1

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
  • 77,844
user27606
  • 109
  • 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. – COOLSerdash Jul 16 '13 at 22:12
  • 2
    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
  • 1
    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
  • 1
    @JohnRos: judging from Wikipedia, the result would be noncentrally $\chi^2$ if all $X_i$ had unit variance but (potentially) different means, in contrast to what you write. 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
  • 1
    @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
  • 2
    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
  • 1
    @Stephan is correct. This question can be considered the "non-central" version of questions about sums of gamma distributions. 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

0 Answers0