$W=\sqrt2R\cos\Phi$$ where $\Phi$ is independent of $R$ and uniform on $(-\frac\pi4,\frac\pi4)$, or uniform on $(0,\frac\pi4)$.
Can you continue?