Solution
= Solution
For $0\leq s\leq1$, independence gives
$$
G_{X+Y}(s)
=\mathbb E[s^{X+Y}]
=\mathbb E[s^Xs^Y]
=\mathbb E[s^X]\,\mathbb E[s^Y].
$$
Therefore
$$
\boxed{G_{X+Y}(s)=G_X(s)G_Y(s)}.
$$
= Solution
For $0\leq s\leq1$, independence gives
$$
G_{X+Y}(s)
=\mathbb E[s^{X+Y}]
=\mathbb E[s^Xs^Y]
=\mathbb E[s^X]\,\mathbb E[s^Y].
$$
Therefore
$$
\boxed{G_{X+Y}(s)=G_X(s)G_Y(s)}.
$$