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The function $f(t) = \frac{t^{\alpha-1}e^{-t}}{\Gamma(\alpha)}, \ \ 0 < t < \infty$ is a pdf. But Why is the gamma family defined as $f(x| \alpha, \beta) = \frac{1}{\Gamma(\alpha) \beta^{\alpha}} x^{\alpha-1}e^{-x/ \beta}, \ \ 0 < x < \infty, \ \ \ \alpha >0, \ \ \beta >0$

I know you make the substitution $x = \beta t$. But why do this?

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    Because the parameter $\beta$ has practical significance. To paraphrase your question for a more familiar situation, why is the normal family defined using the parameter $\sigma$ when a substitution brings us to $\sigma=1$? Because $\sigma$ describes something important about the distribution.2012-01-28

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André's comment is right; the parameter $\beta$ has a particular meaning in certain settings.

Here's an example. One common use of the gamma distribution is that if $T$ is the amount of time until the $n$th arrival in a Poisson process with rate $\lambda$, then the pdf of $T$ is

$f_T(t) = \frac{ t^{n-1}\lambda^n e^{-\lambda t}}{\Gamma(n)}, \ \ t> 0.$

(Poisson processes are often used to model arrivals and departures in queuing systems.)

Since $\lambda$ is a rate, it represents the average number of arrivals per unit time.

If we rewrite the pdf with $\beta = 1/\lambda$ we get the form in your question,

$f(t| n, \beta) = \frac{1}{\Gamma(n) \beta^n} t^{n-1}e^{-t/ \beta},\ \ t > 0.$ Given the interpretation of $\lambda$ above, $\beta$ represents the average amount of time between arrivals; i.e., the average interarrival time.

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It's a scale parameter. If the probability density of the random variable $X$ is $x\mapsto \text{constant}\cdot x^{\alpha-1} e^{-x}$ for $x>0$ and $Y= \beta X$ then the probability density of $Y$ is $x\mapsto\text{constant} \cdot x^{\alpha-1} e^{-x/\beta}$ for $x>0$.

The other really important bit of "intuition" about the gamma distribution is that if the densities of two independent random variables are $x\mapsto\text{constant} \cdot x^{\alpha_1-1} e^{-x/\beta}\text{ for }x>0$ and $x\mapsto\text{constant} \cdot x^{\alpha_2-1} e^{-x/\beta}\text{ for }x>0$ then the density of the sum of those two random variables is $x\mapsto\text{constant} \cdot x^{\alpha_1+\alpha_2-1} e^{-x/\beta}\text{ for }x>0.$ In a sense, that explains why the "${}-1$" follows the "$\alpha$".