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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 $X_1^2+X_2^2+\dots+X_k^2$?

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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.

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.

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