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Assuming $1D$ random walk: For the two random walkers($W_1, W_2$) to meet at some time point $N$, the number of left steps taken by $W_1$ should equal the number of steps taken by $W_2$ The number of sequences of lefts and rights possible for each walker is $2^N$. Now, in order for $W_1$ and $W_2$ to meet, they must have taken $0\ or\ 1\ or\ 2\ or\ 3\ or\ ...


2

For every $n$, the random variable $(q/p)^{S_n}$ is bounded either by $(q/p)^n$ or by $(p/q)^n$ depending on if $p>q$ or $q<p$. At any case the random variable $(q/p)^{S_n}$ is bounded by the sum of both unless it is infinite, which is the case here. Once again $E[X]\leq \max X$ is used. Added: If $q>p$ the expectation is still finite as long as ...


2

Assume that the random walk is defined as $$ X_n = \sum_{k = 1} ^ n \xi_k $$where $(\xi_a)_{a\ge 1}$ are iid and such as $P(\xi_1 = \pm 1) = 1/2$. $$1_{T_0 = m,X_0=0} = 1_{T_0 = m,X_0=0, X_1 = 1}+ 1_{T_0 = m,X_0=0, X_1 = -1} $$ Now using the Markov property, and as $T_0$ depends in a deterministic way on $(X_a)_{a\ge 1}$, you get $$ T_0 |(X_0=a, X_1 = ...


1

To compute $P(j\cap i)$, the probability of reaching $i$ having reached $j$ as minimum, consider playing two games after each other: $G$ Is the game that is won by starting from zero, ending at $j$ without having reached $i$ $H$ is the game that is won by starting from $j$, ending at $i$ without having reached $j-1$ For the first one we have $$ ...



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