The random variable $X$ with parameters $a>0$ and $\lambda$ is gamma distributed with
$$f_X(x)=\frac{1}{\Gamma(a)}\lambda^ax^{a-1}e^{-\lambda x}, \quad x\ge 0.$$
a) Find the MGF for $X$. Hint:
$$\int_0^{\infty}\frac{1}{\Gamma(a)}(\lambda-t)^ax^{(a-1)}e^{-(\lambda-1)x} \ dx=1.$$
a) Use the result above to show that the sum of two independent gamma distributed random variables with the same $\lambda$ are also gamma distributed.
My solution:
a) We have that
$$M_X(t)=E[e^{tX}]=\frac{\lambda^a}{\Gamma(a)}\int_0^{\infty}x^{a-1}e^{-\lambda x}e^{tx} \ dx= \frac{\lambda^a}{\Gamma(a)}\int_0^{\infty}x^{a-1}e^{-(\lambda-t) x} \ dx,$$
Using the hint, I can replace the last integral above with $\Gamma(a)/(\lambda-t)^a,$ so
$$M_X(t)=\frac{\lambda^a}{\Gamma(a)}\cdot\frac{\Gamma(a)}{(\lambda-t)^a}=\left(\frac{\lambda}{\lambda-t}\right)^a.$$
b) Let $X\sim\Gamma(a,\lambda)$ and $Y\sim\Gamma(b,\lambda).$ Using the above we have that
$$M_{X+Y}(t)=M_X(t)M_Y(t)=\left(\frac{\lambda}{\lambda-t}\right)^a\cdot \left(\frac{\lambda}{\lambda-t}\right)^b=\left(\frac{\lambda}{\lambda-t}\right)^{a+b} \Rightarrow X+Y\sim\Gamma(a+b,\lambda),$$
where the second equality followed by independence.
Question: Is it really true that if $X$ and $Y$ are independent, then it follows that $M_{X+Y}(t)=M_X(t)M_Y(t)?$ Or Is the answer just a lucky coincidence? I haven't seen this rule anywhere in my book.
It is really true (at least provided all exist) and I’m very surprised you can’t find this fact in the book. We have $$ E(e^{t(X+Y)})= E(e^{tX}e^{tY})=E(e^{tX})E(e^{tY})$$