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Based on my understanding, Chi square distribution and noncentral Chi-squared distribution are used for special Gaussian distribution. However, when the Gaussian is not standard and with no unit variance, how could I calculate the distribution of this Gaussian squared?

Mathematically, $X \sim N(\mu,\sigma^2)$. Let $Y=X^2$, what is the distribution of Y?

linda
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2 Answers2

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Note that $X_1=\frac{X-\mu}{\sigma^2}$ follows the standard normal distribution, and so $X_1^2$ has the $\chi^2$ distribution.

Now, $X_1^2=(X^2-2\mu X+\mu^2)/\sigma^4$. Thus you can get the distribution of $X^2$ in terms of distributions of $X_1^2$, $X$ and a constant.

I do not know though, if the distribution of $X^2$ has any standard name or not.

voldemort
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  • Thank you for your answer! However, if I want to calculate out the exact distribution of $X_1^2$+X in this case, can I convolve these two distributions? Seems like I can't, right?..Since these two are not independent... – linda Sep 07 '14 at 03:34
  • @linda: These two aren't independent... calculating the exact distribution would be hard I think. – voldemort Sep 07 '14 at 04:31
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Y is distributed as a scaled, non-central chi-square random variable (https://en.wikipedia.org/wiki/Noncentral_chi-squared_distribution), with one degree of freedom.

In particular, using the notation of the wikipedia page, we would say $Y/\sigma^2 \sim \chi^{\prime 2}_1 (\mu^2/\sigma^2)$

Brian
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