# Random walk with reset?

Is there such a random walk, that "good times" it just looks like a random walk, while when "bad moment" comes, it will reset => jump to zero, afterwards continue doing random walk again? Thanks!

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A weighted random walk? projecteuclid.org/… – user996522 Mar 10 '12 at 12:47
I guess... if you can define "good times" and "bad moments", you can quite possibly tell a computer how to recognize when it happens and tell it to reset. – Arthur Mar 10 '12 at 12:55
Would a (possible infinite) Markov chain do what you want? – Johannes Kloos Mar 10 '12 at 15:28

Sure, for example let us consider the case $\mathbb Z$. The standard random walk $X$ on $\mathbb Z$ is a stochastic process with integer values $0,\pm1, \pm2,\dots$ such that $$\mathsf P\{X_{k+1} = i+1|X_{k} = i\} = \mathsf P\{X_{k+1} = i-1|X_{k} = i\} = 1/2.$$ There are several methods to modify it in order to have jumps at "bad times". All those I describe here are particular examples of Markov Chains mentioned by Johannes.

1. Time-dependent case: let $\{\tau_n\}_{n\geq0}$ be a countable, strictly increasing sequence of random times $$\tau_0<\tau_1<\dots$$ such that $\tau_n\in\mathbb N$. Define the distribution of $X_{k+1}$ given $X_k$ as above if $k\notin\{\tau_n\}_{n\geq 0}$ and $X_{k+1} = 0$ if $k\in \{\tau_n\}_{n\geq 0}$. A simple example is given by purely deterministic sequence $\{\tau_n\}_{n\geq 0}$ which can be e.g. periodic: $\tau_n = n\cdot\tau_1$.

2. Time-independent (state-dependent case): in the previous case the sequence of times for a process to jump to zero might be independent from the process itself $X$, while bringing heterogeneity into distributions: the distribution of the next state $X_{k+1}$ was depending both on the current state $X_k$ and the time $k$.

On the other hand, such problem can be avoided if we consider the process $Y_{k}$ given by $$\mathsf P\{Y_{k+1} = i+1|Y_{k} = i\}=\mathsf P\{Y_{k+1} = i-1|Y_{k} = i\} = 1/2 - \delta_i$$ and $$\mathsf P\{Y_{k+1} = 0|Y_k = i\} = 2\delta_i$$ where $\delta_i\in[0,1/2]$ for all $i\in \mathbb Z$.

In that case, we can denote $\sigma_0 = \inf\{k\geq 0: Y_k = 0\}$ and recursively $$\sigma_{n+1} = \inf\{k>\sigma_n:Y_k = 0\}.$$ So $\sigma_n$ is the $n$-th time when $Y$ jumps to zero. In all other good times the process "looks just like a random walk" as you requested.

I hope that this answer gives some insight into methods which can be used to solve your problem.

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Let $(\xi_k)_{k\geqslant1}$ denote an i.i.d. sequence of $\pm1$ random variables and $\mathfrak T$ a random subset of $\mathbb N$, independent on $(\xi_k)_{k\geqslant1}$. In other words, $\mathfrak T=\{\tau_i\mid i\geqslant1\}$, where $(\tau_i)_{i\geqslant1}$ is an increasing sequence of positive integer random variables, independent on $(\xi_k)_{k\geqslant1}$. Define $$X_n=\sum\limits_{k=1}^n\xi_k\cdot[\mathfrak T\cap[k,n]=\varnothing].$$ Then $(X_n)_{n\geqslant1}$ behaves as a random walk starting again from position $0$ at each time $n$ in $\mathfrak T$.

An equivalent formulation is to consider the last reset time before $n$, that is, $$\varrho_n=\max\{0\}\cup(\mathfrak T\cap[1,n]),$$ and to define $$X_n=\sum\limits_{k=1+\varrho_n}^n\xi_k.$$

The simplest case might be when the random variables $\tau_i-\tau_{i-1}$ are i.i.d. and geometric with parameter $p$ (with $\tau_0=0$), hence $\mathrm P(\tau_i-\tau_{i-1}=k)=p(1-p)^{k-1}$ for every $k\geqslant1$ and every $i\geqslant1$. Then the events $[k\in\mathfrak T]$ are i.i.d. with probability $p$, for $k\geqslant 1$, hence $(X_n)_{n\geqslant1}$ is a random walk which jumps from $i$ to $i+1$ with probability $\frac12(1-p)$, from $i$ to $i-1$ with probability $\frac12(1-p)$, and from $i$ to $0$ with probability $p$.

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