How to show that $f(x; \alpha, \beta) = \frac{1}{\beta^{\alpha} \Gamma(\alpha)}x^{\alpha-1}e^{-x/\beta}$is a pdf?

Question

I have some problems trying to prove the following problem:

A continuous random variable $X$ is said to have a gamma distribution with parameters $\alpha > 0$ and $\beta > 0$ if it has a pdf given by: $$f(x; \alpha, \beta) = \frac{1}{\beta^{\alpha} \Gamma(\alpha)}x^{\alpha-1}e^{-x/\beta}$$ if $x>0$, or $0$ otherwise.

Given that apparently this is a pdf by definition, I do not know how to prove it is a pdf. My guess is to check if I take the integration of the distribution the value should be $1$. Is this correct? Is that enough?

Answer 1

$f(x)$ is a pdf if:

$f(x) \geq 0$ for all x. And,

$\int_{-\infty}^{\infty} f(x) dx = 1$

Answer 2

The following seems to be a typical definition of pdf for a basic probability course. I got it from this course site.

You will want to show that the function you have satisfies these conditions.

Answer 3

I'm showing you a detailed solution since this is not your typical PDF problem.

$f \geq 0$; I'm leaving this to you.

As for showing that it integrates to $1$, the $\Gamma$ distribution is usually motivated by means of the $\Gamma$ function; by definition, $$\Gamma(\alpha) = \int_{0}^{\infty}x^{\alpha-1}e^{-x}\text{ d}x\text{.}$$ Notice that showing $$\int_{0}^{\infty}\dfrac{1}{\beta^{\alpha}\Gamma(\alpha)}x^{\alpha-1}e^{-x/\beta}\text{ d}x=1$$ is equivalent to showing $$\int_{0}^{\infty}x^{\alpha-1}e^{-x/\beta}\text{ d}x = \Gamma(\alpha)\beta^{\alpha}\text{.}$$ Well, $$\Gamma(\alpha)\beta^{\alpha} = \int_{0}^{\infty}\beta^{\alpha}x^{\alpha-1}e^{-x}\text{ d}x\text{.}$$ This doesn't seem to lead me anywhere, so I'm going to try combining $x$ and $\beta$ into a single exponent by multiplying by $\beta/\beta$: $$\Gamma(\alpha)\beta^{\alpha} = \int_{0}^{\infty}\dfrac{\beta}{\beta}\cdot\beta^{\alpha}x^{\alpha-1}e^{-x}\text{ d}x = \int_{0}^{\infty}\beta\cdot\beta^{\alpha-1}x^{\alpha-1}e^{-x}\text{ d}x = \beta\int_{0}^{\infty}(x\beta)^{\alpha-1}e^{-x}\text{ d}x\text{.}$$ This seems to suggest the substitution $y = x\beta$, or $x = y/\beta$. Then $\text{d}y = \beta\text{ d}x$, or $\text{d}x = \dfrac{\text{d}y}{\beta}$, and (note the limits of integration don't change) $$\Gamma(\alpha)\beta^{\alpha} = \int_{0}^{\infty}y^{\alpha-1}e^{-y/\beta}\text{ d}y$$ (notice the $\beta$ cancel out). $y$ is just a dummy variable, so we have shown $$\Gamma(\alpha)\beta^{\alpha} = \int_{0}^{\infty}x^{\alpha-1}e^{-x/\beta}\text{ d}x$$ or $$\int_{0}^{\infty}\dfrac{1}{\beta^{\alpha}\Gamma(\alpha)}x^{\alpha-1}e^{-x/\beta}\text{ d}x=1$$ as desired.