The distribution of the limit random variable

Question

Produce a sequence of random variables $\left\{X_{n}\right\}_{n \geq 0}$ as follows: Let $X_{0}=q$ with probability 1, where $q \in(0,1)$ is some constant. For $n \geq 1,$ let $X_{n}=X_{n-1}^{2}$ with probability $1 / 2$ and $X_{n}=2 X_{n-1}-X_{n-1}^{2}$ with probability $1 / 2 .$ Prove that $\left\{X_{n}\right\}_{n \geq 0}$ converges almost surely, and find the distribution of the limit random variable.
Thanks for your help!

Answer 1

In order to talk about almost sure convergence we need the sequence $\{X_n\}_{n\geq 0}$ to live on the same probability space. So to formalize your question let $(\Omega,\mathcal{F}, \mathbb{P})$ be a probability space and let $\{Z_n\}_{n\geq 1}$ be an i.i.d sequence such that $\mathbb{P}(Z_n=1)=1/2=\mathbb{P}(Z_n=0)$. Your sequence can now be defined as: $$ \begin{align}X_0:&=q\\ X_n:&=\mathbf{1}_{\{Z_n=1\}}X_{n-1}^2+\mathbf{1}_{\{Z_n=0\}}(2X_{n-1}-X_{n-1}^2) \end{align} $$

Can we use some symmetry of the random walk from $X_{n-1}$ to $X_{n}$? (hint: martingale convergence theorem)

We now assume $X_n\longrightarrow X$ a.s. on $\Omega$. Is it possible for some $\omega\in\Omega$ that $X(\omega)\in(0,1)$? What does this tell us about the distribution of $X$?