$\mathbf{P}\left(2<\left|X\right|+\left|Y\right|<3\right)$ where the random variables are independent $N(0,1)$?

Question

If $X$ and $Y$ are independent $N(0,1)$ random variables, then how can I calculate the following probability: $$\mathbf{P}\left(2<\left|X\right|+\left|Y\right|<3\right)?$$ I calculated the density function of $\left|X\right|$ and $\left|Y\right|$. I tried to write the density functions of there sums, but I couldn't continue where I had to integrate $e^{-t^{2}}$. I had to acknowledge this path I followed leads nowhere to solve this problem. Is there any easier way? Anyway, I got $$f_{\left|X\right|+\left|Y\right|}\left(y\right)=\frac{4}{\pi}e^{-\frac{y^{2}}{2}}e^{\frac{1}{4}y^{2}}\sqrt{2\pi}\frac{1}{\sqrt[4]{2}}\left[\Phi\left(\frac{1}{\sqrt{2}}y\right)-\frac{1}{2}\right] $$ if it means something, but I'm not sure it is a good solution. Any easier way?

Answer 1

Hint: Consider the integral $I=\int_a^b f'(x)f(x)\,dx$. Then by integration by parts with $dv=f'(x)\,dx$ and $u=f(x),$ we have $$ I=f(b)^2-f(a)^2-I $$ so $I=\frac{f(b)^2-f(a)^2}{2}$.

Your problem is very similar; try performing integration by parts. I didn't check your pdf of $|X|+|Y|$ too closely, but it's definitely at least the correct form.