I have the following martingale: $M_0 = x \in ] 0,1 [$, $M_n = \begin{cases} M_n^2\quad wp\quad 1/2\\ 2M_n -M_n^2 \quad wp\quad 1/2\end{cases}$
defined with respect to the filtration $(\mathcal{F_n},n\in\mathbb{N})$.
The exercise asks me to compute the following recursive process: $$ A_0=0, A_{n+1}= A_n + \log \mathbb{E}(\exp(M_{n+1}-M_n)|\mathcal{F_n}) $$According to my own analysis this process is predictible and increasing. \begin{align} \log \mathbb{E}(\exp(M_{n+1}-M_n)|\mathcal{F_n}) &= \log\frac{\mathbb{E}(\exp(M_{n+1})|\mathcal{F_n})}{\exp(M_n)} \\ &=\log \exp(1/2\cdot M_n^2+1/2\cdot (2M_n-M_n^2)-M_n) \\ &= 1 \end{align}
My conclusion is that $A_n = n\quad \forall n \in\mathbb{N}.$
Then it asks me if, given that $X_n = \exp(M_n - A_n)$ it is possible to say that $\exists X_{\infty} : \mathbb{E}(X_{\infty}|\mathcal{F_n})= X_n$.
Out of curiousity I proved that $X_n$ is a martingale provided that $M_n$ is bounded. And I also managed to prove that $M_n$ in this case is $\in ]0,1[$. So, for the existence of $X_{\infty}$ I would need that $\sup_n\mathbb{E}(X_n^2)<+\infty$. $$\mathbb{E}(X_n^2)= \mathbb{E}(\exp(2M_n-2A_n)) \leq \mathbb{E}(\exp(2M_n)) = \exp(2\cdot M_{n-1}) \leq \exp(2).$$
This should prove the existence of $X_{\infty}$, right? It was not necessary to compute the value $A_n$? As long as it was positive I could always upperbound the martingale, right? Also, could you explain to me why that $X_n$ is a martingale? It seems related to Doob's decomposition of $\exp(M_n)$ which is always a sub-martingale if $M_n$ is a martingale but I don't understand why that construction always leads me to a martingale...