Let t_n be a sequence such that t_n\sim n as n\rightarrow\infty. Let (\omega_0,...,\omega_{2n}) be a trajectory of a simple symmetric random walk on \mathbb{Z}. How do you find the limit of the following conditional probabilities:
\lim_{n\rightarrow\infty}\mathbb{P}(a\leq\frac{\om  ega_{t_n}}{\sqrt{n}}\leq b|\omega_0=\omega_{2n}=0)
Where a and 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:

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