Derivation of chi-squared pdf with one degree of freedom from normal distribution pdf How can we derive the chi-squared probability density function (pdf) using the pdf of normal distribution?
I mean, I need to show that 
$$f(x)=\frac{1}{2^{r/2}\Gamma(r/2)}x^{r/2-1}e^{-x/2} \>, \qquad x > 0\>.$$
 A: 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.
A: 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
$\frac{r}{2}$.
