# Sum of Independent Folded-Normal distributions

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?

• Yuck. I ... doubt it. – dfeuer Jun 25 '13 at 3:40
• Identical variances for $X$ and $Y$? Or different? Zero means? Or different? – wolfies Jun 25 '13 at 17:09
• There is a problem with the title:$$Sum.of.Independent.Half Normal.distributions$$ If $X$~$N(\mu, \sigma^2)$, then $|X|$ has a folded Normal distribution (not a half-Normal (unless $\mu = 0$)). Accordingly, the question is actually seeking the sum of independent folded Normals … not the sum of independent half-Normals. – wolfies Jun 29 '13 at 10:38
• Okay thanks I fixed it – Mark Jun 30 '13 at 14:44
• @wolfies : There's such a thing as \text{} within MathJax: $$\text{Sum of Independent Half-Normal distributions}$$ – Michael Hardy Jul 24 '15 at 13:23

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 $\alpha$.
• @Cemre $\Phi(x)$ is the cumulative probability distribution function (CDF) of the standard normal random variable. – Dilip Sarwate Jan 16 '14 at 19:25
• @Swistack Circular symmetry means the joint pdf looks the same and has the same formula if the axes are rotated by any arbitrary angle $\theta$. So rotate the axes by $\pi/4$ which makes the sides of the square parallel to the axes. – Dilip Sarwate Mar 24 at 18:07