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Consider a bounded one dimensional random walk, with end points as 0 and N. If we start from x, where x is an interior point of {0,1,2,...,N}, what is the expected number of steps(m(x)) required to reach 0 or N. The probability of going to the left or right side from any interior point is equal(=1/2). Is it possible to prove that m(x) is finite and find a recursive equation(involving m(x-1) and m(x+1))?

Note: Actually, this question is a follow up to the harmonic nature of the probability (p(x)) of starting from an interior point and reaching the bounds, where I proved $p(x)=\frac{p(x-1)+p(x+1)}{2}$.

Ajay Shanmuga Sakthivasan
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    The recursive relation for $m$ is derived in an almost identical way to that of $p$. In particular, $m(x) = 1 + (m(x-1) + m(x+1))/2$. The solution is $m(x) = x(N-x)$. – user217285 Jul 15 '18 at 19:04
  • $m(x)$ is almost surely finite, i.e. with probability $1$ – Henry Jul 15 '18 at 19:28
  • Do all the questions and answers on the site about random walks really don't answer this question? For instance, you could improve your question by explaining what's causing you trouble in applying the answer to this question to your problem. – joriki Jul 15 '18 at 19:45
  • @joriki The answer to the question you have mentioned doesn't show how the recursive equation is derived. I'd like to know how the recursive equation is derived. (And I have mentioned in my question that I need to know how the recursive equation is derived). – Ajay Shanmuga Sakthivasan Jul 16 '18 at 05:51
  • @Nitin Is it possible for you to derive the recursive equation? Thanks. – Ajay Shanmuga Sakthivasan Jul 16 '18 at 05:52
  • @AjayShanmugaSakthivasan: No, you didn't mention in the question that you needed to know how it's derived. You asked whether it's possible to find one. That's answered by the answer I linked to. If you're going to ask about something that's been discussed a gazillion times on this site and all that knowledge is at your fingertips, you should be precise about which specific aspect you still find lacking in the existing answers. – joriki Jul 16 '18 at 06:10
  • no. how did you derive the recursion for $p(x)$? – user217285 Jul 16 '18 at 06:11
  • @Nitin So I assumed that I start from some interior point x. I then used the law of total probability, which says that the probability of reaching the boundary equals the product of probability of going to the left and probability of reaching the boundary given that he starts from the left point plus the product of probability of going to the right and probability of reaching the boundary given that he starts from the right point. (The probability of going to left or right = 1/2). This directly translates to the equation I derived – Ajay Shanmuga Sakthivasan Jul 16 '18 at 06:21
  • ok so do the same thing with $m$. – user217285 Jul 16 '18 at 17:21
  • @Nitin Okay, even if use law of total expectation here, where does the 1 come from? – Ajay Shanmuga Sakthivasan Jul 17 '18 at 05:07
  • Well, what does law of total expectation give you? – user217285 Jul 17 '18 at 05:11
  • @Nitin $m(x)=\frac{m(x-1)+m(x+1)}{2}$ – Ajay Shanmuga Sakthivasan Jul 17 '18 at 06:35
  • Just consider a random walk on ${0,1,2}$ to see where your expression fails. Completely spell out the law of total expectations, being careful with your notation. – user217285 Jul 17 '18 at 07:07
  • Anyhow, let's not keep this going on. If you're standing at $x$, then your expected exit time will be $(1+m(x-1))$ if you hit $x-1$ first and $(1+m(x+1))$ if you hit $x+1$ first. – user217285 Jul 17 '18 at 07:09
  • @Nitin Alright, got it. Thank you for being patient. – Ajay Shanmuga Sakthivasan Jul 17 '18 at 07:31

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