I have a random walk on some interval $[0, N]$ with probability $p$ of taking a $+1$ step, probability $(1-p)$ of taking a $-1$ step, and where we have that $p>(1-p)$. Initially the boundaries are reflecting, i.e. if the walker lands at $0$ or $N$, then it will remain in place until it takes a $+1$ or $-1$ step with probability $p$ and $(1-p)$, respectively. Let $\pi$ be the stationary distribution of this walk ( Finding the exact stationary distribution for a biased random walk on a bounded interval).

Now, imagine that we allow the walk to proceed for an arbitrary number of time steps until the walker's probability distribution on the interval approaches the stationary distribution. We then "flip a switch" and specify that the boundary $0$ will be fully adsorbing and that the walker will have a probability of $f(x)$ of being moved to $0$ at any step. More specifically, we flip a coin to decide $p$ and $q$ as we normally would, then flip another coin to decide whether to adsorb with probability $f(x)$. Here, $f(x)$ is some function that depends on the position, $x$, of the walker.

What is our mean time until adsorption? Is there a technical term for what $\pi$ is after we "flip the switch" (some kind of quasi-stationary distribution)?

For example, is it correct to assume that:

(1) Ignoring $f(x)$, that the probability of adsorbing at $(x=0)$ is $\approx \pi(0)$?

(2) That the probability the walker will be moved to $0$ at any given step can be approximated as:

$P[k \to 0]=\sum_{k=1}^{N} k(f(k)\pi(k))$

And thus that we can approximate the adsorption time as: $\mu(ads)=[(1-P[k \to 0])(1-\pi(0))]^{-1}$

I'm pretty sure, however, that the above approximation is wrong.

Update - I would be happy for a solution that only looks at absorbance at $0$ without consideration of $f(x)$. Hopefully there can an analytic solution with this simplification?

  • $\begingroup$ Why does $f(x)$ enter suddenly? Are you changing the rules of the game and allow the walker to directly jump from $x$ to 0 with probability $f(x)$? And are the +1 and -1 processes still in place? $\endgroup$ – Fabian Jan 6 '13 at 6:39
  • $\begingroup$ $Fabian Yes, I'm suddenly changing the rules of the game to allow for any kind of adsorption (the point is just to say that we can assume that the probability distribution for our walker is the stationary distribution). The two kinds of adsorption I'm allowing are: f(x) where we go from some position k = [1, N] to '0', and adsorption at '0' independent of f(x). $\endgroup$ – PlacidLake Jan 6 '13 at 6:46
  • $\begingroup$ @Fabian The step probabilities remain the same before and after the switch is flipped and adsorption is allowed. $\endgroup$ – PlacidLake Jan 6 '13 at 6:52
  • $\begingroup$ But the probabilities should change (at least how I understand your problem): If the walker is at position 2. Then it has the probability $p$ to go to position 3, the probability $1-p$ to go to position 1, an the probability $f(2)$ to go directly to position 0. Unless $f(2)=0$ the probabilities don't add up to 1. $\endgroup$ – Fabian Jan 6 '13 at 7:02
  • $\begingroup$ @Fabian Sorry, you're right. The way I want to handle this is to "flip a coin" to determine whether the walker takes a $+1$ or a $-1$ step according to the previous probabilities, then flip another coin to determine if the walker should be adsorbed with probability $f(x)$. $\endgroup$ – PlacidLake Jan 6 '13 at 7:06

This is not (yet) an answer: I just try to formalise the question asked by the OP

The stationary distribution before the adsorption process sets in can be obtained from the balance equations $p \pi(k) = (1-p)\pi(k+1)$. It is given by $$\pi(k) = \alpha^k \pi(0)= \frac{(1-\alpha) \alpha^k}{1-\alpha^{1+N}} ,\qquad \alpha = \frac{p}{1-p}.$$ This distribution serves as the start for the new random walk. The new random walk is given by the rate equations $$\begin{align}P_{i+1}(0) &= (1-f_1)(1-p) P_i (1) + \sum_{k=1}^N f_k P_i(k)\\ P_{i+1}(1)&= (1-f_2)(1-p) P_i(2)\\ P_{i+1} (2\leq j \leq N-1)&= (1-f_{j+1})(1-p) P_i (j+1) + (1-f_{j-1})p P(j-1) \\ P_{i+1} (N) &= (1-f_{N-1})p P_{i}(N-1) +(1-f_{N})p P_i(N) \end{align}$$ where $P_i(k)$ denotes the probability to be at step $i$ at position $k$. The initial condition is $P_0(k)=\pi(k)$.

You are asking for the mean absorbtion time given by $$\mu=\sum_{i=0}^\infty i P_{i}(0) .$$

We define the transition matrix $M$ via $\mathbf{P}_{i+1} = M \mathbf{P}_i$ where we collect the probabilities to a vector $\mathbf{P} =(P_0, \ldots, P_N)$. We have also the initial vector of probabilities $\boldsymbol{\pi}$. With this notation, we can write $$\mu= \sum_{j=0}^\infty e_0 j M^j \boldsymbol{\pi} =e_0 \frac{M}{(I-M)^2} \boldsymbol{\pi} $$

  • $\begingroup$ The initial stationary distribution was answered in a previous question of mine: math.stackexchange.com/questions/262530/… $\endgroup$ – PlacidLake Jan 6 '13 at 7:35
  • $\begingroup$ Yes, I'm asking for the above mean adsorption time. Sorry, I misread earlier. $\endgroup$ – PlacidLake Jan 6 '13 at 7:42
  • $\begingroup$ $P_i(0)$ is the probability of adsorption at time step $i$ and so the sum is what I consider the mean time of adsorption. If you have a different random process in mind or a different definition of mean time of adsorption could you please try to write it down in your question more explicitly? (my answer is the best shot I could obtain reading your question several times) $\endgroup$ – Fabian Jan 6 '13 at 7:43
  • $\begingroup$ Sorry, I misread you earlier. Your statement for the mean adsorption time is what I have in mind. $\endgroup$ – PlacidLake Jan 6 '13 at 7:44
  • $\begingroup$ @PlacidLake: I am not sure it has an explicit analytical result. But I will think about it later... $\endgroup$ – Fabian Jan 6 '13 at 7:48

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