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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.

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  • $\begingroup$ Are the $X_i$'s jointly distributed? If so, their sum will be Gaussian, so $R$ is a folded Gaussian. $\endgroup$
    – user169852
    Sep 5, 2018 at 17:53
  • $\begingroup$ They are not independent. I heard the professor say that absolute value of a gaussian is Rayleigh. Is it somehow related to Rayleigh distribution? $\endgroup$
    – S. Khan
    Sep 5, 2018 at 17:55
  • $\begingroup$ 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. $\endgroup$
    – user169852
    Sep 5, 2018 at 17:57
  • $\begingroup$ 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? $\endgroup$
    – S. Khan
    Sep 5, 2018 at 18:01
  • $\begingroup$ 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. $\endgroup$
    – user169852
    Sep 5, 2018 at 18:03

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