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 a hint towards the solution:


