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I want to find the support of a non-central chi-squared distribution ($99.9 \%$ of the energy). For example, If I have a Gaussian distribution with parameters $\mu$ and $\sigma$, I know $99.9 \%$ of the energy is in $[\mu - 3\sigma, \mu + 3\sigma]$. How can I find a similar support for non-central chi-squared distribution with degrees of freedom $\nu$ and non-centrality parameter $\lambda$?

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    $\begingroup$ By "energy" do you mean "probability"? Assuming that's the case, could you tell us how you computed the answer for the Gaussian distribution--and then explain why the same method would not work for a non-central $\chi^2$ distribution? $\endgroup$ – whuber Sep 7 '15 at 16:17
  • $\begingroup$ For Gaussian distribution, I did not compute the answer but it is known (see tables in en.wikipedia.org/wiki/Normal_distribution for F(μ + nσ) − F(μ − nσ) for different n). $\endgroup$ – bassir Sep 7 '15 at 21:14

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