We define the stochastic process $(X_n)_{n \geq 0}$ with $X_0 = x \in [0, 1]$, and: \begin{align*} X_{n+1} = \begin{cases} X_n^2, &\text{ with probability $\frac{1}{2}$} \\ 2X_n - X_n^2, &\text{ with probability $\frac{1}{2}$} \end{cases} \end{align*} I want to show that $\mathrm{P}[\lim_{n \to \infty} X_n = 1] = x$. This is a homework problem, and I was given the following hint:
Use Martingale Convergence Theorem. Under what conditions is the expectation of limit equals that of martingale under what conditions?
My attempt involves first showing that $(X_n)_{n \geq 0}$ is indeed a martingale. One can check that $0 \leq X_n \leq 1 \; \forall n$, so we can apply both Martingale Convergence Theorem and Dominated Convergence Theorem to conclude that $X_n \to X$ for some random variable $X$, and $\mathrm{E}[X] = \lim_{n \to \infty}\mathrm{E}[X_n] = x$. I'm stuck on where to proceed here.
I conjecture that $X \in \{0,1\}$ almost surely, which would thus imply the result.
Any help is appreciated.
By Martingale Convergence Theorem $X =\lim X_n$ exist a.e. Taking limit in the definition we see that $X=X^{2}$ or $X=2X-X^{2}$. The second equation is same as the first so $X=X^{2}$ a.s. This means $X=0$ or $1$ a.s. Also $P(\lim X_n=1)=P(X=1)$. Since $X$ is $0-1$ valued $EX=P(X=1)$. Hence $P(\lim X_n=1)=EX=EX_0=x$