Distribution of the sum of squared independent normal random variables.

Solution 1:

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{equation} \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} \end{equation}

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.

Solution 2:

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.