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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 $f$,  ​
world/robbins_monro.1616806159.txt.gz · Last modified: 2022/03/16 01:37 (external edit)