Deriving of the chi-square density function

1. Jul 21, 2009

jone

According to my textbook, to derive the chi-square density function, one should perform three steps. First we consider a standard i.i.d. Gaussian random vector $\mathbf{X} = [X_1 \cdot \cdot \cdot X_n]^T$ and its squared magnitude
$||\mathbf{X}||^2 = \sum_{i = 1}^nX_i^2$.

1. For n = 1, show that the PDF of $||X_1||^2$ is
$f_{X^2_1}(x) = \frac{1}{\sqrt{2\pi x}}\exp\left(-\frac{x}{2}\right)$

I did step 1 using the CDF-method, so I don't need any help with that.

2. For any n, show that the PDF satisfies the recursive relation
$f_{X_{n+2}^2}(x) = \frac{x}{n}f_{X_n^2}(x)$

I was thinking this could be proven by induction and this is where I'm stuck. To prove by induction I first show the recursive relation for n = 1:
$f_{X_3^2}(x) = xf_{X_1^2}(x) = \squrt{\frac{x}{2\pi}}\exp\left(-\frac{x}{2}\right)$.
I tried to do that using the CDF method:
$P(X_1^2 + X_2^2 + X_3^2 < x) = \int_{x_1^2 + x_2^2 + x_3^2<x}\frac{1}{(2\pi)^{3/2}}\exp\left(-\frac{x_1^2 + x_2^2 + x_3^2}{2}\right)dx_1dx_2dx_3$

Changing to spherical coordinates and integrating I get
$P(X_1^2 + X_2^2 + X_3^2 < x) = \frac{2}{\sqrt{2\pi}}\int_0^{\sqrt{x}}\exp\left(-\frac{r^2}{2}\right)r^2dr$

Differentiating back gives
$f_{X_3^2}(x) = \frac{2x}{\sqrt{2\pi}}\exp\left(-\frac{x}{2}\right)$
But this is not what the recursive relation gives, and I don't see why. Although I didn't obtain the correct result, next I have to show that
$f_{X_{m+1+2}^2}(x) = \frac{x}{m+1}f_{X_{m+1}^2}(x)$.
I'm not sure I know how to prove that. Maybe there is another way to show the recursive relation.

3. Using the recursive relation as well as the PDF:s for n = 1 and n = 2, find the density for n ≥ 3.