Tutorials for Chi-square Distribution 1

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Before we discuss the chi-square distribution, we here talk about the gamma distribution at first.

Definition 1: A random variable $X$ is said to have a gamma distribution with parameters $\alpha,\beta$ if its probability density function is given by:

$f(x)=\frac{x^{\alpha-1}e^{-\frac{x}{\beta}}}{\beta^\alpha\Gamma (\alpha)},(\alpha,\beta>0,x\geq 0).$

where,

$E(X)=\alpha\beta~\text{and}~D(X)=\sigma ^2=\alpha\beta^2.$

A brief note on the gamma function:

The quantity $\Gamma (\alpha)$ is known as the gamma function and it is equal to:

$\Gamma (x)=\int_{0}^{\infty}x^{\alpha-1}e^{-x}~dx.$

Useful result:

$\Gamma (1)=1,~\Gamma (\frac{1}{2})=\sqrt{\pi}.$

(1) If we set $\alpha=1,\beta=\frac{1}{\lambda}$ we get $f(x)=\lambda e^{-\lambda x}$ . We see that the exponential distribution is a special case of the gamma distribution.

(2) $\Gamma(\alpha+1)=\alpha\Gamma(\alpha).$ When $\alpha$ equals to a natural integer $n$ , then we have

$\Gamma(n+1)=n\Gamma(n)=n!.$

(3) Proof for gamma distribution:

$E(X)=\int_{0}^{+\infty}xf(x)~dx=\int_{0}^{+\infty}\frac{x^\alpha e^{-\frac{x}{\beta}}}{\beta^\alpha \Gamma(\alpha)}~dx$

$=\frac{1}{\beta^\alpha \Gamma(\alpha)}\int_{0}^{+\infty}x^\alpha e^{-\frac{x}{\beta}}~dx=\frac{1}{\beta^\alpha \Gamma(\alpha)}\cdot \beta^{\alpha+1}\int_{0}^{+\infty}(\frac{x}{\beta})^\alpha e^{-\frac{x}{\beta}}~d\frac{x}{\beta}$

$=\frac{1}{\beta^\alpha \Gamma(\alpha)}\cdot \beta^{\alpha+1}\cdot \Gamma(\alpha+1)=\alpha\beta.$

$E(X^2)=\frac{1}{\beta^\alpha \Gamma(\alpha)}\cdot \beta^{\alpha+2}\cdot \Gamma(\alpha+2)=\beta^2\alpha(\alpha+1).$

$D(X)=E(X^2)-[E(X)]^2=\alpha\beta^2.$

(4) Moment generating function of the $X\sim \Gamma(\alpha,\beta)$ random variable:

$M_X(t)=(1-\beta t)^{-\alpha}.$

Proof.

$M_X(t)=E(e^{tX})=\int_{0}^{+\infty}e^{tx}\frac{x^{\alpha-1}e^{-\frac{x}{\beta}}}{\beta^\alpha \Gamma(\alpha)}~dx$

$=\frac{1}{\beta^\alpha \Gamma(\alpha)}\int_{0}^{+\infty}x^{\alpha-1}e^{-x\frac{1-\beta t}{\beta}}~dx.$

let:

$y=x\frac{1-\beta t}{\beta}\Rightarrow x=\frac{\beta}{1-\beta t}y, dx=\frac{\beta}{1-\beta t}dy.$

therefore:

$M_X(t)=\frac{1}{\beta^\alpha \Gamma(\alpha)}(\frac{\beta}{1-\beta t})^\alpha \int_{0}^{+\infty}y^{\alpha-1}e^{-y}~dy=(1-\beta t)^{-\alpha}.$

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