# Tag Info

## Hot answers tagged random-walk

2

Counterexample Let $$S_n := \sum_{j=1}^n Y_j, \qquad n \in \mathbb{N} \tag{1}$$ a simple random walk on $\mathbb{Z}$, i.e. $Y_j \sim \frac{1}{2} (\delta_1+\delta_{-1})$ independent identically distributed random variables. By Stirling's formula, we have $$\mathbb{P}(S_{2n}=0) = 2^{-2n} {2n \choose n} \sim \frac{1}{\sqrt{\pi n}} \qquad \text{and} \qquad ... 2 As you noted, away from zero the bias is uniformly negative. This suffices to guarantee recurrence, that is, that the process hits zero with full probability. In a nutshell, choose N\gt2(k+1)c, and replace the dynamics on b\geqslant N by b\to b+k with probability c/N and b\to b-1 with probability 1-c/N. Starting from any b(0)\geqslant N, this ... 1 Yes, As long as there are white balls and black balls, the probability of increasing whites W/(B+W) * B/(B+W-1) is equal to probability of increasing blacks B/(B+W) * W/(B+W-1), so the process is as likely to go toward W as to go toward B. But once we reach a state of only one color the probabilities are 1 (for no change) and 0 (for change). The ... 1 For every |z|\leqslant1, the generating function u(x)=E_x(z^\tau) for the random walk starting at x is the unique solution of the integral identity$$ u(x)=z\cdot\int_\mathbb Ru(x+y)g(y)\mathrm dy, $$where g is the standard normal density, with the boundary condition that u(x)=1 for every x\geqslant C. 1$$\sum_{n=2N}^{\infty}p_{ii}^{(n)} = \sum_{n=N}^\infty {2n \choose n} p^n (1-p)^n$$Via the Stirling formula:$${2n \choose n}\sim \frac{2^{2n}}{\sqrt{\pi n}}$$so$${2n \choose n} p^n (1-p)^n \sim \frac{(4p(1-p))^{n}}{\sqrt{\pi n}}$$As p\neq \frac 12, 0\le 4p(1-p)<1 and the series$$\sum {2n \choose n} p^n (1-p)^n and ...

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