# Help with understanding the $\chi^2$-distribution

I'm studying statistics and there's one part in my book I can't understand. I tried to make as good translation as I can of the problematic part...here goes:

Chi squared $\chi^2$ distribution

Let $Z_{1}, ..., Z_{v}$ be $v$ independent $N(0,1)-$ distributed random variables and

$\chi_{v}^2 = Z_{1}^2 + ... + Z_{v}^2$.

If $v = 1$ the cumulative distribution function $F_{\chi_{1}^2}(x) = F_{1}(x)$ is the following:

$F_{1}(x) = P(Z_{1}^2 \leq x) = P(-\sqrt{x} \leq Z_{1} \leq \sqrt{x}) = 2\Phi(\sqrt{x}) - 1,$ where $x \geq 0$ and $\Phi(x)$ is the CDF of the normal distribution . By taking the derivative we get the probability density function to be

$f_{1}(x) = x^{-\frac{1}{2}}\phi(\sqrt{x}) = C_{1}x^{-\frac{1}{2}}e^{-\frac{1}{2}x}$, where $x \geq 0$ and $C_{1} = \frac{1}{\sqrt{2\pi}}$ is the normalizing constant.

Let's take a closer look of the sum $Z_{1}^2 + Z_{2}^2$ when $v = 2$. Both $Z_{1}^2$ and $Z_{2}^2$ are normally distributed. By calculating the CDF $P(Z_{1}^2 + Z_{2}^2 \leq x)$ as the integral of the product of density functions of $Z_{1}^2$ and $Z_{2}^2$ and then by taking the derivative we get

(1) $f_{2}(x) = \displaystyle\int_{0}^{x} f_{1}(x-u)f_{1}(u)du = C_{1}^2\displaystyle\int_{0}^{x}(x-u)^{-\frac{1}{2}}u^{-\frac{1}{2}}e^{-\frac{1}{2}x}du$

By making the substitution $u = xt$, we get easily

(2) $f_{2}(x) = C_{2}e^{-\frac{1}{2}x}$, where $C_{2}$ is again the normalizing constant.

Now my question is: "How did the writer of this book get the (1)-part of the $f_{2}(x)$-function. I didn't understand that part. Can someone give me more detailed steps or something? Thank you for any help

P.S. If any part is unclear or something I can provide more information on the part

-

You can write the probability in term of expectation, with indicator functions:

$$P(Z_1^2 + Z_2^2 \leq x) = E[ 1_{Z_1^2 + Z_2^2 \leq x}]$$

$$= \int_0^{+\infty} \int_0^{+\infty} 1_{u+v\leq x} f_1(u) f_1(v) du dv$$

$$= \int_0^{+\infty} \int_0^{+\infty} 1_{u \leq x-v} f_1(u) du f_1(v) dv$$

$$= \int_0^{+\infty} F_1(x-v) 1_{v \leq x} f_1(v) dv$$

$$= \int_0^{x} F_1(x-v) f_1(v) dv$$

I added the indicator function $1_{v \leq x}$ in the 3rd line because $F_1(x-v)$ is only defined where $x \geq v$.

Also, you know that $f_2(x) = \frac{d}{dx} P(Z_1^2 + Z_2^2 \leq x)$

So:

$$f_2(x) = \int_0^{x} f_1(x-v) f_1(v) dv$$

$$f_2(x) = C_1^2 \int_0^{x} \sqrt{u(x-u)} dv e^{-\frac{1}{2}x}$$

$$f_2(x) = C_1^2 \int_0^{1} \sqrt{t(1-t)} dt e^{-\frac{1}{2}x}$$

and we get the same result as the author.

-
Thank you for your effort! =) – jjepsuomi Feb 17 '13 at 20:07