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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?

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    $\begingroup$ A function is a "PDF" (probability distribution function) if it the derivative of a "CDF" (cumulative probability function). In particular that means a PDF is always positive, increasing, and its integral is 1. $\endgroup$ – user247327 Mar 22 '16 at 16:49
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    $\begingroup$ Attention! PDF is NOT increasing! It is only non-negative and its integral must be 1. It can be easily seen: Integral of positive and increasing function could not be convergent. $\endgroup$ – Joseph Mar 22 '16 at 17:25
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    $\begingroup$ Just look at the extension to the file name $\endgroup$ – Yeah.. Mar 22 '16 at 17:59
  • $\begingroup$ @Yeah.. Hahahahaha, in an ideal world, yours could be an answer too. (+1). $\endgroup$ – Em. Mar 22 '16 at 18:36
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The following seems to be a typical definition of pdf for a basic probability course. I got it from this course site. enter image description here

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

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  • $\begingroup$ (+1) Actually, if the first two are satisfied, $f$ is a PDF. Then we can use $(3)$ to compute the probability that the random variable with this PDF is in the set $A$. $\endgroup$ – robjohn Mar 22 '16 at 17:53
  • $\begingroup$ Yes, thanks. In fact, in $(3)$ it is stated in this manner. $\endgroup$ – Em. Mar 22 '16 at 17:58
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$f(x)$ is a pdf if:

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

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

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  • $\begingroup$ (+1) this is correct if the variable is defined on all of $\mathbb{R}$. $\endgroup$ – robjohn Mar 22 '16 at 17:56
  • $\begingroup$ Indeed, I sacrificed generality for simplicity. $\endgroup$ – Doug M Mar 22 '16 at 18:00
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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.

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  • $\begingroup$ directly : $\int_0^\infty x^{\alpha-1} e^{-x/\beta}dx = \beta^a \int_0^\infty y^{\alpha-1} e^{-y} dy = \Gamma(a) \beta^{\alpha}$ with the change of variable $y = x/\beta$ $\endgroup$ – reuns Mar 22 '16 at 18:22

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