world:random-walk
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world:random-walk [2020/04/22 18:01] rdouc [Comments] |
world:random-walk [2023/04/18 14:58] (current) rdouc [Statement] |
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| <WRAP center round box 80%> | <WRAP center round box 80%> | ||
| - | Let $(U_i)$ be iid Rademacher random variables, i.e. $U_i=1$ or $-1$ with probability $1/2$ and set $S_i=\sum_{j=1}^iU_j$ the associated partial sum. Define $\Delta=\inf\set{t>0}{S_t=0}$. Show that $S_n$ returns to 0 with probability one. What is the law of $\Delta$? | + | Let $(U_i)$ be iid Rademacher random variables, i.e. $U_i=1$ or $-1$ with probability $1/2$ and set $S_i=\sum_{j=1}^iU_j$ the associated partial sum. Define $\Delta=\inf\set{t>0}{S_t=0}$. Show that $S_n$ returns to $0$ with probability one. What is the law of $\Delta$? |
| </WRAP> | </WRAP> | ||
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| ====== Other method ====== | ====== Other method ====== | ||
| - | Note that, setting $U_i$ such that $X_i=2U_i-1$, we have $\PP(S_{2n}=0)=\PP(\sum_{i=1}^{2n} U_i=n)=\frac{(2n)!}{(n!)^2} \lr{\frac14}^n \sim \frac{1}{\sqrt{n\pi}}$. THis implies that | + | Note that, setting $U_i$ such that $X_i=2U_i-1$, we have that $(U_i)$ are iid Bernoulli random variables with success probability $1/2$. Then, $\PP(S_{2n}=0)=\PP(\sum_{i=1}^{2n} U_i=n)=\frac{(2n)!}{(n!)^2} \lr{\frac14}^{2n} \sim \frac{1}{\sqrt{n\pi}}$ where we have used the Stirling equivalence. This implies that |
| + | \begin{equation} | ||
| + | \label{eq:one} | ||
| + | \infty=\sum_{n=1}^\infty \PP(S_{2n}=0)=\PE[\sum_{n=1}^\infty \indi{0}(S_{2n})]=\PE[\sum_{k=1}^\infty \indi{T^k<\infty}]=\sum_{k=1}^\infty \PP(T^k<\infty) | ||
| + | \end{equation} | ||
| + | where $T^k$ is the time index of the $k$-th visit of $(S_n)$ to $0$. By convention, we set $T^1=T$. We now show by induction that for all $k\geq 1$, | ||
| + | \begin{equation} \label{eq:induc} | ||
| + | \PP(T^k<\infty)=\PP(T<\infty)^k\eqsp. | ||
| + | \end{equation} | ||
| + | The case $k=1$ obviously holds. Now, assume that \eqref{eq:induc} holds for some $k\geq 1$. Then, | ||
| + | \begin{align*} | ||
| + | \PP(T^{k+1}<\infty)&=\PP(T^{k}<\infty,T^{k+1}<\infty)=\sum_{m=1}^\infty \sum_{n=1}^\infty\PP(T^{k}=m,T^{k+1}=m+n)\\ | ||
| + | &=\sum_{m=1}^\infty \sum_{n=1}^\infty\PP(T^{k}=m,\forall t\in[1:n-1],\ \sum_{\ell=1}^{t}X_{m+\ell}\neq 0,\sum_{\ell=1}^{n}X_{m+\ell}=0)\\ | ||
| + | &=\sum_{m=1}^\infty \sum_{n=1}^\infty\PP(T^{k}=m)\underbrace{\PP(\forall t\in[1:n-1],\ \sum_{\ell=1}^{t}X_{m+\ell}\neq 0,\sum_{\ell=1}^{n}X_{m+\ell}=0)}_{\PP(T=n)}\\ | ||
| + | &=\PP(T^k<\infty) \PP(T<\infty) | ||
| + | \end{align*} | ||
| + | and by the induction assumption, we get $\PP(T^{k+1}<\infty)=\PP(T<\infty)^{k+1}$. | ||
| + | Plugging \eqref{eq:induc} into \eqref{eq:one} yields | ||
| $$ | $$ | ||
| - | \infty=\sum_{n=1}^\infty \PP(S_{2n}=0)=\PE[\sum_{n=1}^\infty \indi{0}(S_{2n})] | + | \infty=\sum_{k=1}^\infty \PP(T<\infty)^k |
| $$ | $$ | ||
| + | Since $\PP(T<\infty) \in[0,1]$, this implies that $\PP(T<\infty)=1$. | ||
world/random-walk.1587571262.txt.gz · Last modified: 2022/03/16 01:37 (external edit)
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