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Let $\zeta = (\zeta_{0},\zeta_{1}, \ldots,\zeta_{2n} )$ be the trajectory of a simple symmetric random on $\mathbb{Z}$. Find $\lim_{n \to \infty}{P (a \leq \frac{\zeta_{n}}{\sqrt{n}} \leq b\ | \ \zeta_{0} = \zeta_{2n}=0 })$, where $a$ and $b$ are fixed.

I can't figure how to compute $P (a \leq \frac{\zeta_{n}}{\sqrt{n}}\leq b, \zeta_{0} = \zeta_{2n}=0 )$. I know how to do $P (a \leq \frac{\zeta_{n}}{\sqrt{n}} \leq b)$ with Moivre-Laplace and how to get the asymptotics for the probability of return at the $2n$-step... So, I need your help..

Thank you!

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The walk has to go from $0$ to $\zeta_n$ and then from $\zeta_n$ to $0$. You know the asymptotic probability density for a random walk of length $n$ to travel by $c\sqrt n$. The probability of doing that twice is just the square, and the conditional probability you want is the integral of that square from $a$ to $b$ over the integral of that square from $-\infty$ to $\infty$.

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    I'm still lost... doing it as above gives that the denominator is $~ 1/ \sqrt{\pi n}$, whereas the numerator is $O(1/n)$, so the conditional probability should be zero... I don't think this is correct; or at least we're wrong about the numerator being $\sum_{z}{P^{2}(\zeta_{n} / \sqrt{n} = z)}$...2012-10-17