Question

Verify that the gamma density function integrates to$1$.

Short answer

To establish the assertion, integrate$F$over$\mathrm{ℝ}$and analyze the definition of the Gamma function.

Step-by-step solution

Find The density function.

Suppose that$X~$Gamma $(\alpha ,\lambda )$The density function is

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

For $x\ge 0$ otherwise it is equal to zero. We want to prove that

${\int }_{\mathrm{ℝ}}f\left(x\right)dx=1$

We have that,

localid="1649661055597" ${\int }_{\mathrm{ℝ}}f\left(x\right)dx={\int }_0^{\infty }\frac{{\lambda }^{\alpha }}{\Gamma \left(\alpha \right)}{x}^{\alpha -1}{e}^{-\lambda x}dx$

$=\frac{{\lambda }^{\alpha }}{\Gamma \left(\alpha \right)}{\int }_0^{\infty }{x}^{\alpha -1}{e}^{-\lambda x}dx$ (1)

Evaluate the integral process.

In order to evaluate the integral, make the substitution $s=\lambda x$that yields$x=\frac{s}{\lambda }$and $ds=\lambda dx$we have that

localid="1649661029116" ${\int }_0^{\infty }{x}^{\alpha -1}{e}^{-\lambda x}dx={\int }_0^{\infty }{\left(\frac{s}{\lambda }\right)}^{\alpha -1}{e}^{-s}\frac{ds}{\lambda }$

$=\frac{1}{{\lambda }^{\alpha }}{\int }_0^{\infty }{s}^{\alpha -1}{e}^{-s}ds=\frac{\Gamma \left(\alpha \right)}{{\lambda }^{\alpha }}$

If that plug $\left(1\right)$, we end up with

localid="1649661004300" $\frac{{\lambda }^{\alpha }}{\Gamma \left(\alpha \right)}{\int }_0^{\infty }{x}^{\alpha -1}{e}^{-\lambda x}dx$

$=\frac{{\lambda }^{\alpha }}{\Gamma \left(\alpha \right)}\times \frac{\Gamma \left(\alpha \right)}{{\lambda }^{\alpha }}=1$

Hence, we have proved the claim.