Probability of $0\leq X\leq Y$ for two standard Gaussian random variables random variables We are given two independent standard Gaussian random variables $X\sim N(0, \sigma_x^2), Y\sim N(0, \sigma_y^2)$. Compute $Pr(X-Y\leq 0 \cap X\geq 0)$.
Here is what I did so far:
Denote by $\mathbb{1}(x)$ a function which is $1$ if $0\leq x \leq y$ and $0$ else. Moreover let $F(x)$ be th CDF of $X$ and $G(y)$ the CDF of $Y$.
$Pr(X-Y \leq 0 \cap X\geq 0)=Pr(0\leq X \leq Y)=\int \int \mathbb{1}(x) dF(x)dG(y)=\int \int_0^y F'(x) dx dG(y)=\int \frac{1}{2} erf(\frac{y}{\sqrt{2\sigma_x^2}})dG(y)  $ But now I don't know which bound do apply to the integral? $0$ to $\infty?$
EDIT: Please note that $X$ and $Y$ have different variances.

As $X$ and $Y$ are independent,
$ \displaystyle f(x,y) = \frac{1}{2 \pi \sigma_x \sigma_y} \ e^{- \left(\dfrac{x^2}{2\sigma_x^2} + \dfrac{y^2}{2\sigma_y^2}\right)}$
$X - Y \leq 0 \cap X \geq 0 \implies 0 \leq X \leq Y$,
Use change of variable,
$x = r \sqrt{2} \ \sigma_x \cos\theta, y = r \sqrt{2} \ \sigma_y \sin\theta$
Jacobian $|J| = 2  r  \sigma_x  \sigma_y$
$Y \geq X \geq 0 \implies \dfrac{\sigma_x}{\sigma_y} \leq \tan \theta$.
So, $ \ \arctan \left(\dfrac{\sigma_x}{\sigma_y}\right) \leq \theta \leq \dfrac{\pi}{2}$.
$0 \leq r \leq \infty$
$ \displaystyle P(0 \leq X \leq Y) = \int_{\arctan(\sigma_x / \sigma_y)}^{\pi/2} \int_0^{\infty} |J| \ f(r,\theta) \ dr \ d\theta$
Can you take it from here?
