Let $\displaystyle t_n$ be a sequence such that $\displaystyle t_n\sim n$ as $\displaystyle n\rightarrow\infty$. Let $\displaystyle (\omega_0,...,\omega_{2n})$ be a trajectory of a simple symmetric random walk on $\displaystyle \mathbb{Z}$. How do you find the limit of the following conditional probabilities:

$\displaystyle \lim_{n\rightarrow\infty}\mathbb{P}(a\leq\frac{\om ega_{t_n}}{\sqrt{n}}\leq b|\omega_0=\omega_{2n}=0)$

Where $\displaystyle a$ and $\displaystyle b$ are fixed numbers?

Using independence of the increments of the random walk and Moivre-Laplace Theorem (e.g. Central Limit Theorem in discrete case), I find an expression of the form:

$\displaystyle \frac{1}{\sqrt{2\pi}}\int_a^be^{-x^2}dx$

Which I am far from certain about. I would appreciate if anyone could come out with a proper proof.

Thanks for your help.