TITLE: Help with proof of expected value of gamma distribution
QUESTION [14 upvotes]: I am struggling with this proof of the expected value for the gamma distribution.
I need help with the step indicated by the red arrow. Could someone please break it down for me.

Thanks.

REPLY [8 votes]: The argument is direct if one knows that every gamma function is a PDF. Namely, call $$g_a(x)=\frac{\lambda^a}{\Gamma(a)}x^{a-1}\mathrm e^{-\lambda x},$$ then, for every positive $a$, $g_a$ is a PDF hence $$\int_0^\infty g_a(x)\mathrm dx=1.$$ In your case, $f_X=g_\alpha$ and you are after $$E(X)=\int_0^\infty xf_X(x)\mathrm dx=\int_0^\infty xg_\alpha(x)\mathrm dx.$$ The key observation is that, for every positive $a$, $$xg_a(x)=\frac{\Gamma(a+1)}{\Gamma(a)\lambda}g_{a+1}(x)=\frac{a}{\lambda}g_{a+1}(x),$$ hence $$\int_0^\infty xg_a(x)\mathrm dx=\frac{a}{\lambda}\int_0^\infty g_{a+1}(x)\mathrm dx=\frac{a}{\lambda},$$ in particular, $$E(X)=\frac{\alpha}{\lambda}.$$ I believe this is the argument the red arrow is referring to.