Proving that the $\chi^{2}$ Distribution is a PDF

Question

The $\chi^{2}$-distribution is given by $$f(x; k)\equiv \frac{1}{2^{\frac{k}{2}} \Gamma\left(\frac{k}{2}\right)}x^{\frac{k}{2}- 1}e^{-\frac{x}{2}}.$$

If the $\chi^{2}$-distribution is a probability-density function (PDF), then it must hold that $$ \int_{-\infty}^{+\infty} f(x; k) \, dx = 1. $$ However, when trying to prove this, I am stuck:

$$\int_{-\infty}^{+\infty} f(x;k) \, dx = \int_{-\infty}^{+\infty}\frac{1}{2^{\frac{k}{2}}\Gamma\left(\frac{k}{2}\right)}x^{\frac{k}{2}- 1}e^{-\frac{x}{2}} \, dx = \frac{1}{2^{\frac{k}{2}}\Gamma\left(\frac{k}{2}\right)}\int_{-\infty}^{+\infty}x^{\frac{k}{2}-1}e^{-\frac{x}{2}} \, dx$$

Looking only at the last integral, we get: $$\int_{-\infty}^{+\infty}x^{\frac{k}{2}-1}e^{-\frac{x}{2}}dx = \int_{-\infty}^{0} x^{\frac{k}{2}-1}e^{-\frac{x}{2}}dx + \underbrace{\int_{0}^{\infty} x^{\frac{k}{2}-1}e^{-\frac{x}{2}}dx}_{= \Gamma\left(\frac{k}{2}\right)}$$

However, I am not yet sure on what to do with the first term. Any feedback would be appreciated.

Answer 1

As pointed out in the comments, you don't integrate in $(\infty,0]$ since the chi squared distribution isn't defined there. You also have written

$$\underbrace{\int_{0}^{\infty} x^{\frac{k}{2}-1}\color{red}{e^{-\frac{x}{2}}}\mathrm{d}x}_{= \Gamma\left(\frac{k}{2}\right)}$$

Which is wrong. The integral definition of the Gamma function for $\operatorname{Re}(z)>0$ is $$\Gamma(z)=\int_0^\infty x^{z-1}\color{red}{e^{-x}}\mathrm{d}x$$ The exponent should be $-x$, not $-x/2$. With a change of variable $t=x/2$ we can write $$\int_0^\infty x^{k/2-1}e^{-x/2}\mathrm{d}x=2^{k/2}\int_0^\infty t^{k/2-1} e^{-t}\mathrm{d}t=2^{k/2}\Gamma(k/2)$$

Answer 2

The integral is

$$\frac{1}{\Gamma\left(\frac{k}{2}\right)}\int_0^{\infty}\left(\frac{1}{2}\right)^{k/2}x^{k/2-1}e^{-x/2}dx$$

Substituting

$$\frac{x}{2}=y$$

you get

$$\frac{1}{\Gamma\left(\frac{k}{2}\right)}\underbrace{\int_0^{\infty}y^{k/2-1}e^{-y}dy}_{=\Gamma\left(\frac{k}{2}\right)}=1$$

Answer 3

The density of the chi-square distribution is $$ f(x) = \begin{cases} \displaystyle \frac{1}{2^{k/2} \Gamma(k/2)}x^{k/2- 1}e^{-x/2} & \text{for } x>0, \\[6pt] 0 & \text{for } x<0. \end{cases} $$ And so $$ \int_{-\infty}^{+\infty} f(x)\,dx = \int_{-\infty}^0 f(x)\, dx + \int_0^{+\infty} f(x) \, dx = 0 + \int_0^{+\infty} f(x)\, dx. $$

And then $$ \int_0^{+\infty} \left( \frac x 2 \right)^{k/2-1} e^{-x/2} \left(\frac{dx} 2 \right) = \int_0^{+\infty} u^{k/2-1} e^{-u} \, du = \Gamma\left( \frac k2 \right). $$