I see this question on Stochastic Processes by Ross.

Let $X_n$ be a Markov process for which $X_0$ is uniform on (0, 1) and,

$$ X_{n+1}= \begin{cases} \alpha X_n + 1 - \alpha,& \text{w.p. } X_n,\\ \alpha X_n, & \text{w.p. }1-X_n. \end{cases} $$

where $0<\alpha<1$. Discuss the limiting properties of the sequence $\lim_{n\to\infty} X_n$.

I was able to show $E[X_{n+1}\mid X_n]=X_n$, so the sequence is a martingale. Also since it is bounded between 0 and 1, a limit exists by the martingale convergence theorem. Also the mean of this limit is 0.5, as $E[X_0]=0.5$ and the mean is constant for a martingale. Can we precisely say what the limit is? Is there anything else we could say about the limit?


1 Answer 1


Denote $X_\infty$ the limit of the process. Let $x\in(0,1)$ and define $A_x$ as $$A_x\equiv\left\{\omega\in\Omega:\lim_{n\rightarrow\infty}X_n(\omega)=x\right\}$$ Now, fix $\omega\in A_x$ and $\varepsilon>0$. By definition $$\exists N\in\mathbb{N}, \forall n\geq N:\quad X_n(\omega)\in (x-\varepsilon, x+\varepsilon)\equiv I$$ In particular, for all $n\geq N$, $X_{n+1}(\omega)$ must lie in the set $$(\alpha(x-\varepsilon),\alpha(x+\varepsilon))\cup(\alpha(x-\varepsilon)+1-\alpha,\alpha(x+\varepsilon)+1-\alpha)\equiv J$$ Notice that if we take $\varepsilon$ small enough then $I\cap J=\emptyset$, which is impossible since $X_{n+1}(\omega)\in I\cap J$. This implies that for all $x\in(0,1)$, the set $A_x$ is empty. Therefore, $X_\infty$ a.s. takes values in $\left\{0,1\right\}$, and its distribution can be inferred from the equality $\mathbb{E}X_\infty=1/2$.


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