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9

Update: I originally got the probability is $1-2/\pi$, but there seemed to be mistake taking the wrong order of limits. Taking the right order I got $2-4/\pi$. Denote $f_n(\mathbf{x})$ the number of ways to end at $\mathbf{x} = (x,y)$ after $n$ steps. Denote $g_n(\mathbf{x})$ the number of ways to end at $\mathbf{x}$ after $n$ steps for the first time. ...

4

An excursion away from 0 starts with step to $1$ or a step to $-1$. By the evident symmetry, it's enough to consider the first case. For the r.w. started at $1$, the chance to hit $n$ before $0$ is well known to be $1/n$. Thus, $$P[M\ge n] = 1/n$$ and indeed $E[M]=+\infty$.

3

Here are some explanations: Note that $$\mathbb{P}\{N>n(b-a)\}\leq \mathbb{P}\{S_{n(b-a)}\in (a,b), S_{(n-1)(b-a)}\in (a,b),\ldots,S_{(b-a)}\in (a,b)\}.$$ Now $$\mathbb{P}\{S_{n(b-a)}\in (a,b), S_{(n-1)(b-a)}\in (a,b),\ldots,S_{(b-a)}\in (a,b)\} =\mathbb{P}\{S_{n(b-a)}\in (a,b)\,| \,S_{(n-1)(b-a)}\in (a,b),\ldots,S_{(b-a)}\in (a,b)\} \mathbb{P}\{S_{(n-... 3 Assuming the \epsilon_t are iid, you can simply use$$Var(X_t)=Var(\sum \epsilon_i)=\sum Var(\epsilon_i)$$1 You're close. Let E_j be the expected number of steps to get to state j from state 1. Then$$E_j=E_{j-1}+\frac12(1)+\frac12(1+E_j)\implies E_j=2E_{j-1}+2 In order to get to state $j$, we must first get to state $j-1$. There's no reason to multiply by $\frac12$ here. Then half the time, we finish in one more transition, and half the time we begin ...

1

Without loss of generality, let $i=0$. Note that you are dealing with a first passage time (i.e., the first time your random walk hits $0$), and therefore, it is not sufficient to just consider the current state at time $2n$. In the second equation (the one not marked), you are calculating the probability of the following: in the first $2n$ steps, what is ...

1

Here is a statement that may not be completely rigorous: We know that the probability of a simple random walk revisiting 0 is 1. This means in the whole random sequence, there has to be a 0 except the first one. Then, starting at that point, it will be a new simple random walk. So it would return 0 once again. Continuing this process, we conclude that a ...

1

I read the following intuitive statement somewhere: After N steps in any dimension, you are at average distance $sqrt(N)$ from the origin. In 3 dimensions therefore you can reach $N^{3/2}$ points, which is growing faster than N, so you cannot on average visit all the points. In 2d however this calculation gives that you can visit N points, which is (just) ...

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