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I did not have a rigorous proof here, but the following analysis should provide some reasoning to your quesiton. If we denote $\delta_i=x_i-x_{i-1}$, we can define $$\Delta_M^{(N)}\equiv \frac{1}{N-M}\sum_{i=1}^{N-M}i\delta_{N-i+1}=\frac{1}{N-M}\sum_{i=M+1}^N x_i -x_M$$ Because $\delta_i \sim (i.i.d.) (0,\sigma^2)$, $$... 1 Let a_n be the probability that the particle eventually gets absorbed when starting from n. That is, a_0=1 and for n\gt0,$$ a_n=pa_{n+2}+(1-p)a_{n-1}\tag{1} $$Let's try to solve the recurrence$$ a_n=\frac1pa_{n-2}-\frac{1-p}{p}a_{n-3}\tag{2} $$To do this, we need to find the roots of the associated polynomial$$ ...

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To begin with, consider the difference between distance and displacement. The displacement is a vector, the distance is a nonnegative number. The expected displacement of a symmetric random walk is always $0$, and thus is not an interesting quantity to look at. The expected distance is typically positive, and grows with time; we may be interested in how ...

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If we make the boundaries at $0$ and $N$ reflecting, rather than absorbing, then your random walk is the Ehrenfest urn model. It is well-known that this Markov chain has invariant distribution $\pi_i={N\choose i}/2^N$ for $0\leq i\leq N$. In particular, the expected time to return to the origin, starting there, is $\mathbb{E}_0(T_0)=1/\pi_0=2^N$. This ...

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