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!