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The probability to hit $1$ after $1$ step is $\frac{1}{2}$, encoded by the string $+$; the probability to hit $1$ after $3$ steps is $\frac{1}{8}$, encoded by the string $-++$; the probability to hit $1$ after $5$ steps is $\frac{1}{16}$, encoded by the strings $-+-++$ and $--+++$. To compute the probability to hit $1$ after $2k+1$ steps, we just have to ...


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These are not problems on a circle but on the linear interval of integers $$L=\{0,1,2,\ldots,12\},$$ where the hand $12$ is represented by both states $0$ and $12$, and every other hand by the state with its number. Starting from $k$ in $L$, the mean number of steps $t_k$ that the symmetric simple random walk needs to hit $0$ or $12$ is $$t_k=k(12-k),$$ ...


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You are confusing the measure on path space with Lebesgue measure. The "almost everywhere" refers to the former: almost every individual path can be taken to be continuous everywhere. Indeed, the Wikipedia page you link to says that Brownian motion is "almost surely everywhere continuous". In other words, if $\mathbb{P}$ is Wiener measure on a suitable ...


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For the symmetric case where , ℙ(St+1−St=1)=0.5,andℙ(St+1−St=−1)=0.5, the solution is $\frac{T-\frac{[T]}{2}}{2^{T}}$.


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Hint: If the event $A=\{X_{n+1} \in B\}$ is in $\sigma(X_1, \ldots, X_n)$ then $A$ is independent of $A$ hence... (The exercise uses only the independence of the random variables $X_n$, not that they are identically distributed.)


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The subject is huge and called random walks in random environment. The lecture notes by Ofer Zeitouni at Saint Flour summer school in 2001 are a classical introduction but there are other good texts also available on the web. A perhaps counterintuitive result one encounters very soon in the theory is that, in the model you describe, if each site $i$ in ...


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if you are allowed to wait forever: 1. There are lots of ways to do this. Suppose we ask the question hit 1 or -n first and stop when this happens. Then since hitting time is a finite stopping time, the value of the random walk stopped at the hitting is still a martingale and so has expectation 0. So probability of hitting $1$ is $n$ times that of ...


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Without loss of generality, suppose that the first step is to $1$ (rather than $-1$). Now you have a fair random walk with a walker at position $1$, and you're interested in the probability that it hits $N = 2^n$ before it hits $0$. You must have learnt a formula for this in class.


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Think of it like this; On your first step you go onto 1 (say). Now the probability you hit 2 before 0 is a half (you either go left or right on your next step). Now we want to know whether we hit 4 next or 0, and we are in the middle of both, so again equal likelihood and the probability is 1/2. So both steps is a half squared and we get a quarter. This ...


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This is just an answer to the first question on when the random walk converges to the uniform distribution. See Theorem 2.1 and Proposition 2.3 in this article by Laurent Saloff-Coste: Consider a left-invariant random walk on a finite group $G$, and let by $\Sigma \subset G$ the set of steps (right multiplications) which have non-zero probability. (This ...


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Consider some nonnegative integers $i$ and $j$. To reach $i+j$ starting from $0$, one must first reach $i$ starting from $0$ then reach $i+j$ starting from $i$. The probability of the first event is $d_i$. By the Markov property and the invariance of the dynamics by the translations of $\mathbb Z$, the property of the second event conditionally on the ...



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