Let $X$ and $Y$ be independent, normally distributed random variables.
How is $|X| + |Y|$ distributed?
Is it known to be $|Z|$, where $Z$ is distributed normally?
$\endgroup$ 51 Answer
$\begingroup$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 answerand 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 $\alpha$.
