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I am modeling a biological process using the Moran model which effectively describes a 1D random walk.

Starting at $i = 1$, the transition probabilities are:

$p_{i, i+1} = \frac{ri}{ri+N} \frac{N-i}{N}$,

$p_{i, i-1} = \frac{N-i}{ri+N-i} \frac{i}{N}$,

$p_{i, i} = 1 - p_{i, i+1} - p_{i, i-1}$,

where if $r > 1$ the random walk is biased towards larger values and the boundaries are $i = 0$ and $i = N$.

Now, given that the random walk doesn't get absorbed at $i = 0$, I'm interested in the expected number of steps until it is absorbed at $i=N$.

Unfortunately, I can't seem to find a derivation of this in the literature and can't derive it myself. Thanks for your help.

Dahlai
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  • You don't give transition probabilities for state $0$. Does that mean if you're in state $0$ you automatically go to state $1$ on the next step? Or is there some (positive) probability of staying in state $0$ on each step? – paw88789 Jun 24 '22 at 12:38
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    Assume you start from 1 . This's a time-homogeneous Markov Chain . You have $N$ states in total : $1,..,N$ . Your only absorbing state is $N$ . Build a usual $N\times N$ transition matrix , then remove the final row & column to obtain $(N-1)\times (N-1)$ matrix $Q$ . Your solution is vector $k=[k_i] , i =1,..,N-1$ where $k_i$ is the expected absorption time starting from $i$ . This can be solved from $(I-Q)k = 1_{N-1}$ where $I$ is identity matrix and $1_{N-1}$ is the all-1 vector . If possible you may get $k=(I-Q)^{-1}1_{N-1}$ . – C.C. Jun 24 '22 at 13:43
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    For more info : here – C.C. Jun 24 '22 at 13:43
  • @C.C thanks for pointing out the corresponding wiki. I just googled the wrong terminology. – Dahlai Jun 27 '22 at 13:59

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