# Distribution of the sum of squared independent normal random variables.

The sum of squares of $k$ independent standard normal random variables $\sim\chi^2_k$

I read here that if I have $k$ i.i.d normal random variables where $X_i\sim\mathcal{N}(0,\sigma^2)$ then $X_1^2+X_2^2+\dots+X_k^2\sim\sigma^2\chi^2_k$. How do I go about obtaining the pdf?

If I have $k$ independent normal random variables where $X_i\sim\mathcal{N}(0,\sigma_i^2)$ then what is the distribution of $X_1^2+X_2^2+\dots+X_k^2$?

• What you are looking for is the Noncentral Chi-Squared Distribution. – AnonSubmitter85 Jan 30 '14 at 3:17
• @AnonSubmitter85 No - it is not a noncentral Chisquare. The non central Chisquare is composed using $X_i \sim N(\mu_i,1)$ random variables ... whereas this question requires changing variance term $\sigma_i^2$ – wolfies Aug 15 '14 at 12:30
• For the second portion,please see mathoverflow.net/questions/89779/… – upol94 Nov 20 '14 at 20:50

Let's answer the first one. If you know the PDF for $Z$, say $f_{Z}\left(z\right)$, then $f_{c\cdot Z}\left(c\cdot z\right)$ is found from the probability definition:
$$\begin{split} \text{Pr}\left\{c\cdot Z < z \right\} &= \text{Pr}\left\{Z < \cfrac{z}{c} \right\} = F_{Z}\left(\cfrac{z}{c}\right) \quad \text{so} \\ \cfrac{d}{dz}\left[F_{Z}\left(\cfrac{z}{c}\right)\right] &= \cfrac{1}{c} f_{Z}\left(\cfrac{z}{c}\right) \end{split}$$
So, applied to a chi-square, just scaled the PDF for $\chi_{N}^{2}$ by $\cfrac{1}{\sigma^{2}}$ and scale it's argument by the same $\cfrac{1}{\sigma^{2}}$ and plot it.
To answer the first part, remember that the $\chi_k^2$ distribution is (as a special case of the gamma) $\Gamma (k/2,2)$, so by the properties of the gamma distribution $\sigma\chi_k^2$ is equivalent to the $\Gamma(k/2,2\sigma)$ and from here is just a case of plugging in to the known gamma pdf to obtain your desired pdf. For part to go from the fact that $X_i^2$ has also a gamma distribution and then find the sum of independent gamma distributions.