# Show that $P(\lim\limits_{n\to \infty} X_{n}=0 \operatorname{or} 1)=1$ and if $X_{0}=\theta$ then $P(\lim\limits_{n\to \infty} X_{n}=1)=\theta$

Let $$X_{n} \in [0,1]$$ be adapted to $$\mathcal{F}_{n}$$ and $$\alpha, \beta > 0$$ where $$\alpha + \beta =1$$. Further:

$$P(X_{n+1}=\alpha + \beta X_{n} \vert \mathcal{F}_{n})=X_{n}$$ or $$P(X_{n+1}= \beta X_{n} \vert \mathcal{F}_{n})=1-X_{n}$$

Show that $$P(\lim\limits_{n\to \infty} X_{n}=0 \operatorname{or} 1)=1$$ and if $$X_{0}=\theta$$ then $$P(\lim\limits_{n\to \infty} X_{n}=1)=\theta$$

I honestly do not know where to begin:

$$E[1_{\{X_{n+1}=\alpha + \beta X_{n}\}} \vert \mathcal{F}_{n}]=X_{n}$$

$$E[1_{\{X_{n+1}=\beta X_{n}\}} \vert \mathcal{F}_{n}]=1-X_{n}$$

Maybe the tower property could help:

$$E[E[1_{\{X_{n+1}=\alpha + \beta X_{n}\}}\vert \mathcal{F}_{n+1}] \vert \mathcal{F}_{n}]=X_{n}$$

• Get rid of the indicator functions and just compute $E[X_{n+1}|F_n]$. Commented Oct 4, 2019 at 22:47

• Since $$(X_n)_{n\geq 0}$$ is a bounded martingale, it converges almost surely. Let $$X_{\infty}$$ denote its a.s-limit. We know that $$\mathbb{P}(X_{\infty} \in [0, 1]) = 1$$.

• Now we claim that

$$\mathbb{E}[X_{n+1}(1-X_{n+1})] = (1-\alpha^2)\mathbb{E}[X_n(1-X_n)] \tag{1}$$

holds. Indeed, write $$A_{n+1} = \{ X_{n+1} = \alpha + \beta X_n \}$$. Then by noting that

$$\mathbb{P}(\{ X_{n+1} = \alpha + \beta X_n\} \cup \{ X_{n+1} = \beta X_n\}) \stackrel{\text{(tower)}}{=} \mathbb{E}[X_n + (1-X_n)] = 1,$$

we may write $$X_{n+1} = \alpha \mathbf{1}_{A_{n+1}} + \beta X_n$$. From this, we get

\begin{align*} \mathbb{E}[X_{n+1}(1-X_{n+1})] &= \mathbb{E}[X_{n+1}] - \mathbb{E}[X_{n+1}^2] \\ &= \mathbb{E}[X_n] - \mathbb{E}[\alpha^2 \mathbf{1}_{A_{n+1}} + 2\alpha\beta X_n \mathbf{1}_{A_{n+1}} + \beta^2 X_n^2] \\ &\stackrel{\text{(tower)}}{=} \mathbb{E}[X_n] - \mathbb{E}[\alpha^2 X_n + 2\alpha\beta X_n^2 + \beta^2 X_n^2] \\ &= (1-\alpha^2)\mathbb{E}[X_n(1-X_n)] \end{align*}

as desired.

• From $$\text{(1)}$$, we get $$\mathbb{E}[X_n(1-X_n)] \to 0$$ as $$n\to\infty$$. Then by the dominated convergence theorem,

$$\mathbb{E}[X_{\infty}(1-X_{\infty})] = \lim_{n\to\infty} \mathbb{E}[X_n(1-X_n)] = 0.$$

Since $$X_{\infty}(1-X_{\infty})$$ a.s. non-negative, the above computation shows that $$X_{\infty}(1-X_{\infty}) = 0$$ a.s., and so, we get $$\mathbb{P}(X_{\infty} \in \{0, 1\}) = 1$$ as desired.

• Finally, by noting that $$X_{\infty}$$ is $$\{0,1\}$$-valued almost surely, we can write

$$\mathbb{P}(X_{\infty} = 1) = \mathbb{E}[X_{\infty}] = \lim_{n\to\infty} \mathbb{E}[X_n] = \mathbb{E}[X_0] = \theta.$$