# 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. Jan 30, 2014 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 $\sigma_i^2$ Aug 15, 2014 at 12:30
• For the second portion,please see mathoverflow.net/questions/89779/… Nov 20, 2014 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:
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 you have $$X_i^2=\sigma^2 Z_i^{2}$$ and so $$\sigma^2\chi_k^2$$ is equivalent to the $$\Gamma(k/2,2\sigma^2)$$ 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.