# Tag Info

## Hot answers tagged random-walk

3

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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