# What is the distribution of "absolute value of sum of N gaussian random variables with mean 0 and variance non-zero"?

I know that for N = 2, we have $\sqrt{X^{2} + Y^{2}} \sim \mathrm{Rayleigh}(\sigma)$, where $X \sim \mathcal{N} (0,\sigma^{2})$ and $Y \sim \mathcal{N} (0,\sigma^{2})$ but what about the following? $$R = \left| \sum_{i=1}^{N} X_{i} \right|$$ where each $X_{i} \sim \mathcal{N} (0,\sigma^{2})$.

Please guide me on how to proceed? I searched the web but found nothing. Thanks in advance.

• Are the $X_i$'s jointly distributed? If so, their sum will be Gaussian, so $R$ is a folded Gaussian.
– user169852
Sep 5, 2018 at 17:53
• They are not independent. I heard the professor say that absolute value of a gaussian is Rayleigh. Is it somehow related to Rayleigh distribution? Sep 5, 2018 at 17:55
• The absolute value of a complex zero-mean Gaussian (which is merely $\sqrt{X^2 + Y^2}$ where $X$ and $Y$ are the real and imaginary parts) is Rayleigh. The absolute value of a real Gaussian has a folded Gaussian or folded normal distribution.
– user169852
Sep 5, 2018 at 17:57
• So if each $X_{i} = a + n_{i}$, where $a \sim \mathcal{N} (0,\sigma_{a}^{2})$ and $n_{i} \sim \mathcal{N} (0,\sigma_{n}^{2})$. Then it's not complex, right? But because of the correlation, we don't have a folded distribution either? Sep 5, 2018 at 18:01
• The correlation is irrelevant. $X_i$ is the sum of two zero-mean Gaussians, hence $X_i$ is a zero-mean Gaussian. Then $\sum X_i$ is the sum of zero-mean Gaussians, hence is a zero-mean Gaussian. Therefore $R = \left|\sum X_i\right|$ is a folded Gaussian.
– user169852
Sep 5, 2018 at 18:03