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Fix $\alpha , \beta > 0$ with $\alpha + \beta = 1$ and consider the sequence of random variables defined by $X_0 = \theta \in (0,1)$ and

$$P(X_n = \alpha + \beta X_{n-1} \mid X_{n-1} , ... , X_0) = X_n , \qquad P(X_n = \beta X_{n-1} \mid X_{n-1} , ... , X_0) =1- X_n.$$

I am trying to prove that $P(\lim_{n\rightarrow \infty} X_n = 1) = \theta$ and $P(\lim_{n\rightarrow \infty} X_n = 0) = 1-\theta$. It seems that there should be a way to do this using the result that $P(E \mid \mathcal F_n) \rightarrow P(E \mid \mathcal F)$, whenever $\mathcal F_n \uparrow \mathcal F$ is an increasing sequence of $\sigma$ algebras. But, I am not seeing a solution. Any suggestions?

By the way, this is an exercise on page 224 of Durrett's Probability: Theory and Examples.

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For both probabilities, it should be $X_{n-1}$ and $1 - X_{n-1}$ on the right hand sides. – Learner Nov 24 '12 at 5:50
Can you use Martingale theorems? This looks like a convergent Martingale which converges to some $X_\inf$ which is a $Bin(\theta)$ . – yoki Nov 24 '12 at 6:49
up vote 4 down vote accepted
  1. $(X_n)_n$ is a martingale: First observe that $\mathbb{P}[X_n = \alpha+\beta \cdot X_{n-1}]+\mathbb{P}[X_n = \beta \cdot X_{n-1}] = 1$. Hence $1_A + 1_B = 1$ almost surely where $A:= [X_n = \alpha+\beta \cdot X_{n-1}]$, $B:=[X_n = \beta \cdot X_{n-1}]$. Using the tower property, we find $$\begin{align*} \mathbb{E}(X_n \mid \mathcal{F}_{n-1}) &= \mathbb{E}(X_n \cdot (1_A+1_B) \mid \mathcal{F}_{n-1}) \\ &= \mathbb{E} \left( (\alpha+\beta \cdot X_{n-1}) \cdot 1_A \mid \mathcal{F}_{n-1} \right)+\mathbb{E} \left( \beta \cdot X_{n-1} \cdot 1_B \mid \mathcal{F}_{n-1} \right) \\ &= (\alpha+\beta \cdot X_{n-1}) \cdot \underbrace{\mathbb{P}(A \mid \mathcal{F}_{n-1})}_{X_{n-1}} + \beta \cdot X_{n-1} \cdot \underbrace{\mathbb{P}(B \mid \mathcal{F}_{n-1})}_{1-X_{n-1}}=X_{n-1}. \end{align*}$$ Since $(X_n)_n$ is $L^2$-bounded (as $0 \leq X_n \leq 1$), there exists $Z$ such that $X_n \to Z$ in $L^2$ and almost surely.
  2. Similar proof shows $$\mathbb{E}((X_n-X_{n-1})^2)=\alpha^2 \cdot \mathbb{E}(X_{n-1} \cdot (1-X_{n-1})). \tag{1}$$ (Same idea as in 1., use $\alpha+\beta=1$.)
  3. $X_n \to Z$ in $L^2$ implies $$\alpha^2 \mathbb{E}(X_{n-1} \cdot (1-X_{n-1})) \stackrel{(1)}{=} \mathbb{E}((X_n-X_{n-1})^2) \to 0 \qquad (n \to \infty)$$ as well as $$\alpha^2 \mathbb{E}(X_{n-1} \cdot (1-X_{n-1}))\to \alpha^2 \cdot \mathbb{E}(Z \cdot (1-Z)). \qquad (n \to \infty)$$ Hence, $$\mathbb{E}(Z (1-Z)) = 0.$$ Since $0 \leq Z \leq 1$ (by pointwise convergence of $(X_n)_n$ and $0 \leq X_n \leq 1$), we conclude $Z \cdot (1-Z)=0$ almost surely. Moreover, we have $\mathbb{E}Z=\mathbb{E}X_0=\theta$ as $(X_n)_n$ is a martingale. Consequently, $$\mathbb{P}[Z=1]=\theta \qquad \qquad \mathbb{P}[Z=0]=1-\theta.$$
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