Help with proof of expected value of gamma distribution

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.

Solution 1:

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.

Solution 2:

As Daniel was saying, set $t = \lambda x$ and substitute in the integral to get

$$ \int_0^\infty \left( \frac{t}{\lambda} \right)^{(\alpha+1) - 1} e^{-t} \frac{dt}{\lambda} \quad=\quad \frac{1}{\lambda^{\alpha+1}} \int_0^\infty t ^{(\alpha+1) - 1} e^{-t} dt \quad=\quad \frac{\Gamma(\alpha+1)}{\lambda^{\alpha+1}} $$

since by definition of the $\Gamma$ function we have

$$ \Gamma(z) := \int_0^\infty t^{z-1}e^{-t}dt $$