Exercise 5.5.7 in Durrett's "Probability Theory and Examples 4ed" states: Let $X_n \in [0 , 1] $ be adapted to $\mathcal{F_n}$. Let $\alpha$, $\beta>0$ with $\alpha+\beta=1$ and suppose: $$P(X_{n+1}=\alpha+\beta X_n \mid\mathcal{F_n})=X_n \qquad\qquad P(X_{n+1}=\beta X_n \mid\mathcal{F_n})=1-X_n$$ Show $P\left(\lim_{n}X_n=0 \: \text{or}\: 1\right)=1 $ and if $X_0=\theta$ then $P\left(\lim_{n}X_n=1\right)=\theta $.

I know that, first of all I should show that $X_n$ is martingale with respect to $\mathcal{F_n}$, i.e: $$E[X_{n+1}\mid \mathcal{F_n}]=X_n \qquad \forall n \in \mathbb{N}$$ \begin{align}E[X_{n+1}\mid \mathcal{F_n}]&=E[X_{n+1}[I(X_{n+1}=\alpha+\beta X_n)+I(X_{n+1}=\beta X_n)]\:\mid \mathcal{F_n}]\\&=E[(\alpha+\beta X_n)I(X_{n+1}=\alpha+\beta X_n)+\beta X_nI(X_{n+1}=\beta X_n)\:\mid \mathcal{F_n}]\\&=(\alpha+\beta X_n)E[I(X_{n+1}=\alpha+\beta X_n)\:\mid \mathcal{F_n}]+\beta X_nE[I(X_{n+1}=\beta X_n)\:\mid \mathcal{F_n}]\\&=(\alpha+\beta X_n)P[X_{n+1}=\alpha+\beta X_n\:\mid \mathcal{F_n}]+\beta X_nP[X_{n+1}=\beta X_n\:\mid \mathcal{F_n}]\\&=(\alpha+\beta X_n)X_n+\beta X_n(1-X_n)=\alpha X_n+\beta X_n^2+\beta X_n-\beta X_n^2=X_n(\alpha+\beta)\\&=X_n\end{align} So $X_n$ is a martingale. Now I am confused to use which Theorem and why I can use it to reach what the problem requested. I would be thankful if anyone could explain the rest of solution with detail. Thanks in advance.

  • $\begingroup$ The second thing you want to show can be written as $θ=Ε[Χ_0]\overset{!}=P(\lim_n X_n=1)=1\cdot P(\lim_n X_n=1)+0\cdot P(\lim_n X_n=0)=E[\lim_n X_n]$. So, this make me think: Can you perhaps introduce some appropriate stopping time and use the optional stopping theorem? (not sure though) $\endgroup$ – Jimmy R. Dec 18 '15 at 9:43
  • $\begingroup$ Or Doob's convergence theorem $\endgroup$ – Jimmy R. Dec 18 '15 at 9:46

As you already stated, the sequence $(X_n)_n$ is a positive martingale, thus it converges almost surely. Moreover, it is uniformly integrable, thus it also converges in $L_1$. So, denote by $X$ its limit and write $$ B_n = \{ X_n = \alpha + \beta X_{n-1} \}; $$ $$ B = \limsup_n B_n = \{ B_n, \text{occurs } i.o.\} $$

Take $\omega \in B$ and suppose $(X_n(\omega))_n$ converges to $X(\omega)$. Now, extract a subsequence, $(X_{n_k}(\omega))_k$, from $(X_n(\omega))_n$ such that $X_{n_k}(\omega) = \alpha + \beta X_{n_k-1}(\omega)$ pair-wisely. Well, this subsequence must converges to $X(\omega)$, right? So, it is a Cauchy sequence, i.e., there exists $k_0(\omega)$ such that, for all $k_1,k_2 \ge k_0$ we have $$|X_{n_{k_1}}(\omega) - X_{n_{k_2}}(\omega)| \le \varepsilon.$$

But, choosing $k_1$ large enough, $X_{n_{k_1}}(\omega) = \alpha + \beta X_{n_{k_1}-1}(\omega) $, thus

$$ |X_{n_{k_1}}(\omega) - X_{n_{k_1} - 1}(\omega)| = |\alpha + \beta X_{n_{k_1}-1}(\omega) - X_{n_{k_1}-1}(\omega)| \ge \alpha - \alpha X_{n_{k_1}-1}(\omega) $$ since $\alpha + \beta = 1$. Thus, our subsequence should converge to $1$. Similar argument to the occurrence of $B_n^c$ i.o. implies that $X_n \longrightarrow 0 $ if $\omega \in \limsup B_n^c$.

Thus, $X = 1$ in $\liminf_n B_n$ and equals $0$ in $\liminf_n B_n^c$. This proves $X \in \{ 0,1 \}$

The last part is @Stef 's comment combined to convergence in $L_1$.

$$ E[X] = 1\cdot P(X=1) = \lim_nE[X_n] = \theta. $$

Honestly, I would love to see an argument more probabilistic instead of this analytic I just gave. Maybe using Levy's 0-1 Law ? If you have another approach, please, share with me.

Hope this help!


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