Finding sequence $a_n$ such that $a_n(X_n-n)$ converges in distribution

Letting $$N$$ be a standard normal random variable and $$X_n$$ such that for $$x\geq 0$$: $$P(X_n \geq n+x) = P(N \geq n+x \vert N \geq n)$$. I'm asked to find deterministic $$a_n$$ such that $$Y_n = a_n(X_n-n)$$ converges in distribution to some finite positive random variable $$Y$$.

First of all we can have the distribution of $$X_n - n$$: Notice from the formula given, $$X_n - n \geq 0$$ a.s., and for $$x \geq 0$$, $$\displaystyle{P(X_n-n \geq x) = \frac{P(N \geq n+x)}{P(N \geq n)}}$$. By Gaussian variable has no atoms, $$X_n - n$$ has no atoms and thus $$P(X_n-n \leq x) = 1-P(X_n-n \geq x) = \frac{P(n \leq N \leq n+x)}{P(N \geq N)}$$

I'm thinking of trying to make the CDF $$F_n(y) = P(Y_n \leq y)$$ converges to some other CDF. Because we want the distributional limit to be positive, then $$a_n >0$$. Then I found $$P(Y_n \leq y) = P(a_n(X_n-n)\leq y) = P(X_n-x \leq \frac{y}{a_n}) = \frac{\int_{n}^{n+\frac{y}{a_n}}e^{-\frac{x^2}{2}}dx}{\int_{n}^{\infty}e^{-\frac{x^2}{2}}dx}$$, and to make it more convenient to guess such $$a_n$$, I further used weighted mean value theorem to get:$$F_n(y) = P(Y_n \leq y) = \frac{e^{\frac{-\epsilon_n^2}{2}}\frac{y}{a_n}}{\int_{n}^{\infty}e^{-\frac{x^2}{2}}dx}$$ where $$\epsilon_n \in (n,n+\frac{y}{a_n})$$. It seems that we should choose $$a_n$$ such that it cancels the denominator and yet should have a limit. But I'm stuck here. Any help or hint would be appreciated.

Fix some $$t\in \mathbb R$$. If $$t\leq 0$$, then as you noticed $$P(a_n (X_n-n). Assume next that $$t>0$$ and note that $$P(a_n (X_n-n) where $$\Phi$$ denotes the cdf of the standard normal. By a classical result, letting $$\phi$$ denote the pdf of the standard normal, the following estimate holds as $$x\to \infty$$: $$\frac{1-\Phi(x)}{\phi(x)} = \frac 1x - \frac 1{x^3} + O\Big(\frac 1{x^5}\Big),$$ hence $$1-\Phi(x) = \frac{\phi(x)}{x} + o(\frac{\phi(x)}{x})$$, which rewrites $$1-\Phi(x)\sim \frac{\phi(x)}{x}$$.
Therefore $$\frac{1-\Phi(n+\frac t{a_n})}{1-\Phi(n)}\sim_{n\to\infty}\frac{n}{n+\frac t{a_n}} \exp\Big(-\frac 12 t(2\frac n{a_n} + \frac t{a_n^2})\Big).$$
This suggests letting $$a_n=n$$, which yields $$\lim_n \frac{1-\Phi(n+\frac t{n})}{1-\Phi(n)} = \exp(-t).$$
Thus for every $$t\in \mathbb R$$, $$P(n (X_n-n)
Let $$Y$$ denote a random variable with exponential distribution of parameter $$1$$. We have established that $$P(n (X_n-n). This is enough to conclude that $$n (X_n-n)$$ converges in distribution to $$Y$$.
Note the technicality that we dealt with $$P(n (X_n-n) instead of the usual $$P(n (X_n-n)\leq t)$$. This makes no difference for convergence since $$P(Z_n, thus we have convergence in distribution of $$-Z_n$$ to $$-Y$$, and hence of $$Z_n$$ to $$Y$$ by the continuous mapping theorem.