I have a system where a random walker is moving on $\mathbb{Z}$. However, at each point in $\mathbb{Z}$, there is a probability $q$ that an escape route exists along which the walker can escape. I want to find the expected time that the walker stays moving along the line before it escapes. At first I thought this would follow a $\operatorname{Geom}(p)$ distribution, where p is: \begin{align*} p &= \mathbb{P}(\text{leaving } \mathbb{Z})\\ &= \mathbb{P}(\text{an escape exists})\mathbb{P}( \text{taking escape} | \text{an escape exists})\\ &= \frac{1}{3}q. \end{align*}

With the factor of $1/3$ coming from the fact that with an escape route, a walker at $k$ could escape, move to $k+1$, or move to $k-1$, each with equal probability. However when I compared this to computer simulations it doesn't work very well. I think the issue is that the 'escape routes' are only determined once at the beginning (i.e. 5 has an escape route with probability $q$ - determined initially before the walker starts walking - but the above treats it as though the existence of a shortcut is determined anew each time you visit 5).

Could someone help me figure out what this distribution should be or if I can overcome this problem in the work I've done so far?

  • $\begingroup$ It is not clear to me how you get $q/3$ in the first place. $\endgroup$ – Florian Ingels Apr 16 at 13:48
  • $\begingroup$ Yes sorry, $q$ is the probability that the escape exists and if an escape exists at $k$ then the walker can either go to $k-1,k+1$ or escape, all with equal probability. That's what gives the $1/3$ term I think. I'll edit now $\endgroup$ – Fahrenheit997 Apr 16 at 13:51
  • $\begingroup$ Ok. Then, surely, your problem rely on the determination once for all of the possibles escapes points. Maybe you should try consider a simpler model in the first time, with the escape points already fixed, and see how the walker behave. $\endgroup$ – Florian Ingels Apr 16 at 14:00
  • $\begingroup$ If the walker visits an integer for the first time, the probability of escape during the next step is $q/3$. But if the walker visits an integer for the second time, the probability of escape is less than $q/3$ because the fact that the walker visited this integer before but didn't escape means that this integer is less likely to have an exit. If this is the 100th time the walker visits a particular integer, that integer almost certainly doesn't have an exit so the probability of escape is almost zero. $\endgroup$ – PhiNotPi Apr 17 at 1:39

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