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2

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 ...


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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 ...


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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 ...


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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 ...


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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 $2/3$ ...


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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 ...


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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 ...


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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 ...


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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 $0$.


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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} ...



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