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I am looking for the derivation of the density and CDF of the sum

$ Y=\sum_{i=1}^{N}(X_i)^2 $

where $X_i \sim \mathcal{N}(0,\sigma^2)$. This problem has been addressed in this question and in this question however the explanation provided in the comment refers to a section of the Wikipedia page of the Noncentral chi-squared distribution that apparently has been changed since the comment was posted. Any reference to papers or textbooks is very appreciated.

kjetil b halvorsen
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Arrigo Benedetti
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  • Those other quetions deal with noncentral chi-square distributions. Yours is not about a noncentral chi-square distribution, but only about a scaled chi-square distribution, so it's simpler than those. – Michael Hardy Dec 21 '19 at 05:16

1 Answers1

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Note: $k=N$ throughout the answer.

It is well-known that the standard chi-squared distribution with $k$ degrees of freedom has probability density function $$ f_{\sum_{i=1}^{N}Z_i^2}(z)= \frac{z^{\frac{k}{2}-1} e^{-\frac{z}{2}}}{ 2^{\frac{k}{2}}\Gamma(\frac{k}{2})} \mathbb{I}_{z>0}.$$

We may write $Y=\sum_{i=1}^{N}(\sigma Z_i)^2=\sigma^2 \sum_{i=1}^{N}Z_i^2$, where the $Z_i$'s are independent and identically distributed standard normal variables. ($Z_i=\frac{X_i}{\sigma}$ for all i.)

By the change of variables formula applied to the monotonic function $g(y)=\sigma^2 y$, we have, $$f_{Y}(y)= \frac{1}{\sigma^2}f_{\sum_{i=1}^{N}Z_i}(\frac{y}{\sigma^2})= \frac{1}{\sigma^2} \frac{ (\frac{y}{\sigma^2})^{\frac{k}{2}-1} e^{-\frac{\frac{y}{\sigma^2}}{2}}}{ 2^{\frac{k}{2}}\Gamma(\frac{k}{2})} \mathbb{I}_{y>0}.$$ See https://en.wikipedia.org/wiki/Probability_density_function for a reminder on the change of variables formula.

Mickybo Yakari
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