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I'm studying basic probability and in my lecture notes, it shows how the Gamma function results from the convolution of two exponential random variables. To introduce the gamma function it shows that for a random variable $X \sim \mathrm{Gamma}(n,\lambda)$:

$$ \int_{-\infty}^\infty f(x) \ dx = \int_{- \infty}^\infty \frac{\lambda^n x^{n-1} e^{-\lambda x}}{(n-1)!} \ dx = \int_{- \infty}^\infty \frac{\lambda (\lambda x)^{n-1}e^{- \lambda x}}{(n-1)!} \ dx = 1$$

and so by setting $ z = \lambda x $...

$$ (n-1)! = \int_0^\infty z^{n-1} e^{-z} \ dz$$

The step $ \int_{- \infty}^{\infty} \frac{\lambda(\lambda x)^{n-1}e^{- \lambda x}}{(n-1)!} \ dx = 1$ is left unexplained and I am confused as to how my lecturer is obtaining this result.

If they are taking for granted that the Gamma mass function is well-defined and that the unit measure axiom is satisfied , how can I prove this for myself and that this reasoning makes sense? Otherwise, what has my lecturer done in this step? Is there a way to easily integrate this function?

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    $\begingroup$ You may try integration by part and develop a reduction formula for this gamma integral (reducing the power of $x$ inside). $\endgroup$ – BGM Dec 28 '15 at 17:09
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If the question is simply how to prove that $$ \int_{-\infty}^\infty \frac{(\lambda x)^{n-1}e^{- \lambda x}}{(n-1)!} (\lambda\, dx) = 1, $$ then I would do that by proving that $$ \int_0^\infty z^{n-1} e^{-z}\,dz = (n-1)!\,. $$ To do that, integrate by parts repeatedly until you've got $$ (n-1)(n-2)(n-3)\cdots 1 \int_0^\infty e^{-z}\,dz. $$ In the course of that, you will need to know that $z^c e^{-z}\to0$ as $z\to\infty$. That can be established by L'Hopital's rule. (There is also an intelligent way to establish that limit, but it seems many prefer to do things like that under anesthesia.)

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    $\begingroup$ That makes sense. For some reason the proof of the recursive property of the gamma function is later in the lecture notes. Thanks very much! $\endgroup$ – kw3rti Dec 28 '15 at 17:37

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