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1) To be looked at is the event that for each direction the numbers of steps taken in that direction is even. This however under the extra condition that the total number of steps is even (we are looking at times of the form $2n$). If $X_i$ denotes the number of steps in direction $i\in\{1,\dots,d\}$ then: $$P(X_1\text{ is even}\wedge\cdots\wedge X_d\text{ ... 2 Hint: Consider a walker moving on an infinite one-dimensional uniform lattice (i.e. a line split into discrete points). Suppose the walker starts at the origin (x=0) and then moves a short distance \delta either left or right in a short time \Delta t. The motion is assumed to be completely random, so the probabilities of moving both left and right are ... 2 P( \max \{X_1,...,X_M \} \geq X_0)=P(X_i \geq X_0 for some i) = 1-P( \text{each }X_i < X_0). Now P( \text{each }X_i < X_0)=\sum_{k=0}^n P(\text{each }X_i<X_0 \vert X_0=k) \cdot P(X_0 =k) =\sum_{k=0}^n P( \text{each } X_i<k \text{ and } X_0=k) = \sum_{k=1}^n \big[ P(X_1<k) \big]^M \cdot P(X_0=k). Note the k=0 term is 0. 2 There was likely a typo in your question, as the sum is over an index i that does not appear anywhere. I think the true equation is intended to be:$$ u_b = \sum_{i=1}^{\infty} P[S_1S_2\cdots S_i \neq 0, S_i=b] You can prove that by defining $N_b$ as the random number of times we hit $b$ before returning to $0$ and then taking expectations of the ...

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For the loss of $k$ to kick in, he needs to win $k-a$ times. If he does that, he will never go broke (except maybe on the round he would quit because of the $k$ losses). He needs to win those $k-a$ within the first $2k-a$ games. So compute the chance he goes broke in less than $2k-a$ games and the expected length of a game in that scenario. This gives ...

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On a finite-dimensional cubic grid, it's possible to calculate the probability that an ant starting at $P$ lands at point $Q$ after $n$ steps, since there are at any time only finitely many positions the ant can occupy, although I don't know a closed form for the probability. On a cubic grid with uncountably many dimensions: Unless $P = Q$ and $n = 0$, the ...

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Setting $X^i=Y^i - \frac{1}{2} d_i$ and $g_i=\frac{1}{2}d_i - t_i$ gives you $A : X^i\leq -g_i$ and $B : X^i\geq g_i$. you can rewrite : $X^i = x+\sum_{l=1}^i \xi_i$ where $\xi_i$ are i.i.d r.v. taking in your case $\frac{1}{2}$ and $-\frac{1}{2}$ values with proba $p$ and $1-p$ If I choose $\mu$ such that $pe^{\frac{\mu}{2} }+(1-p)e^{-\frac{\mu}{2}}=1$, $... 1 xivaxy, I don't think that's correct. Suppose$X_n = X_{n-1}$with probability .5, and$X_n = 0$with probability .5. We set$X_0 \ne 0$. Let$N$be the first$n$such that$X_n = 0$.$N$is a stopping time because$1_{\{N=n\}}$is a function of$(X_1,...,X_n)$only. Then we have$\mathbb{E}[X_n 1_{\{N \ge n\}}] = \mathbb{E}[X_{n}]$because$X_n \ne 0$... 1 You move by$\sqrt{h}$so that the variance of a step is proportional to$h$. This ensures that the variance after a fixed finite time is some fixed finite number. If you move by asymptotically less than$\sqrt{h}$then the variance at a fixed finite time would go to zero as$h\$ went to zero; if you move by asymptotically more then the variance would blow up ...

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