Let $Z_n$ be a branching process, with $Z_0 = 1$. Let the Progeny N be distributed as $P(N=0)=P(N=2)=\frac{1}{2}$.
I want to solve the following problems:
1) Show that $E[Z_{n}] \to 1$, as $n \to \infty $
2) Show that $Z_n$ cannot converge in probability to 1 as $n \to \infty$
3) Show that given $Z_{n-1} >0$
$P(Z_n = 0) = E[(\frac{1}{2})^{Z_{n-1}}]$
Progress:
1) So I start of by computing the probability generating function of N which is simply $g_N(t) = \frac{1}{2} + \frac{1}{2}t^2$. From there I compute the expectation of N which is $g'_N(1) = 1$.
Then I use the idea of conditioning to compute the expected value of $Z_N$.
$E[Z_N] = \sum_{i=0}^{\infty} P(Z_{N-1} = i)E[Z_N|Z_{n-1} = i]$ (*)
Since $Z_n$ can be written as $Z_n = N_1 + N_2 +...+ N_{Z_{n-1}}$ I can write $E[Z_n|Z_{n-1} = i]$ as $E[Z_n|Z_{n-1} = i] = E[\sum_{i=1}^{Z_{n-1}} N_i|Z_{n-i} = k] = kE[N] = k$. Now (*) becomes:
$E[Z_n] = \sum_{i=0}^{\infty} kE[Z_{n-1}]$. As you can see the right hand side is simply the expected value for $E[Z_{n-1}]$ and using recursion I obtain.
$E[Z_n] = 1^nE[X_0] = 1^n$ and we can see that $1^n \to 1$ as $n \to \infty$
Although it kind of proves the problem, I am unsure if this is a correct way to prove it.
2) For this problem I used Chebyshev's inequality, but I am unsure if it's the correct path to take.
I computed the variances of $Z_n$ and $N$ and got the solution: $var(Z_n) = nVar(N)$. Then using Chebyshev's inequality I concluded:
$P(|Z_n - 1|> \epsilon) \le \frac{nVar(N)}{\epsilon} \to \infty$ as $n \to \infty$
I don't know if this is enough to assume that $Z_n$ does not converge in probability to 1. Would be greatful if someone could lead me to the right direction on this.
3) Since $Z_n = N_1 + N_2 +...+ N_{Z_{n-1}}$ I assume that you can write as following:
$P(Z_n = 0) = g_{Z_n}(0) = (g_N(0))^{Z_{n-1}} = (P(N=0))^{Z_{n-1}} = (\frac{1}{2})^{Z_{n-1}}$
However, this didn't prove what the problem stated. so far I don't know if this is correct or I am suppose to prove that $(\frac{1}{2})^{Z_{n-1}} = E[(\frac{1}{2})^{Z_{n-1}}]$
If anyone could help me out I'd very much appreciate it!
