If we take this definition for the gamma function:

$$ \Gamma(\alpha)=\int_{0}^{\infty}x^{\alpha-1}e^{-x}\mathrm dx \ \text{ for } \ \alpha>0,$$

and we substitute $\lambda y$ for $x$. Why do we get: $$\Gamma(\alpha)=\lambda^{\alpha}\int_{0}^{\infty}y^{\alpha-1}e^{-\lambda y}\mathrm dy$$ and not: $$\Gamma(\alpha)=\lambda^{\alpha - 1}\int_{0}^{\infty}y^{\alpha-1}e^{-\lambda y}\mathrm dy$$

See (https://www.probabilitycourse.com/chapter4/4_2_4_Gamma_distribution.php)

  • 11
    $\begingroup$ Don't forget about the dy term. $\endgroup$
    – Chris
    Mar 25 '18 at 18:59

When substituting, $\lambda y = x$ means $\lambda\mathrm dy = \mathrm dx$. That is where the extra $\lambda$ comes from.


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