Verify that the gamma density function integrates to 1

gamma density

Answered by LanceJ 2 years ago

The **gamma density function** is defined as

$$f(x)=\frac{\lambda e^{-\lambda x}(\lambda x)^{\alpha-1}}{\Gamma(\alpha)}$$

for $x \ge 0$

to show the integral is equal to 1 utilize the definition of the **gamma function**,

$$\int_0^\infty \frac{\lambda e^{-\lambda x}(\lambda x)^{\alpha-1}}{\Gamma(\alpha)} dx$$

let $u=\lambda $x and $du = \lambda d$x

$$=\int_0^\infty \frac{e^{-u}u^{\alpha-1}}{\Gamma(\alpha)}$$

$$=\frac{1}{\Gamma(\alpha)}\int_0^{\infty}e^{-u}u^{\alpha-1}$$

$$=\frac{1}{\Gamma(\alpha)}\Gamma(\alpha)=1$$

where I have used the definition of the gamma function $\int_0^{\infty}e^{-u}u^{\alpha-1}=\Gamma(\alpha)$

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Asked: 2 years ago