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I will follow the outline that I described in my first comment. Let's start with random walk. Fix $\alpha \in \mathbf{R}$ and consider the process $(M_j)$ defined as $$M_j = (\cosh{\alpha})^{-j}\cosh{(\alpha S_j)}$$ We want to show that $(M_j)$ is a martingale with respect to the natural filtration $(\mathcal{F}_j)$ generated by the constituent random ...
Let's study the expected time it takes for the symmetric random walk to reach $+1$, assuming it begins at $0$. Let $S$ denote the number of steps we take until we reach $+1$ that is $S_1 = \min\{n > 0; X_n = +1\}$ where $X_n$ is your symmetric random walk. Note that $S_1$ is an odd number. $$P(S_1 = 1) = \frac{1}{2}\\ P(S_1 = 3) = \frac{1}{2^3} \\ P(S_1 ... 2 Here's the proof for a simple random walk, which can be generalized further. Hopefully it's clear that T_{N(k)} as a function of k changes exactly one coordinate for each k and moreover |T_{N(k+1)}-T_{N(k)}|_\infty=1. More importantly, T_{N(k+1)}| is a Markov process. So it suffices to verify that each step is uniformly random: the probability of a ... 1 This is covered in Feller's book, Introduction to Probability Theory and its Applications, Section III.5, Changes of Sign, Theorem 1. He assumes the path length, L say, is odd: if L is even, the probability is the same as that for L-1 because there can be no change of sign on an even-numbered step. So, for path length 2n+1, the probability of ... 1 Not a complete answer, but this might give you some ideas about that probability: If the jumps arrive with exponential waiting time, your process can be expressed as a compound Poisson process:$$ X_t = \sum^{N_t}_{i=1}J_i $$where N_t is a Poisson process with intensity \lambda=1 and J_i are iid random variables which are 1 with probability 2/3 ... 1 Let P(i) denote the probability he'll get to \1,000,002 if he starts from \i. We know that$$P(0)=0\;\;\;P(1,000,002)=1$$You are asked to find P(1,000,000). Imagine he has \i and places a bet. He either gets to state (i+1) or to state (i-1) and we have:$$P(i) = \frac 13 P(i+1)+\frac 23 P(i-1)$$It is easier to start near the "ruin ... 1 Then after the mosquito goes to 0, the mosquito must immediately go back to 1 next move, i.e. the whole situation restarts. Consider the case that when the mosquito goes from 1 to 4 without passing 0 as a success, and the case when the mosquito goes from 1 to 0 without passing 4 as a failure. You already calculated the two probabilities as ... 1 For any random walk on \mathbb{R} there are only four possibilities. Exactly one of the following happens with probability one. S_n = 0 for all n S_n \to \infty S_n \to -\infty -\infty = \liminf S_n < \limsup S_n = \infty This is because \limsup S_n is an exchangeable random variable, meaning reordering finitely many of the X_i doesn't ... 1 I may be missing something on the conditions: what about$$A=\begin{pmatrix} 0 & 1 & 0\\ 1& 0 & 0\\ 0&0&0\end{pmatrix}?$$You have A^3 = A, so its powers cannot converge to 0. 1 Let n_i,n_{-i},i=1,\dots,d denote a move of a (simple symmetric) RW in i-th and the opposite direction, respectively. Then$$P\{X_{2n}=0\}=P\{n_1=n_{-1},\dots,n_d=n_{-d}\}=\sum_{n_1+n_{-1}\cdots+n_d+n_{-d}=2n}1\{n_1=n_{-1},\dots,n_d=n_{-d}\}\binom{2n}{n_1,n_{-1}\dots,n_d,n_{-d}}\left(\frac{1}{d}\right)^{2n}=(2d)^{-2n} \binom{2n}{n} ...