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When trying to understand my professors notes, I came upon this piece and don’t understand the step he took.

$\frac{(\sum x_i+2)^{n+1}}{\Gamma (n+1)} \int_0^\infty\theta^{n+1} e^{-\theta \sum x_i+2}\text{d}\theta$

To

$\frac{(\sum x_i+2)^{n+1}}{\Gamma (n+1)} \frac{\Gamma (n+2)}{(\sum x_i+2)^{n+1}} $

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    $\begingroup$ Google Gamma distribution and Gamma function. $\endgroup$ – Xi'an Jan 19 at 16:21
  • $\begingroup$ Thanks, I got the $\Gamma (\Theta) = \Theta^{n+1}*e^{-\Theta} $. But why do we have divided by the term above? $\endgroup$ – ToTom Jan 19 at 16:34
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    $\begingroup$ That's not what the Gamma function is: in your comment you have written down part of the density for a Gamma$(n+2)$ distribution. $\endgroup$ – whuber Jan 19 at 17:04
  • $\begingroup$ Hint: try a change of scale in the original integral. $\endgroup$ – Xi'an Jan 19 at 17:26

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