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world:robbins_monro [2021/03/27 01:49] rdouc [Statement] |
world:robbins_monro [2023/02/10 18:17] (current) rdouc [Proof of $\mathbb{P}(A=0)=1$.] |
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- $\lrcb{\delta/2 < |A| <\delta} \subset \lrcb{W_\infty\eqdef \lim_n W_n=\infty}$ | - $\lrcb{\delta/2 < |A| <\delta} \subset \lrcb{W_\infty\eqdef \lim_n W_n=\infty}$ | ||
- $\PP(W_\infty=\infty)=0$. | - $\PP(W_\infty=\infty)=0$. | ||
- | To get the second property (2), note that $M_n\leq S_n$ so that $\sup_{n\in\nset} \PE[M_n^+]\leq \sup_{n\in\nset} \PE[S_n^+]<\infty$. Finally, $(M_n)$ is a martingale, with a positive part which is uniformly bounded in $L^1$. Therefore, $(M_n)$ converges $\PP$-a.s. And since $W_n=S_n-M_n$, it also implies that $(W_n)$ converges a.s. so that $\PP(W_\infty=\infty)=0$. | + | To get the second property (2), note that $M_n\leq S_n$ so that $\sup_{n\in\nset} \PE[M_n^+]\leq \sup_{n\in\nset} \PE[S_n^+]<\infty$. Finally, $(M_n)$ is a martingale, with a positive part which is uniformly bounded in $L^1$. Therefore (see for example [[https://wiki.randaldouc.xyz/doku.php?id=world:martingale| some convergence properties for martingales]], $(M_n)$ converges $\PP$-a.s. And since $W_n=S_n-M_n$, it also implies that $(W_n)$ converges a.s. so that $\PP(W_\infty=\infty)=0$. |
We now turn to the first property (1). Let $\omega \in \lrcb{\delta/2 < |A| <\delta}$. Then, there exists $N(\omega)$ such that for all $n\geq N(\omega)$, $X_n\in B$ where $B=\set{x\in\Xset}{\delta/2 <(x-x_*)^2<\delta}$. Moreover, by continuity of $f$, | We now turn to the first property (1). Let $\omega \in \lrcb{\delta/2 < |A| <\delta}$. Then, there exists $N(\omega)$ such that for all $n\geq N(\omega)$, $X_n\in B$ where $B=\set{x\in\Xset}{\delta/2 <(x-x_*)^2<\delta}$. Moreover, by continuity of $f$, |