= Solution
{c}
The <moment-generating function> is $M_X(t)=\mathbb E e^{tX}$ on the real values where this expectation is finite, normally including a neighborhood of zero. Direct substitution gives
$$
\boxed{M_{aX+b}(t)=e^{bt}M_X(at)}
$$
whenever the right side is finite.
For the given <Gamma distribution> and $t<\lambda$,
$$
M_X(t)=\frac{\lambda^m}{(m-1)!}
\int_0^\infty x^{m-1}e^{-(\lambda-t)x}dx
=\left(\frac\lambda{\lambda-t}\right)^m,
$$
using the <Gamma function> after the scaling $u=(\lambda-t)x$. At $t\geq\lambda$ the positive integral diverges.
<independence> multiplies <moment-generating functions>. Thus
$$
M_{\sum_iY_i}(t)=\left(\frac\lambda{\lambda-t}\right)^{2n}.
$$
Uniqueness of a <moment-generating function> on a neighborhood of zero identifies
$$
\boxed{\sum_{i=1}^nY_i\sim\operatorname{Gamma}(2n,\lambda),}
$$
where $\lambda$ is the rate parameter, as in the printed <probability density function>.
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