I'm looking for a direct proof of the following identity:

Let $W_n$ be a simple random walk with $W_0=0$. For all $x>0$ we have $$ \lim _{N\to \infty} \sqrt{N} \cdot \mathbb P \Big( \forall n \le N , \ W_n \le x\sqrt{N} -x\sqrt{N-n} \Big) =\frac{e^{-\frac{x^2}{2}}}{\int _x^\infty e^{-\frac{y^2}{2}}dy}. $$ I have an indirect proof that follows from a specific model that I've been working on related to DLA.

It's not hard to show that the probability in the left hand side decays like $CN^{-\frac{1}{2}}$ (because one can show that it behaves like the probability that a random walk stays negative for $N$ steps). This is the reason for the factor $\sqrt{N}$ inside the limit.

I'm looking for a direct proof of the following identity:

Let $W_n$ be a simple random walk with $W_0=0$. For all $x>0$ we have $$ \lim _{N\to \infty} \sqrt{N} \cdot \mathbb P \Big( \forall n \le N , \ W_n \le x\sqrt{N} -x\sqrt{N-n} \Big) =\frac{e^{-\frac{x^2}{2}}}{\int _x^\infty e^{-\frac{y^2}{2}}dy}. $$ I have an indirect proof that follows from a specific model that I've been working on related to DLA.

It's not hard to show that the probability in the left hand side decays like $C/\sqrt{N}$ (because one can show that it behaves like the probability that a random walk stays negative for $N$ steps). This is the reason for the factor $\sqrt{N}$ inside the limit.