= Solution
The rates in part d turn the generating-function equation into
$$
\partial_tF=\lambda\bigl[(z-1)+(z^{-1}-1)\bigr]z\partial_zF
=\lambda(z-1)^2\partial_zF,\qquad F(z,0)=z.
$$
For the <distribution of a critical linear birth-death population>, the <method of characteristics> gives $\dot z=-\lambda(z-1)^2$ and $\dot F=0$. Setting $w=(1-z)^{-1}$ gives $\dot w=-\lambda$, so the initial characteristic coordinate is $w_0=w+\lambda t$. Hence
$$
\boxed{F(z,t)=1-\frac{1-z}{1+\lambda t(1-z)}.}
$$
It has the correct initial value and $F(1,t)=1$. With $a=\lambda t$, rewrite the denominator as $(1+a)[1-az/(1+a)]$ and expand its <geometric series>. The <probabilities> are
$$
\boxed{P_0(t)=\frac{a}{1+a},\qquad
P_n(t)=\frac{a^{n-1}}{(1+a)^{n+1}}\quad(n\geq1).}
$$
At $t=0$ this is understood by continuity: $P_1=1$ and all other <probabilities> vanish. Summing the positive-state <probabilities> gives $1/(1+a)$, which together with $P_0$ verifies normalization directly.
There is eventual extinction with <probability> one, because $P_0(t)\to1$. Nevertheless $\langle n\rangle=F_z(1,t)=1$ at every finite time, while $\operatorname{Var}(n)=2\lambda t$. There is no contradiction: the increasingly rare surviving populations have conditional <mean> $\langle n\mid n>0\rangle=1+\lambda t$. This is the distinction between a constant ensemble <mean> and survival of a typical realization.
\Image[/past-exam-of-the-theoretical-physics-major-option-course-of-the-university-of-oxford/2017-critical-birth-death-distribution.png]
{title=Extinction probability and surviving-population distribution of the <critical branching process>}
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