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3

Dan Uznanski has already sketched the solution in a comment; I'll work it out in more detail. Regard one cell as stationary and let the other cell make two steps at a time. Turn the grid by $\frac\pi4$. Then the walk can be decomposed into two independent simple symmetric one-dimensional random walks. We want the probability that these walks end up at ...


2

My answer here gives an explicit formula for $\mathbb{E}\left|\sum_{i=1}^{2n} \epsilon_i\right|$. You can get the lower bound $$\mathbb{E}\left|\sum_{i=1}^{2n} \epsilon_i\right|\geq \sqrt{n}, \tag1$$ if you combine (1) with the bound at the link there.


2

We will make four cases based on the first two steps in our path: RR: For the next six steps, we are allowed to take only one U step, if we take it at all. If we take it, then there are $\binom{6}{1}$ ways for the six steps and $\binom{8}{1}$ ways for the next eight steps. If we don't take it, there is only one way to choose the six steps (all Rs) and then ...


2

For this particular question, it i s probably easiest just to write a table of the number of ways of reaching each square moving up or right from the bottom left. Each value is the sum of the value below it and the value to its left. 1 9 17 25 33 41 49 65 130 1 8 8 8 8 8 8 16 65 1 7 8 49 1 6 ...


2

For $k\in\left\{ 0,1,\dots,8\right\} $ let $p_{k}$ denote the probability that $A$ will win the whole thing if $A$ starts with $k$ dollars. Then $p_{0}=0$, $p_{8}=1$ and $p_{k}=\frac{2}{3}p_{k+1}+\frac{1}{3}p_{k-1}$ for $1\leq k\leq7$. Do you understand why? Now find a solution for these equations. You are actually looking for $p_3$.


2

Without an additional regularisation (say with colored noise), the integral you are trying to compute is strongly divergent. The integrand is dominated by $\xi^2=\left(\frac{dW}{dt}\right)^2$ which is positive and of order $\frac{1}{dt}$ and there is thus no hope to give it a meaning in the stochastic integral sense. You are just integrating infinite ...


1

This is not true. For instance, you have different options for the future whether you're at $(3,4)$ or $(5,0)$, and you can tell which one you're at by looking at the past, so the process is not memoryless.


1

$P(T_n=t)=P(\sum_{i=1}^n S_i=t)$ so this is essentially the $n$-fold convolution of $S_1,S_2,...,S_n$. Not sure if a neat answer exists. $E(T_n)=E(\sum_{i=1}^nS_i)=\sum_{i=1}^nE(S_i)=\sum_{i=1}^n[1\times p-2\times(1-p)]=n(3p-2)$ $Var(T_n)=Var(\sum_{i=1}^nS_i)=\sum_{i=1}^nVar(S_i)=nVar(S_1) $ due to iid $S_i$. Find $Var(S_1)=E(S_1^2)-(E(S_1))^2$. Simplify. ...


1

Note that $\frac{S_i+2}{3}$ is distributed as $B(1,p)$, giving that $\frac{T_n+2n}{3}$ is distributed as $B(n,p)$. Thus: $$\mathbb{E}\left(\frac{T_n+2n}{3}\right)=np, \mathbb{V}\left(\frac{T_n+2n}{3}\right)=np(1-p)$$ from which you can derive: $$\mathbb{E}(T_n)=n(3p-2), \mathbb{V}(T_n)=9np(1-p)$$ Furthermore, knowledge of the binomial distribution gives ...


1

Let X_n be the random variable corresponding to the number of +1 steps in n turns. $X_n$ will follow Binomial Distribution with parameters $n$ and $p$. If $k$ is the number of +1 steps. Then, $n-k$ is the number of +2 steps. Then, $k+2(n-k) =t$. On solving, we get $k = 2n - t$. Thus, $P(T_n = t) = P(X_n = 2n-t) = {n \choose 2n-t}p^{2n-t}(1-p)^{t-n}$. Now, ...


1

In general it is not. The transition matrix for the random walk is $D^{-1}W$ (which is row stochastic) and this matrix is similar to $D^{-1/2}WD^{1/2}$ if $G$ has no isolated vertices.


1

For i), take one step from the origin, without loss of generality to $1$, and then consider the walk to end if it reaches either $0$ or $m$. A recurrence relation for the probability of ending at $m$ shows that this probability increases linearly from $0$ at $0$ to $1$ at $m$ and hence is $\frac1m$ at $1$. For ii), note that the walk almost surely does not ...



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