Sum of Independent Folded-Normal distributions

For $\alpha > 0$, $$F_{|X|+|Y|}(\alpha) = P\{|X|+|Y|\leq \alpha\} = P\{(X,Y) \in A\}$$ where $A$ is a square region with vertices $(\alpha,0), (0,\alpha), (-\alpha, 0), (0,-\alpha)$.

• Assume that $X$ and $Y$ have $0$ mean and identical variance $\sigma^2$. Then, since the joint density of $X$ and $Y$ has circular symmetry, we can rotate the square about the origin so that the sides are parallel to the axes and at distance $\alpha/\sqrt{2}$ from them. Consequently, $$F_{|X|+|Y|}(\alpha) = P\{(X,Y) \in A\} = \left[\Phi\left(\frac{\alpha}{\sqrt{2}\sigma}\right) - \Phi\left(\frac{-\alpha}{\sqrt{2}\sigma}\right)\right]^2 = \left[2\Phi\left(\frac{\alpha}{\sqrt{2}\sigma}\right) - 1\right]^2.$$ Can you get the density of $Z$ from this? (Hint: think of the chain rule for differentiation from basic calculus, and remember that you know the derivative of $\Phi(x)$) Note that $Z$ is not the absolute value of a normal random variable.

• More generally, for arbitrary independent normal random variables, we have that $$F_{|X|+|Y|}(\alpha) = P\{(X,Y) \in A\} = \int_{-\alpha}^0\int_{-\alpha-x}^{\alpha+x}f_X(x)f_Y(y)\mathrm dy \mathrm dx + \int_0^{\alpha}\int_{x-\alpha}^{-x+\alpha}f_X(x)f_Y(y)\mathrm dy\mathrm dx.$$ Rather than computing the integrals and then differentiating with respect to $\alpha$ to find the density of $|X|+|Y|$, one can directly differentiate the integrals with respect to $\alpha$. If you don't remember the details of how to do so, see the comment following this answer and remember that when you are differentiating the outer integral (the one with respect to $x$), the integrand (a.k.a. the value of the inner integral) is also a function of $\alpha$.