= Solution
{c}
For the <geometric distribution>, $\mu=1/p$, so the <probability generating function> of the <size-biased distribution> is
$$
G_{\widetilde Y}(s)=\frac{p^2s}{(1-qs)^2}.
$$
If $X_1,X_2$ are <independent and identically distributed> with <geometric distribution> of parameter $p$, the <probability generating function of an independent sum> gives
$$
G_{X_1+X_2-1}(s)=s^{-1}\left(\frac{ps}{1-qs}\right)^2
=\frac{p^2s}{(1-qs)^2}.
$$
At zero, interpret the expression by its continuous extension. The <probability generating functions> agree, hence
$$
\boxed{\widetilde Y\overset d=X_1+X_2-1,\qquad X_1,X_2\sim\operatorname{Geom}(p)\text{ independently}.}
$$
One can also check the <probability mass functions> directly: for $k\ge1$, there are $k$ positive integer pairs with sum $k+1$, each having <probability> $p^2q^{k-1}$. Thus both sides assign <probability> $kp^2q^{k-1}$ to $k$.
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