A 1D random walk with absorbing barriers at 0 and N. Can we determine the mean time before an absorption event takes place at either barriers given that the probability to move left/right is $0.5$ everywhere except at barriers, where then the probability to be absorbed in next jump is $p$ and the probability to jump away from the barrier is $q?$ where $p>q$.

In the simpler case where p and q are also equal to 0.5, the mean time is known, $t_a=a(N-a)$ where $a$ is the starting site of the walker and 0 and N are the positions of the two barriers.

  • $\begingroup$ There is a standard renewal theory technique way to get this mean: given the transition probability matrix $P$, the mean time $u(x)$ to hit any set $A$ starting from $x$ satisfies $((P-I)u)(x)=-1$ for $x \not \in A$ and $u(x)=0$ for $x \in A$. I haven't done the calculation to see whether this system of linear equations can be solved in closed form in your example. This system comes from the total expectation formula, conditioning on all possible values of the process at the second step, and the Markov property. $\endgroup$ – Ian Mar 20 '17 at 17:01
  • $\begingroup$ Do you mean $t_a = a(N-a)$ when there is no reflection at the boundaries, i.e., $p=1, q=0$? $\endgroup$ – ChargeShivers Mar 20 '17 at 17:28
  • $\begingroup$ @user929304 In that case, I don't think $t_a = a(N-a)$. Can you explain how you got it? $\endgroup$ – ChargeShivers Mar 20 '17 at 18:09
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    $\begingroup$ The usual Markov decomposition after one step yields $t_1=1+qt_2$, $t_{N-1}=1+qt_{N-2}$, and, for every $2\leqslant k\leqslant N-2$, $t_k=1+\frac12t_{k-1}+\frac12t_{k+1}$. These conditions for $2\leqslant k\leqslant N-2$ imply that $t_k=uk+v-k^2$ for some $(u,v)$ independent of $k$. The conditions on $t_1$ and $t_{N-1}$ imply that $u+v-1=1+q(2u+v-4)$ and $u(N-1)+v-(N-1)^2=1+q(u(N-2)+v-(N-2)^2)$. The solution of this system is $u=N$ and $v=((q/p)-1)(N-2)$, hence $t_a=a(N-a)+((q/p)-1)(N-2)$ for every $1\leqslant a\leqslant N-1$. $\endgroup$ – Did Mar 20 '17 at 20:29
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    $\begingroup$ "could I write up here an answer based on your comment" Absolutely, excellent idea. $\endgroup$ – Did Mar 21 '17 at 17:55

Many interesting points have already been pointed out in the comments, amongst others by user Did, which in essence already contain an answer to the post, so I decided to write them up here as an answer for future readers.

Given that the absorbing barriers are placed at $0$ and $N,$ we have the following important transition probabilities: $N-1\to N-2$ happens with probability $q$ and $N-1\to N$ with probability $p$ ($q+p=1$), similarly, the transition $1\to 2$ happens with probability $q$ and $1\to 0$ with probability $p.$ All remaining transitions happen with probability $1/2.$ Thus we are only modifying the behaviour at the barriers, and asking how favouring absorption influences the mean time before an absorption event takes place.

Now we perform a Markov decomposition after one step of the walk, this yields the mean time before absorption $t_1 = 1+qt_2$ starting the walk from $1,$ $t_{N-1}=1+qt_{N-2}$ starting from $N-1$ and for every other starting point $2\leqslant k \leqslant N-2 ,$ the corresponding mean time is $t_k = 1+\frac{1}{2}(t_{k-1}+t_{k+1}).$ The latter condition implies that $t_k$ ought to take the form of $$t_k=uk+v-k^2\tag{1}$$ for some couple $(u,v)$ independent of $k.$ Moreover, the former conditions on $t_1$ and $t_{N-1}$ imply the following relations: $$ \begin{align} u+v-1 =& 1+q(2u+v-4) \tag{2} \\ u(N-1)+v-(N-1)^2 =& 1+q(u(N-2)+v-(N-2)^2). \tag{3} \end{align} $$ The equations $(1), (2)$ and $(3)$ together form a system of equation that can be solved, yielding: $$u=N\qquad v=((q/p)-1)(N-2)$$ hence the modified mean time before absorption is given by:

$$ t_a = a(N-a) + ((q/p)-1)(N-2) $$ for every starting point $a$ such that $1\leqslant a\leqslant N-1.$


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