= Solution
For the <Poisson process>, $q_n=g$ and $r_n=0$, so
$$
\partial_tF=g(z-1)F,\qquad F(z,0)=1.
$$
This is an <ordinary differential equation> in time for each fixed $z$, with solution
$$
\boxed{F(z,t)=e^{gt(z-1)}=e^{-gt}\sum_{n=0}^{\infty}\frac{(gt)^n}{n!}z^n.}
$$
Consequently
$$
\boxed{P_n(t)=e^{-gt}\frac{(gt)^n}{n!},\qquad n=0,1,2,\ldots.}
$$
The initial condition is interpreted as $P_0(0)=1$, with all other initial <probabilities> zero. Normalization follows from $F(1,t)=1$. The <mean> and <variance> are both $gt$, as expected for independent events arriving at constant rate.
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