Asymptotic conditional distribution of normal variable $X$ is a normal variable $\mathcal{N}(0,1)$, $Y$ is a normal variable $\mathcal{N}(n,n-1)$, independent of $X$. I want to prove that the distribution of $X$ conditionally on $X > Y$ is asymptotically a normal distribution with mean 1 and variance 1.
This is to say that if $X_i, i = 1,2,\cdots$ are i.i.d. standard normal variables and $S_n = \frac{1}{n}\sum_{i=1}^n X_i$, then the distribution of $X_1$ knowing $S_n >1$ is asymptotically $\mathcal{N}(1,1)$.
I try to prove it by direct computation:
\begin{align}
E(f(X) | X > Y) = \dfrac{\int_{-\infty}^{+\infty} f(x) (\int_{-\infty}^{x} e^{-\frac{(y-n)^2}{2(n-1)}} dy)e^{-\frac{x^2}{2}}dx}{\int_{-\infty}^{+\infty} (\int_{-\infty}^{x} e^{-\frac{(y-n)^2}{2(n-1)}} dy)e^{-\frac{x^2}{2}}dx}
\end{align}
which means the condtional density is
\begin{align}
\dfrac{(\int_{-\infty}^{x} e^{-\frac{(y-n)^2}{2(n-1)}} dy)e^{-\frac{x^2}{2}}}{\int_{-\infty}^{+\infty} (\int_{-\infty}^{x} e^{-\frac{(y-n)^2}{2(n-1)}} dy)e^{-\frac{x^2}{2}}dx}
\end{align}
I want to prove the density converges to $\frac{1}{\sqrt{2\pi}}e^{-\frac{(x-1)^2}{2}}$.
I try to use
\begin{align}
\int_{-\infty}^{x} e^{-\frac{(y-n)^2}{2(n-1)}} dy = \sqrt{n-1} \int_{-\infty}^{\frac{x-n}{\sqrt{n-1}}} e^{-\frac{y^2}{2}} dy
\end{align}
and when $x < 0$
\begin{align}
\frac{1}{|x|} - \frac{1}{|x|^3} e^{-\frac{x^2}{2}} \leq \int_{-\infty}^x e^{-\frac{y^2}{2}}dy \leq \frac{1}{|x|}  e^{-\frac{x^2}{2}}
\end{align}
But then I don't know how to proceed.
 A: 
Indeed the limit is normal $(1,1)$. 

To prove this, recall that $X-Y$ is normal $(-n,\sqrt{n})$ hence, considering some standard normal random variable $Z$, one gets $$P(X\gt Y)=P(Z\gt\sqrt{n})=\bar\Phi(\sqrt{n}),
$$
where $\bar\Phi$ denotes the standard normal complementary CDF. On the other hand, $(X,Y)$ is distributed as $(X,n-\sqrt{n-1}\cdot Z)$ where $Z$ is standard normal independent of $X$ hence, for every $x$, $$P(X\gt x,X\gt Y)=P(X\gt x,\sqrt{n-1}\cdot Z\gt n-X)=\int_x^{+\infty}P(\sqrt{n-1}\cdot Z\gt n-x)\varphi(x)\mathrm dx,$$ that is, $$P(X\gt x,X\gt Y)=\int_x^{+\infty}\bar\Phi\left(\frac{n-x}{\sqrt{n-1}}\right)\varphi(x)\mathrm dx,
$$
where $\varphi$ denotes the standard normal PDF. Differentiating the second identity with respect to $x$ and dividing by the first identity one sees that $X$ conditionally on $X\gt Y$ has the density $f_n$, where $$f_n(x)=\bar\Phi\left(\frac{n-x}{\sqrt{n-1}}\right)\cdot\frac1{\bar\Phi(\sqrt{n})}\cdot\varphi(x).
$$
The last ingredient is the exact asymptotics, when $t\to+\infty$, $$\bar\Phi(t)\sim\frac{\mathrm e^{-t^2/2}}{t\sqrt{2\pi}}.
$$
Using this twice in the expression of $f_n(x)$ shows that $f_n(x)\to f(x)$, where $$f(x)=\mathrm e^{x-1/2}\varphi(x)=\varphi(x-1),$$ from which the result follows.
A: I have found a way to answer my question.
Since $X$ and $Y$ are independent normal variance, by simple computation, we can find that $X - Y$ and $X + \frac{1}{n-1}Y$ are independent. Remark that 
\begin{align}
X = \frac{1}{n}(X-Y) + \frac{n-1}{n}(X + \frac{1}{n-1}Y)
\end{align}
$\frac{n-1}{n}(X + \frac{1}{n-1}Y)$ is independent of $X-Y$ and is a normal variable $\mathcal{N}(1, \frac{n-1}{n})$, which converges to $\mathcal{N}(1,1)$
We only need to prove $\frac{1}{n}(X-Y)$ knowing $X-Y >0$ converges to 0 in law, then use Slutsky's theorem 
to conclude.
Now we prove $\frac{1}{n}(X-Y)$ knowing $X-Y >0$ converges to 0 in law:
Firstly, we know that $\frac{1}{n}(X-Y)$ is $\mathcal{N}(-1, \frac{1}{n})$. For a bounded Lipschitz function $f$, we have 
\begin{align}
E(f(\frac{1}{n}(X-Y))| X- Y >0) = \dfrac{\int_{0}^{+\infty}f(\frac{z}{\sqrt{n}})e^{-\frac{(z+\sqrt{n})^2}{2}}dz}{\int_{0}^{+\infty}e^{-\frac{(z+\sqrt{n})^2}{2}}dz}
\end{align}
It's not difficlt to prove the last limit convergs to $f(0)$
for $\epsilon > 0$, $\int_{0}^{\epsilon}e^{-\frac{(z+\sqrt{n})^2}{2}}dz > e^{-\frac{(\epsilon+\sqrt{n})^2}{2}}\epsilon$ and $\int_{\epsilon}^{+\infty}e^{-\frac{(z+\sqrt{n})^2}{2}}dz < \frac{1}{\sqrt{n} + \epsilon}e^{-\frac{(\epsilon+\sqrt{n})^2}{2}}$, when $n$ tends to infinity, we can see all the weight is attributed to the interval $[0, \epsilon]$. Consider the expectation as a weighted average, intuitively, it's ok.
We can complete the proof by dividing both integrals into the above two parts and using continuity and the bound of $f$ 
