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I have done some extensive googling and I haven't been successful. Say I have 3 independent random variables that are normally distributed with different means, that is, $X_i \sim \mathcal{N}(\mu_i,\sigma_i^2)$ for $i=1,2,3$. How is the sum of them, or the square root of the sum of them distributed?

This question stems form a research project where I am analyzing the magnitude of vectors whose components are normally distributed. I realize this is closely related to the Maxwell, Rayleigh and Chi squared distributions, but transforming the variables isn't an option because a reverse transformation will be to hard to derive. For example, say I transform the variables into standard normals and apply the Maxwell distribution to find the 75th percentile of the transformed vector magnitude. Without making approximations, I can't relate this to the 75th percentile of the untransformed vector magnitude.

kjetil b halvorsen
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user27606
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  • If the random variables are *independent*, then the sum $Z=X_{1}+X_{2}+X_{3}$ is also normally distributed: $Z\sim \mathcal{N}(\mu_{1}+\mu_{2}+\mu_{3}, \sigma_{1}^2+\sigma_{2}^{2}+\sigma_{3}^{2})$. See [here](http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables). – COOLSerdash Jul 16 '13 at 22:12
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    If they all have the same mean, the sum of their squares is a scaled non central Chi-Square distribution with the mean as the non-centrality parameter. If they have different means, I know of no nice closed form solution. – JohnRos Jul 16 '13 at 22:14
  • They are independent and they do have different means. – user27606 Jul 16 '13 at 22:19
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    The *answer* to this question http://stats.stackexchange.com/questions/29860/confidence-interval-of-multivariate-gaussian-distribution/61993#61993, and its Wikipedia source (https://en.wikipedia.org/wiki/Hotelling%27s_T-squared_distribution) might be very helpful. – QuantIbex Jul 17 '13 at 00:11
  • I already looked into this. I don't see how it applies to what I'm doing. I guess $y = (x-\mu)^T\Sigma{}^{-1}(x-\mu)$ can be thought of as a norm induced by $\Sigma^{-1}$, the physics of the problem doesn't really dictate working in that vector space. I also won't be able to recover the data I'm interested in if I shift by $\mu$. – user27606 Jul 17 '13 at 16:36
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    @JohnRos: judging from [Wikipedia](https://en.wikipedia.org/wiki/Noncentral_chi-squared_distribution), the result would be noncentrally $\chi^2$ if all $X_i$ had unit variance but (potentially) different means, in contrast to what [you write](http://stats.stackexchange.com/questions/64526/the-distribution-of-the-sum-of-squared-normal-random-variables#comment124482_64526). Are you mistaken, or is Wikipedia? – Stephan Kolassa Jan 06 '16 at 11:10
  • @StephanKolassa: by "scaled non-central" I allowed for non unit variance (assuming the same variance for all X_i s. – JohnRos Jan 06 '16 at 13:32
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    @JohnRos: but you take equal means as the precondition for the sum of squares to be noncentral $\chi^2 – Stephan Kolassa Jan 06 '16 at 14:00
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    Maybe you should state your real problem in less abstract language. You say "physics of the problem doesn't really dictate working in that vector space" What *is* the physics of the problem? tell us, please. – kjetil b halvorsen Oct 06 '17 at 10:09
  • @Stephan is correct. This question can be considered the "non-central" version of [questions about sums of gamma distributions](https://stats.stackexchange.com/questions/72479). Their answers make it clear that the present question is going to be analytically intractable or at best very messy. But approximations are not needed: numerical integration is straightforward. – whuber Oct 06 '17 at 14:27

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