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If $X \sim \mathcal{N}(0,4)$, then I know that $\left(\frac{X}{2}\right)^2$ is has a $\chi^2$ distribution. But $X^2$ itself $\chi^2$ distributed as well?

Also if $X_i \sim \mathcal{N}(0,4)$ where $i \in (1,2,3)$, what is the distribution of $X_1^2+X_2^2+X_3^2$?

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  • $\begingroup$ $X^2$ is a $\chi^2_1$ variable multiplied by $4$ and not a $\chi^2_1$ variable. Similarly, $X_1^2+X_2^2+X^3_2$ is a $\chi^2_3$ variable multiplied by $3$. They all are Gamma variables. $\endgroup$ – Xi'an Dec 29 '16 at 20:57
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I presume that your notation is $X \sim \mathcal{N}(\mu,\sigma^2)$.

The Chi-squared distribution relates to the square of a standard normal variable. Scaling $X$ by diving by 2 will cause the variance to be 4 times smaller, therefore $\frac{X}{2}$ has a distribution $\sim \mathcal{N}(0,1)$.

The chi squared distribution has one degree of freedom which is the number of squared normal distributed variables added together. The distribution of $X_1^2+X_2^2+X_3^2$ has 3 degrees of freedom but these are not standard normal variables.

First scale to make the variables into standard normal to find out their distribution

$X_1^2+X_2^2+X_3^2 = 4 \left( (\frac{X_1}{2})^2+(\frac{X_2}{2})^2+(\frac{X_3}{2})^2 \right)$

From this you can see that the distribution is 4 times a chi squared distribution with 3 degrees of freedom.

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    $\begingroup$ Very good explanation. I think with respect to the second part you omitted the requirement that X_1, X_2 , and X_3 must be independent. Xi'an answer conveys the same information more briefly but also omits the independence requirement. $\endgroup$ – Michael Chernick Dec 29 '16 at 21:44

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