distribution of limit of a Martingale

Question

let $p,X_0\in (0,1)$ and $(X_n)_{n\in\mathbb{N}_0}$ is a stochastic process with values in$[0,1]$. $X_{n+1}=1-p+pX_n$ with probability $X_n$ and $X_{n+1}=pX_n$ with probability $1-X_n$. I have already shown, that $X$ is a martingale and because $(X_n)_{n\in\mathbb{N}_0}$ is uniformly integrable it converges a.s. and in $L^1$ to some random variable $X_\infty$. I want to determine the distribution of this random variable. I was given the hint to look at the process $(X_n(1-X_n))_{n\in\mathbb{N}_0}$. My intuition tells me it will be some kind of Binomial distribution and that I should look at the generating function, but I was not able to compute it. A hint would be appreciated.

Answer 1

Using the hint, we will compute $\mathbb{E}[X_n(1-X_n)]$. First, we will compute \begin{align*} \mathbb{E}[X_{n+1}(1-X_{n+1})|\mathcal F_n] &= X_n(1-p+pX_n)(1-(1-p+pX_n)) + (1-X_n)pX_n(1-pX_n) \\ &= X_n((1-p+pX_n)(p-pX_n) + p(1-X_n)(1-pX_n)) \\ &= X_n(1-X_n)p(1-p+pX_n+1-pX_n) \\ &= X_n(1-X_n)p(2-p), \end{align*}

so \begin{align*} \mathbb{E}[X_{n+1}(1-X_{n+1})] &= \mathbb{E}[\mathbb{E}[X_{n+1}(1-X_{n+1})|\mathcal F_n]] = p(2-p)\mathbb{E}[X_n(1-X_n)] = (p(2-p))^nX_0 \end{align*} by induction.

Now note that $p(2-p)$ attains a maximum value of $1$ at $p=1$, so since $p \in (0,1)$ we have $p(2-p) < 1.$ Thus we have $\lim_{n \rightarrow \infty}\mathbb{E}[X_n(1-X_n)] = 0$, and because $X_n(1-X_n)$ is bounded we have \begin{align*} 0 &= \lim_{n \rightarrow \infty}\mathbb{E}[X_n(1-X_n)] =\mathbb{E}[\lim_{n \rightarrow \infty} X_n(1-X_n)] = \mathbb{E}[X_\infty(1-X_\infty)]. \end{align*} Since $X_\infty \in [0,1]$ we have $0 \le X_\infty(1-X_\infty)$ and because $\mathbb{E}[X_\infty(1-X_\infty)] = 0$ we conclude $X_\infty(1-X_\infty)=0$ and therefore $X_\infty = 1$ or $X_\infty = 0$. Using the fact that $\mathbb{E}[X_\infty] = X_0$, we can find $P(X_\infty = 1)$ and thus also the distribution of $X_\infty$.

Answer 2

By definition, $$\mathbb{E}( X_n(1-X_n))= \underbrace{p(2-p)}_{<1}\mathbb{E}( X_{n-1}(1-X_{n-1}))=\dots$$ Thus, $$X_n(1-X_n) \xrightarrow[ n \rightarrow \infty]{L^1} 0$$ On the other, because $X_n$ is bounded and $X_n \xrightarrow[ n \rightarrow \infty]{} X_{\infty} $ almost surely, $$X_n(1-X_n) \xrightarrow[ n \rightarrow \infty]{L^1} X_{\infty}(1-X_{\infty})$$ Hence $X_{\infty}=0 $ or $X_{\infty}=1$ almost surely. On the other hand, $(X_n)$ is a $L^1$- martingale $$\mathbb{E}( X_{\infty})=\mathbb{E}(X_0)$$ Hence $X_{\infty} \stackrel{\Delta}{=} \mathcal{B}( \mathbb{E}(X_0))$