Branching process: show $P(Z_m>k|Z_n=0) \le (G_n(0))^k, mLet $(Z_n)_{n \ge 0}$ be a branching process with generating function $G_n(s)=E(s^{Z_n})$.
Let $m<n$, show that $P(Z_m>k|Z_n=0) \le (G_n(0))^k, k \ge 0$.
I know that the event $\{Z_n=0\}_{n \ge 0}$ are nested upward, that is $\{Z_i=0\}\subseteq\{Z_j=0\}$ whenever $i <j$, and that $P(Z_n=0)=G_n(0)$. I guess I need to decompose $\{Z_m>k|Z_n=0\}$ into other events, but I don't find a decomposition that gives the result.
 A: Let us first assume that $Z_0=1$ almost surely and let us estimate
$$
\mathrm P(Z_m>k, Z_n=0)=\mathrm E(\mathrm P(Z_n=0\mid Z_m);Z_m>k).
$$
For each integer $i$, $\mathrm P(Z_n=0\mid Z_m=i)=G_{n-m}(0)^i$ hence
$$
\mathrm P(Z_m>k, Z_n=0)=\mathrm E(G_{n-m}(0)^{Z_m};Z_m>k)\leqslant G_{n-m}(0)^{k+1},
$$
and
$$
\mathrm P(Z_m>k\mid Z_n=0)\leqslant\frac{G_{n-m}(0)^{k+1}}{\mathrm P(Z_n=0)}=G_{n-m}(0)^{k}\frac{G_{n-m}(0)}{G_n(0)},
$$
Now the sequence $(G_k(0))_k$ is nondecreasing hence $G_{n-m}(0)\leqslant G_n(0)$ and $G_{n-m}(0)^{k}\leqslant G_{n}(0)^{k}$. This proves the result when $Z_0=1$ almost surely.
If $Z_0=i$ almost surely, with $i\geqslant1$, $G_j(s)=H_j(s)^i$ where $H_j(s)=\mathrm E(s^{Z_j}\mid Z_0=1)$ and $\mathrm P(Z_n=0)=H_n(0)^i$, hence
the same computations yield
$$
\mathrm P(Z_m>k\mid Z_n=0)\leqslant\frac{H_{n-m}(0)^{k+1}}{H_n(0)^i}\leqslant H_{n}(0)^{k+1-i},
$$
which cannot be bounded uniformly over $i$. Thus the result probably does not hold for every initial population.
Edit One can remove the word probably in the last sentence above. To prove this, assume that the result proposed by the OP holds for every starting population $Z_0$. Whether $G_k$ means $G_k(s)=\mathrm E(s^{Z_k}\mid Z_0=i)$ or $G_k(s)=\mathrm E(s^{Z_k}\mid Z_0=1)$, $G_k(0)\le \mathrm P_1(Z_k=0)=q_k$ where the subscript $1$ indicates that $Z_0=1$. Hence we are asked to prove for example that, for every $k$ and $t$,
$$
\mathrm P_{tk}(Z_1>k\mid Z_n=0)\leqslant q_n^k,
$$
where the subscript $tk$ indicates that $Z_0=tk$. When $n\to\infty$, $[Z_n=0]$ converges to $[\mbox{extinction}]$ and $q_n\to q=\mathrm P[\mbox{extinction}]$. Assume that $0<q<1$, that is, assume that $\mathrm E_1(Z_1)>1$ and $\mathrm P_1(Z_0=0)>0$. One would have
$$
\mathrm P_{tk}(\mbox{extinction},Z_1>k)\leqslant q^k\mathrm P_{tk}(\mbox{extinction})=q^{(t+1)k}.
$$
Using the fact that the event of extinction is the least probable when $Z_1$ is large and restricting the LHS to the event $[k<Z_1\le sk]$, one would have
$$
q^{sk}\mathrm P_{tk}(k<Z_1\le sk)\leqslant q^{(t+1)k}.
$$
Now, choose $s$ and $t$ such that $1<t\mathrm E_1(Z_1)<s$. Then $\mathrm P_{tk}(k<Z_1\le sk)\to1$ when $k\to\infty$ by the law of large numbers hence one would have $q^{s}\leqslant q^{t+1}$. If $s<t+1$, this is absurd and the condition that $1<t\mathrm E_1(Z_1)<s<t+1$ can be met for every $t$ such that $1/t$ is between $\mathrm E_1(Z_1)-1$ and $\mathrm E_1(Z_1)$.
