Derivation of chi-squared pdf with one degree of freedom from normal distribution pdf

Solution 1:

The way the question is expressed is a mess, but I'll assume it means this: if $X\sim N(0,1)$, how do you find the pdf of $X^2$? Here's one way. Remember that the pdf of $X$ is $$ \varphi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}. $$ Let $f$ be the pdf of $X^2$. Then $$ \begin{align} f(x) & = \frac{d}{dx} \Pr(X^2 \le x) = \frac{d}{dx} \Pr(-\sqrt{x}\le X\le\sqrt{x}) \\ \\ & = \frac{d}{dx} \frac{1}{\sqrt{2\pi}} \int_{-\sqrt{x}}^\sqrt{x} e^{-u^2/2} \;du = \frac{2}{\sqrt{2\pi}}\frac{d}{dx} \int_0^\sqrt{x} e^{-u^2/2} \;du \\ \\ & = \frac{2}{\sqrt{2\pi}} e^{-\sqrt{x}^2/2} \frac{d}{dx} \sqrt{x} = \frac{2}{\sqrt{2\pi}} e^{-x/2} \frac{1}{2\sqrt{x}} \\ \\ \\ & = \frac{e^{-x/2}}{\sqrt{2\pi x}}. \end{align} $$

Sometimes it might be written as $\dfrac{1}{\sqrt{2\pi}} x^{\frac12 - 1}e^{-x/2}$ so that you can see how it resembles the function involved in defining the Gamma function.

Your title said $1$ degree of freedom. But what you write seems to allow $r$ to be some number other than $1$. If you want to do that, then there's more work to do.

Solution 2:

If $X \sim (\mu, \Sigma) \neq (0, I)$, the result you wish to prove does not hold: even if the random variables are independent but have nonzero means, you get a non-central $\chi^2$ pdf which is not what you are trying to show.

If $X_1, \ldots, X_n$ are independent standard normal random variables, then $X_i^2$ has a Gamma distribution with scale parameter $\frac{1}{2}$ and order parameter $\frac{1}{2}$. Then, $\sum_{i= 1}^n X_i^2 $ is a sum of $n$ independent Gamma random variables each with scale parameter $\frac{1}{2}$ and order parameter $\frac{1}{2}$ and is thus a Gamma random variable with scale parameter $\frac{1}{2}$ and order parameter $\frac{r}{2}$.