Limiting r.v. for converging martingale

Question

I am studying martingales and as an example, it proposes me the following martingale: $$ M_0 = k \in [0,1], M_{n+1} = \begin{cases} M_n^2 & wp\,\, 1/2 \\ 2M_n - M_n^2 & wp \,\, 1/2 \end{cases}$$ And then, after showing that it converges a.s. to a r.v. $M$ it says that $M = 0$ or $1$ with probability $1$ and that this can be proven by contradiction. I tried proving it but I didn't manage to, can you help me?

Answer 1

It can be shown that $0\leq M_n \leq 1$ for all $n \in \{0, 1, 2, ...\}$, and so a bounded convergence theorem for martingales ensures $M_n$ converges to a random variable $M$ with probability 1, and indeed $M$ must take values in $[0,1]$.

Suppose $M_0 = k$ where $k$ is a deterministic constant in $[0,1]$. Here is a proof that the limiting variable $M$ is Bernoulli with $$ P[M=1] = k, P[M=0]=1-k$$ Under the given dynamics we have \begin{align} E[M_{n+1}^2] &= E[M_n^4](1/2) + E[(2M_n-M_n^2)^2](1/2)\\ &= E[M_n^4(1/2) + (1/2)(4M_n^2 + M_n^4 - 4M_n^3)]\\ &= E[M_n^4 -2M_n^3 + 2M_n^2] \end{align} Now if $M_n$ converges with prob 1 to a random variable $M$, and since $M_n \in [0,1]$ for all $n$, it holds that $\lim_{n\rightarrow\infty} E[M_n^k] = E[M^k]$ for $k \in \{2, 3, 4\}$. Taking a limit of the above equality then gives: $$ E[M^2] = E[M^4 - 2M^3 + 2M^2]$$ Rearranging gives $E[M^4 - 2M^3 + M^2] = 0$, which means $$ E[M^2(M-1)^2] = 0$$ On the other hand it is clear that $M^2(M-1)^2\geq 0$, and so it must be that $M^2(M-1)^2=0$ with prob 1. Thus, $P[M \in \{0,1\}]=1$. It follows that the limiting variable $M$ is Bernoulli. It remains only to compute $P[M=1]$.

Since $M_n$ is a martingale we have $E[M_n]=E[M_0]=k$ for all $n \in \{0, 1, 2,...\}$, and so $E[M]=k$. It follows that $P[M=1]=k, P[M=0]=1-k$. $\Box$