Branching Process in critical case: Easy question about generation size

Question

Let us consider a Branching Process with offspring distribution $X$, having mean $\mu:=E[X]$. Let $Z_n$ denote the number of individuals at generation $n$.

We have shown (with help of a martingale argument) in general, that

$E[Z_n]=\mu^n$,

$P[Z_n > 0]=P[Z_n \geq 1] \leq \mu^n$.

Now I want to show that in the critical case

$$ Z_n \overset{P}{\rightarrow} 0, \text{ as } n \rightarrow \infty \tag1$$

$$E[Z_n]=1 \mbox{ for all } n \tag2 $$

(2) is clear. But what about (1)? So I have to show that $P[Z_n>0] \rightarrow 0$? But obviously I cannot use the estimate from above. So can you help me how to show that $Z_n$ converges to 0 in probability?

And can you explain me, why this makes sense or is even possible? How can it be that we always have expectation one even though the probability of being positive goes to 0?

Thank you very much for your help

Answer 1

If $\phi(s)=\sum_{k=0}^\infty p_k s^k$ is the probability generating function of the offspring distribution, then $a_{n+1}=\phi(a_n)$ where $a_n=P(Z_n=0)$ is the probability that the population is extinct at time $n$. The sequence $(a_n)$ is non-decreasing and convergences to $a$, the probability of eventual extinction. If $p_0>0$ and $\mu=1$, then the only solution to $\phi(a)=a$ is $a=1$. This shows that $Z_n\to0$ as $n\to\infty$

As for the intuition, maybe my answer here will help: Martingale not uniformly integrable

Answer 2

There are two stronger statements about this, if you are interested, first is

Kolmogorov's estimate:

If the offspring distribution $O$: $\mathbb{E}O^2<\infty$, then $$ \lim_{n\to\infty}n\mathbb{P}(Z_n>0)=\frac{2}{\sigma^2} $$

Yaglom's limit law:

The limit law of $\{n^{-1}Z_n|Z_n>0\}$ has exponential distribution with parameter $\frac{\sigma^2}{2}$