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The most commonly considered random walk on the line takes one step left or right with equal probability until a barrier is reached (if there are any barriers).

More generally, suppose we fix any probability distribution on the integers with compact support, and consider the random walk that begins at 0 and, at each step, takes an independent random draw from this probability distribution, and moves by that amount. (The random walk in the previous paragraph corresponds to the probability distribution $P(1)=1/2$ and $P(-1)=1/2$.) Suppose also that there is a positive integer $n$ such that there are absorbing barriers at $+m$ and at $-m$ whenever $m\ge n$. Obviously, if the probability distribution is symmetric about 0 then the random walk is equally likely to terminate at a "positive" barrier as at a "negative" barrier.

Does there exist an asymmetric probability distribution such that, for all sufficiently large $n$, the associated random walk is equally likely to terminate at a positive barrier as at a negative barrier?

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  • $\begingroup$ It looks like the $n$ in the set-up is not the same as the $n$ ($N$?) in the question, right? i.e. the range of the displacements are in $[-n,n]$, but the barriers are at $\pm N$ $\endgroup$ Jan 29, 2015 at 15:33
  • $\begingroup$ Maybe it would have been clearer to use different symbols, but yes, I'm asking about a fixed probability distribution that has compact support, and letting the barriers go to infinity. $\endgroup$ Jan 29, 2015 at 16:41
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    $\begingroup$ Could you edit the question so the $n$-s are replaced with the $N$-s in the right places? $\endgroup$ Jan 29, 2015 at 16:48
  • $\begingroup$ O.K., I edited the description. $\endgroup$ Jan 29, 2015 at 17:59
  • $\begingroup$ Would $P(-2)=1/3$, $P(1)=2/3$ be an asymmetric probability distribution? Or by asymmetry do you mean a distribution where the expected amount of movement on the first step is (say) to the right? $\endgroup$ Jan 29, 2015 at 18:20

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