Identity involving the probability that a random walk stays below a curve

Question

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.

Answer 1

Example 2 of arXiv:0704.2826 considers the analogous problem for the continuous-time random walk, in the more general case that the curve has the form $g(t)=a+b\sqrt{T-t}$ with $a+b\sqrt T\geq 0$. The random walk starting at the origin stays below that curve for all $t<T$ with probability $$P(T,a,b)=1-\frac{\int_{-\infty}^{-a/\sqrt T} e^{-y^2/2}dy}{\int_{-\infty}^{b} e^{-y^2/2}dy}.$$ This holds for any $T$, not only in the large-$T$ limit.

The case in the OP corresponds to $a=x\sqrt T$, $b=-x$, and this probability vanishes, the reason being that in this case $g(0)=0$, the boundary intersects the origin at $t=0$ and the continuous-time random walk is never strictly below the boundary.

To make contact with the discrete-time random walk formula of the OP, I take $a=x\sqrt T + \epsilon$, so $g(0)=\epsilon$, and then to first order in the infinitesimal step size $\epsilon$ one has $$P(T,x\sqrt T+\epsilon,-x)=\frac{\varepsilon}{\sqrt T}\frac{e^{-x^2/2}}{\int_{x}^{\infty} e^{-y^2/2}dy}+{\cal O}(\epsilon^2).$$ This reduces to the expression in the OP upon identification of $N=DT/\epsilon^2$, with $D\equiv 1$ the diffusion constant of the random walk. This correspondence between discrete-time and continuous-time random walks only holds in the $T\rightarrow\infty$ limit.