world:robbins_monro
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world:robbins_monro [2021/03/27 01:48] rdouc [Statement] |
world:robbins_monro [2023/02/10 18:17] (current) rdouc [Proof of $\mathbb{P}(A=0)=1$.] |
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| <WRAP center round box 80%> | <WRAP center round box 80%> | ||
| * $f$ is a bounded and continuous function | * $f$ is a bounded and continuous function | ||
| - | * $f(x)(x-x_*)>0$ for all $x\neq \in \Xset$ | + | * $f(x)(x-x_*)>0$ for all $x\neq x_* \in \Xset$ |
| </WRAP> | </WRAP> | ||
| We assume there exist a Markov kernel $K$ on $\Xset \times \Xsigma$ and a sequence $\seq{\gamma_k}{k\in\nset}$ of real numbers such that | We assume there exist a Markov kernel $K$ on $\Xset \times \Xsigma$ and a sequence $\seq{\gamma_k}{k\in\nset}$ of real numbers such that | ||
| Line 59: | Line 59: | ||
| - $\lrcb{\delta/2 < |A| <\delta} \subset \lrcb{W_\infty\eqdef \lim_n W_n=\infty}$ | - $\lrcb{\delta/2 < |A| <\delta} \subset \lrcb{W_\infty\eqdef \lim_n W_n=\infty}$ | ||
| - $\PP(W_\infty=\infty)=0$. | - $\PP(W_\infty=\infty)=0$. | ||
| - | 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.1616806126.txt.gz · Last modified: 2022/03/16 01:37 (external edit)
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