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For $Y\sim \mathrm{Geom}(p)$, its mean is $1/p$, so the size-biased probability generating function isChoose $X_1,X_2$ independent with distribution $\mathrm{Geom}(p)$. Their shifted sum has generating function, initially for $s>0$,with both sides continuous and zero at $s=0$. Equality of probability generating functions provesEquivalently, $X_1+X_2=k+1$ has $k$ possible positive summand pairs, each of probability $p^2(1-p{)}^{k-1}$, matching the size-biased geometric distribution probabilities.

For $0\le s<1$, differentiate the probability generating function term by term and multiply by $s/\mu$:This is the size-biased probability generating function identity. At $s=1$, the finite-mean assumption gives the limiting value one on both sides; at zero, positive support gives value zero.

For the positive-support geometric distribution with success probability $p$, size bias gives $\mathbb{P}(\tilde{Y}=k)=k{p}^2(1-p{)}^{k-1}$. The sum of two independent copies has $k$ positive-integer decompositions at value $k+1$, so its shifted probability is exactly this expression. Its probability generating function is $p^{2}s/(1-(1-p)s{)}^2$, also obtained by the size-biased probability generating function identity.