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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:

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

But it leads to nothing.

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

1 Answer 1

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  • 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

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

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