Is $\{C^{X_n}\}_{n = 1}^{\infty}$ is a martingale? Suppose $\{X_n\}_{n = 1}^{\infty}$ is a Galton-Watson branching process with ($P(X_1 = 1) = 1$ and $X_{n+1} = \Sigma_{i = 1}^{X_n} \theta_{i,n}$, where $\{\theta_{i,n}\}_{n = 1}^{\infty}$ are i.i.d. random variables, such that $P(\theta_{i,n}\in \mathbb{N}) = 1$). Suppose, that $C = \lim_{n \to \infty} P(X_n = 0)$ is the probability of extinction. 
Does that imply, that $\{C^{X_n}\}_{n = 1}^{\infty}$ always a martingale ($P(E(C^{X_n}|C^{X_1}, …, C^{X_{n-1}}) = C^{X_{n-1}}) = 1$)?
If $E\left(\theta_{i,n}\right) \leq 1$, then $C = 1$, and thus $P(C^{X_n} = 1) = 1$, and so this sequence is a martingale. However, I do not know, how to deal with the case, when $E\left(\theta_{i,n}\right) > 1$.
 A: Yes, with respect to the filtration $\left(\mathcal{F}_n\right)_{n\in\mathbb{N}}$ given by $\mathcal{F}_n:=\sigma\left\{X_k\,:\,k\leq n\right\}$ for all $n\in\mathbb{N}$.
Assume $E\left[\theta\right]<\infty$, where $\theta\overset{d}=\theta_{i,n}$ for all $i,n$. 
Let us assume for convenience $\mathbb{P}\left(X_0=1\right)=1$, rather than $\mathbb{P}\left(X_1=1\right)=1$.
Let $f\,:\,\left[0,1\right]\rightarrow \left[0,1\right]$ with $f(s)\overset{\Delta}=E\left[s^{\theta}\right]$ be the generating function of $\theta$, with $f(0)\overset{\Delta}=\lim_{s\downarrow 0}f(s)$ so that $f$ is continuous. Note that $f(0)=\mathbb{P}\left(\theta=0\right)=\mathbb{P}\left(X_1=0\right)$.
A standard result is that $f^{(n)}(s)=E\left[s^{X_n}\right]$, where $f^{(n)}=\underbrace{f\circ f\circ\ldots \circ f}_{n\mbox{ times}}$ is the $n$-fold composition of $f$. 
Define $C_n\overset{\Delta}=\mathbb{P}\left(X_n=0\right)$ for all $n\in\mathbb{N}$, and note that $C_n=f^{(n)}(0)$. Indeed, $f^{(n)}(0)=\lim_{s\downarrow 0}f^{(n)}(s)=\lim_{s\downarrow 0}E\left[s^{X_n}\right]=\lim_{s\downarrow 0}\left(\mathbb{P}\left(X_n=0\right)+\sum_{k=1}^{\infty} s^k\mathbb{P}\left(X_n=k\right)\right)=\mathbb{P}\left(X_n=0\right)$.
In particular, $C_{n+1}=f(C_n)$.
Since $f$ is continuous, $C=\lim_{n\rightarrow\infty}C_n$ is a fix point of $f$, i.e., $f(C)=C$.
As a result, $E\left[C^{X_{n+1}}\left|\mathcal{F}_n\right.\right]=E\left[C^{\sum_{i=1}^{X_n}\theta_{i,n}}\left|\mathcal{F}_n\right.\right]=\left(E\left[C^{\theta}\right]\right)^{X_n}=f(C)^{X_n}=C^{X_n}$, almost surely.
