= Solution
A <Bernoulli random variable> with parameter $1/2$ has $G_Y(s)=(1+s)/2$. Part (i) and the <geometric distribution> formula give
$$
G_X(s)=\frac1{2-(1+s)/2}=\frac2{3-s}
=\frac23\sum_{j=0}^\infty\left(\frac s3\right)^j.
$$
Comparing coefficients in the <probability generating function> identifies
$$
\boxed{\mathbb P(X=j)=\frac23\left(\frac13\right)^j,\quad j=0,1,\ldots.}
$$
Thus \b[$X$ has the zero-based <geometric distribution> with success parameter $2/3$]. The probabilities sum to one, and its mean $1/2$ agrees with the random-sum mean $\mathbb E(N)\mathbb E(Y_1)=1\cdot1/2$.
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